Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 15 Similarity With Applications To Maps Models

Exercise 15A

Question 1. In the figure, given below, straight lines AB and CD intersect at P; and AC // BD. Prove that:
(i) ∆APC and ∆BPD are similar.
(ii) If BD = 2.4 cm AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
Answer:

[Diagram: Two triangles APC and BPD formed by intersecting lines AB and CD at point P, with AC parallel to BD]

(i) Proving ∆APC ~ ∆BPD:
In ∆APC and ∆BPD,
∠APC = ∠BPD (vertically opposite angles)
∠ACP = ∠BDP (alternate angles since AC || BD)
∴ ∆APC ~ ∆BPD (AA criterion for similarity)

(ii) Finding lengths of PA and PC:
Since ∆APC ~ ∆BPD,
\( \frac{PA}{PB} = \frac{PC}{PD} = \frac{AC}{BD} \)

\( \frac{PA}{3.2} = \frac{PC}{4} = \frac{3.6}{2.4} \)

\( \frac{PA}{3.2} = \frac{3.6}{2.4} \)
\( \implies PA = \frac{3.6 \times 3.2}{2.4} = 4.8 \text{ cm} \)

\( \frac{PC}{4} = \frac{3.6}{2.4} \)
\( \implies PC = \frac{3.6 \times 4}{2.4} = 6 \text{ cm} \)

Hence, PA = 4.8 cm and PC = 6 cm.

📝 Teacher's Note: Show students that when two lines cut by a transversal make parallel lines, the triangles formed are similar. Use scissors and paper to cut triangles and compare their angles.

🎯 Exam Tip: Always write "AA criterion" clearly when proving similarity. For calculations, write the ratio first, then substitute values to get marks.

Question 2. In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that:
(i) ∆APB is similar to ∆CPD
(ii) PA × PD = PB × PC
Answer:

[Diagram: Trapezium ABCD with AB parallel to DC, diagonals AC and BD intersecting at point P]

(i) Proving ∆APB ~ ∆CPD:
In ∆APB and ∆CPD,
∠APB = ∠CPD (vertically opposite angles)
∠ABP = ∠CDP (alternate angles since AB || DC)
∴ ∆APB ~ ∆CPD (AA criterion for similarity)

(ii) Proving PA × PD = PB × PC:
Since ∆APB ~ ∆CPD,
\( \frac{PA}{PC} = \frac{PB}{PD} \) (corresponding sides of similar triangles are equal)
\( \implies PA \times PD = PB \times PC \)

📝 Teacher's Note: In a trapezium, the diagonals always create similar triangles. This property is very useful for finding unknown lengths in trapeziums.

🎯 Exam Tip: Write "corresponding sides of similar triangles are proportional" to get full marks. Cross multiply carefully to show the product relationship.

Question 3. P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that:
(i) DP: PL = DC: BL
(ii) DL: DP = AL: DC
Answer:

[Diagram: Parallelogram ABCD with point P on BC, line DP extended to meet AB extended at point L]

(i) Proving DP: PL = DC: BL:
Since AD || BC, that is, AD || BP,
by the Basic Proportionality theorem, we get
\( \frac{DL}{DP} = \frac{AL}{AB} \)
Since ABCD is a parallelogram, AB = DC.
So, \( \frac{DL}{DP} = \frac{AL}{DC} \)

(ii) Proving DL: DP = AL: DC:
Since AD || BC, that is, AD || BP,
by the Basic Proportionality theorem, we get
\( \frac{DP}{PL} = \frac{AB}{BL} \)
Since ABCD is a parallelogram, AB = DC.
So, \( \frac{DP}{PL} = \frac{DC}{BL} \)

📝 Teacher's Note: Basic Proportionality Theorem (also called Thales' theorem) is the key here. When a line cuts two sides of a triangle, it divides them proportionally.

🎯 Exam Tip: Always mention "Basic Proportionality Theorem" by name. Also write that opposite sides of parallelogram are equal to get full marks.

Question 4. In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that:
(i) ∆AOB is similar to ∆COD
(ii) OA × OD = OB × OC
Answer:

[Diagram: Quadrilateral ABCD with diagonals AC and BD intersecting at point O]

(i) Proving ∆AOB ~ ∆COD:
Since AO = 2CO and BO = 2DO,
\( \frac{AO}{CO} = \frac{2}{1} = \frac{BO}{DO} \)
Also, ∠AOB = ∠DOC (vertically opposite angles)
So, ∆AOB ~ ∆COD (SAS criterion for similarity)

(ii) Proving OA × OD = OB × OC:
Since AO = 2CO and BO = 2DO,
\( \frac{AO}{CO} = \frac{2}{1} = \frac{BO}{DO} \)
So, OA × OD = OB × OC.

📝 Teacher's Note: When two triangles have one angle equal and the sides containing that angle in the same ratio, they are similar by SAS criterion.

🎯 Exam Tip: Write "SAS criterion for similarity" clearly. Show the ratio calculation step by step. Don't jump directly to the final answer.

Question 5. In ∆ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that:
(i) CB: BA = CP: PA
(ii) AB × BC = BP × CA
Answer:

[Diagram: Triangle ABC with angle bisector of angle ABC meeting side AC at point P]

In simple words: This question uses angle bisector properties and similar triangles. When an angle bisector cuts the opposite side, it creates special ratios between the sides.

📝 Teacher's Note: Use the angle bisector theorem here. Students often forget that the angle bisector divides the opposite side in the ratio of the other two sides.

🎯 Exam Tip: State "Angle Bisector Theorem" clearly. Draw the diagram neatly and mark all angles. Show that the given angle condition creates similar triangles.

Question (ii).
Answer:
In ∆ABC,
∠ABC = 2∠ACB
Let ∠ACB = x
\( \implies \) ∠ABC = 2∠ACB = 2x
Given BP is bisector of ∠ABC.
Hence ∠ABP = ∠PBC = x.
Using the angle bisector theorem,
that is, the bisector of an angle divides the side opposite to it in the ratio of other two sides.
Hence, CB : BA = CP : PA.

[Diagram: Triangle ABC with point P inside, where BP bisects angle ABC, creating angles marked as x.]

In simple words: When a line cuts an angle in half, it also cuts the opposite side in a special ratio. This ratio is the same as the ratio of the two other sides of the triangle.

📝 Teacher's Note: Draw a triangle and show how the angle bisector works. Tell students to remember that the bisector divides both the angle and the opposite side in related ratios.

🎯 Exam Tip: Always write "angle bisector theorem" clearly. Then write the ratio CB : BA = CP : PA. This gets you marks.

Question 6. In ∆ABC; BM ⊥ AC and CN ⊥ AB; show that:
\( \frac{AB}{AC} = \frac{BM}{CN} = \frac{AM}{AN} \)
Answer:
In ∆ABM and ∆ACN,
∠AMB = ∠ANC .....(BM ⊥ AC and CN ⊥ AB)
∠BAM = ∠CAN .....(common angle)
\( \implies \) ∆ABM ~ ∆ACN .....(AA criterion for Similarity)
\( \implies \frac{AB}{AC} = \frac{BM}{CN} = \frac{AM}{AN} \)
Consider ∆ABC and ∆APB,
∠ABC = ∠APB ....[Exterior angle property]
∠BCP = ∠ABP ....[Given]
∴ ∆ABC ~ ∆APB [AA criterion for Similarity]
\( \frac{CA}{AB} = \frac{BC}{BP} \) ....(Corresponding sides of similar triangles are proportional.)
\( \implies \) AB × BC = BP × CA

[Diagram: Triangle ABC with perpendicular lines BM to AC and CN to AB, meeting at point D inside the triangle.]

In simple words: When two lines are perpendicular to the sides of a triangle, they create similar triangles. Similar triangles have the same angles and their sides are in the same ratio.

📝 Teacher's Note: Use a cardboard triangle and show how perpendicular lines create right angles. This makes similar triangles that have the same shape but different size.

🎯 Exam Tip: Write "AA criterion" clearly. Show that two angles are equal to prove similarity. Then write the ratio of corresponding sides.

Question 7. In the given figure, DE//BC, AE = 15 cm, EC = 9 cm, NC = 6 cm and BN = 24 cm.
(i) Write all possible pairs of similar triangles.
(ii) Find lengths of ME and DM.
Answer:
(i)
In ∆AME and ∆ANC,
∠AME = ∠ANC ....( Since DE || BC that is, ME || NC.)
∠MAE = ∠NAC .....(common angle)
\( \implies \) ∆AME ~ ∆ANC .....(AA criterion for Similarity)

In ∆ADM and ∆ABN,
∠ADM = ∠ABN ....( Since DE || BC that is, DM || BN.)
∠DAM = ∠BAN .....(common angle)
\( \implies \) ∆ADM ~ ∆ABN .....(AA criterion for Similarity)

In ∆ADE and ∆ABC,
∠ADE = ∠ABC ....( Since DE || BC that is, ME || NC.)
∠AED = ∠ACB ....( Since DE || BC.)
\( \implies \) ∆ADE ~ ∆ABC .....(AA criterion for Similarity)

[Diagram: Triangle ABC with line DE parallel to BC, where D is on AB, E is on AC, M is on DE, and N is on BC.]

(ii)
In ∆AME and ∆ANC,
∠AME = ∠ANC ....( Since DE || BC that is, ME || NC.)
∠MAE = ∠NAC .....(common angle)
\( \implies \) ∆AME ~ ∆ANC .....(AA criterion for Similarity)
\( \implies \frac{ME}{NC} = \frac{AE}{AC} \)
\( \implies \frac{ME}{6} = \frac{15}{24} \)
\( \implies \) ME = 3.75 cm

In ∆ADE and ∆ABN,
∠ADE = ∠ABC ....( Since DE || BC that is, ME || NC.)
∠AED = ∠ACB ....( Since DE || BC.)
\( \implies \) ∆ADE ~ ∆ABC .....(AA criterion for Similarity)
\( \implies \frac{AD}{AB} = \frac{AE}{AC} = \frac{15}{24} \) .....(i)
In ∆ADM and ∆ABN,
∠ADM = ∠ABN ....( Since DE || BC that is, DM || BN.)
∠DAM = ∠BAN .....(common angle)
\( \implies \) ∆ADM ~ ∆ABN .....(AA criterion for Similarity)
\( \implies \frac{DM}{BN} = \frac{AD}{AB} = \frac{15}{24} \) ....(from (i))
\( \implies \frac{DM}{24} = \frac{15}{24} \)
\( \implies \) DM = 15 cm

In simple words: When lines are parallel, they create similar triangles. We use the ratios of sides to find missing lengths. All similar triangles have sides in the same proportion.

📝 Teacher's Note: Use parallel rulers or lines on graph paper to show how parallel lines create similar triangles. Students can see the pattern clearly this way.

🎯 Exam Tip: First find all similar triangles using parallel line properties. Then use ratios to calculate missing sides. Show all working clearly.

Question 8. In the given figure, AD = AE and AD² = BD × EC. Prove that: triangles ABD and CAE are similar.
Answer:
In ∆ABD and ∆CAE,
∠ADE = ∠AED .....(Angles opposite equal sides are equal.)
So, ∠ADB = ∠AEC
.....(Since ∠ADB + ∠ADE = 180° and ∠AEC + ∠AED = 180°)
Also, AD² = BD × EC
\( \implies \frac{AD}{BD} = \frac{EC}{AD} \)
\( \implies \frac{AD}{BD} = \frac{EC}{AE} \)
\( \implies \) ∆ABD ~ ∆CAE .....(SAS criterion for Similarity)

[Diagram: Triangle ABC with points D on AB and E on AC, where AD = AE.]

In simple words: When two sides are equal and we have a special ratio condition, we can prove triangles are similar. Similar triangles have the same angles and side ratios.

📝 Teacher's Note: Show students that when AD = AE, triangle ADE is isosceles. This creates equal angles that help prove similarity.

🎯 Exam Tip: Use the given condition AD² = BD × EC to create a ratio. Then apply SAS similarity criterion. Write "SAS criterion" clearly.

Question 9. In the given figure, AB // DC, BO = 6 cm and DQ = 8 cm; find: BP × DO.
Answer:
In ∆DOQ and ∆BOP,
∠QDO = ∠PBO .....(Since AB || DC that is, PB || DQ.)
So, ∠DOQ = ∠BOP .....(vertically opposite angles)
\( \implies \) ∆DOQ ~ ∆BOP .....(AA criterion for Similarity)
\( \implies \frac{DO}{BO} = \frac{DQ}{BP} \)
\( \implies \frac{DO}{6} = \frac{8}{BP} \)
\( \implies \) BP × DO = 48 cm²

[Diagram: Quadrilateral with AB parallel to DC, diagonals intersecting at O, with points P and Q marked.]

In simple words: When two lines are parallel, they create similar triangles with the intersecting lines. We use this to find the product of the segments.

📝 Teacher's Note: Draw parallel lines and show how they create similar triangles. The cross multiplication gives a constant product.

🎯 Exam Tip: Write that triangles are similar due to parallel lines. Then write the ratio and cross multiply to get BP × DO = 48 cm².

Question 10. Angle BAC of triangle ABC is obtuse and AB = AC. P is a point in BC such that PC = 12 cm. PQ and PR are perpendiculars to sides AB and AC respectively. If PQ = 15 cm and

In simple words: This question deals with an isosceles triangle where one angle is more than 90 degrees. We need to use properties of perpendiculars from a point inside the triangle.

📝 Teacher's Note: Show students that an obtuse angle is larger than a right angle. In an isosceles triangle, two sides are equal.

🎯 Exam Tip: Remember that in isosceles triangles, perpendiculars from points on the base create special relationships. Use the given measurements carefully.

PR=9 cm; find the length of PB

Solution:

[Diagram: Triangle ABC with point P inside. Line segments show AC = 12 cm, PQ = 3 cm, PR = 9 cm, and QB = 15 cm. Point Q appears to be on AB, and there are right angle markers at Q and R.]

In \( \triangle ABC \),
AC = AB ... (Given)
\( \Rightarrow \angle ABC = \angle ACB \) ... (Angles opposite equal sides are equal.)
In \( \triangle PRC \) and \( \triangle PQB \),
\( \angle ABC = \angle ACB \)
\( \angle PRC = \angle PQB \) ... (Both are right angles.)
\( \Rightarrow \triangle PRC \sim \triangle PQB \) ... (AA criterion for Similarity)
\( \Rightarrow \frac{PR}{PQ} = \frac{RC}{QB} = \frac{PC}{PB} \)
\( \Rightarrow \frac{PR}{PQ} = \frac{PC}{PB} \)
\( \Rightarrow \frac{9}{3} = \frac{12}{PB} \)
\( \Rightarrow PB = \frac{12 \times 3}{9} = 4 \) cm
\( \Rightarrow PB = 20 \) cm

📝 Teacher's Note: Show students how to identify similar triangles by finding equal angles. The key is spotting that both triangles have a right angle and share angle relationships from the isosceles triangle.

🎯 Exam Tip: Always write the similarity criterion clearly (AA, SAS, or SSS). Set up the ratio equation step by step. Check your final answer by substituting back.

Question 11. State, true or false:
(i) Two similar polygons are necessarily congruent.
(ii) Two congruent polygons are necessarily similar.
(iii) All equiangular triangles are similar.
(iv) All isosceles triangles are similar.
(v) Two isosceles-right triangles are similar.
(vi) Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.
(vii) The diagonals of a trapezium, divide each other into proportional segments.
Answer:
(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) True
(vii) True
In simple words: Similar means same shape but different size. Congruent means exactly same size and shape. All congruent shapes are similar, but not all similar shapes are congruent.

📝 Teacher's Note: Use paper cutouts of different sized but same shaped triangles. Show that similar triangles have equal angles but different side lengths. This makes the concept very clear.

🎯 Exam Tip: Remember: congruent implies similar, but similar does not imply congruent. For isosceles triangles, you need one more condition (like equal angles) to prove similarity.

Question 12. Given = \( \angle GHE = \angle DFE = 90° \), DH = 8, DF = 12, DG = 3x + 1 and DE = 4x + 2. Find; the lengths of segments DG and DE.

Solution:

[Diagram: Triangle EGD with point F on ED and point H on GD. Both triangles GHE and DFE have right angles marked at H and F respectively.]

In \( \triangle DHG \) and \( \triangle DFE \),
\( \angle GHD = \angle DFE = 90° \)
\( \angle D = \angle D \) (Common)
\( \therefore \triangle DHG \sim \triangle DFE \)
\( \Rightarrow \frac{DH}{DF} = \frac{DG}{DE} \)
\( \Rightarrow \frac{8}{12} = \frac{3x - 1}{4x + 2} \)
\( \Rightarrow 32x + 16 = 36x - 12 \)
\( \Rightarrow 28 = 4x \)
\( \Rightarrow x = 7 \)
\( \therefore DG = 3 \times 7 - 1 = 20 \)
DE = \( 4 \times 7 + 2 = 30 \)

📝 Teacher's Note: Students often confuse which triangles are similar. Point out the common angle and the right angles. Then show how to set up the proportion correctly.

🎯 Exam Tip: Always identify the similar triangles first. Write the proportion with corresponding sides in the same order. Solve the equation step by step and substitute back to find the actual lengths.

Question 13. D is a point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that CA² = CB × CD.

Solution:

[Diagram: Triangle ABC with point D on side BC. Line AD is drawn creating triangles ABD and ACD.]

In \( \triangle ADC \) and \( \triangle BAC \),
\( \angle ADC = \angle BAC \) (Given)
\( \angle ACD = \angle ACB \) (Common)
\( \therefore \triangle ADC \sim \triangle BAC \)
\( \therefore \frac{CA}{CB} = \frac{CD}{CA} \)
Hence, CA² = CB × CD

📝 Teacher's Note: This is a classic similarity proof. Emphasize that when we have one given equal angle and one common angle, we can use AA similarity. The proportion gives us the required result directly.

🎯 Exam Tip: State the similarity clearly. Write the proportion with the right letters in the right places. The square relationship comes from cross multiplication of the proportion.

Question 14. In the given figure, ∆ABC and ∆AMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove that ∆ABC ~ ∆AMP
(ii) Find AB and BC.
Answer:
(i) In \( \triangle ABC \) and \( \triangle AMP \),
\( \angle BAC = \angle PAM \) [Common]
\( \angle ABC = \angle PMA \) [Each = 90°]
\( \triangle ABC \sim \triangle AMP \) [AA Similarity]
(ii)
AM=\( \sqrt{AP^2 - PM^2} = \sqrt{15^2 - 12^2} = 11 \)
Since \( \triangle ABC \sim \triangle AMP \),

\( \frac{AB}{AM} = \frac{BC}{PM} = \frac{AC}{AP} \)
\( \Rightarrow \frac{AB}{AM} = \frac{BC}{PM} = \frac{AC}{AP} \)
\( \Rightarrow \frac{AB}{11} = \frac{BC}{12} = \frac{10}{15} \)
From this we can write,
\( \frac{AB}{11} = \frac{10}{15} \)
\( \Rightarrow AB = \frac{10 \times 11}{15} = 7.33 \)
\( \frac{BC}{12} = \frac{10}{15} \)
\( \Rightarrow BC = 8 \) cm

📝 Teacher's Note: First use Pythagoras theorem to find AM. Then use the similarity ratio. Students often forget to find the missing side first before applying similarity ratios.

🎯 Exam Tip: Always find any missing sides using Pythagoras first. Then set up the similarity ratio. Calculate each unknown side separately using the common ratio.

Question 15. Given : RS and PT are altitudes of ∆ PQR prove that:
(i)∆PQT ~ ∆QRS,
(ii) PQ × QS = RQ × QT.
Answer:

[Diagram: Triangle PQR with altitudes RS from R to PQ at point T, and PT from P to QR at point S.]

(i)
In \( \triangle PQT \) and \( \triangle QRS \),
\( \angle PTQ = \angle RSQ = 90° \)(Given)
\( \angle PQT = \angle RQS \) (Common)
\( \triangle PQT \sim \triangle RQS \) (By AA similarity)
(ii)
Since, triangles PQT and RQS are similar.
\( \therefore \frac{PQ}{RQ} = \frac{QT}{QS} \)
\( \Rightarrow PQ \times QS = RQ \times QT \)

📝 Teacher's Note: Altitude means a perpendicular line from a vertex to the opposite side. Both triangles share angle Q and have right angles, making them similar by AA.

🎯 Exam Tip: Identify the right angles created by altitudes. Mark the common angle clearly. The product relationship comes directly from cross multiplication of the similarity ratio.

Question 16. Given : ABCD is a rhombus, DPR and CBR are straight lines. Prove that: DP × CR = DC × PR.

Solution:

[Diagram: Rhombus ABCD with diagonals intersecting. Points P and R are external points forming straight lines DPR and CBR through the rhombus.]

In \( \triangle DPA \) and \( \triangle RPC \),
\( \angle DPA = \angle RPC \) (Vertically opposite angles)
\( \angle PAD = \angle PCR \) (Alternate angles)
\( \triangle DPA \sim \triangle RPC \)
\( \therefore \frac{DP}{PR} = \frac{AD}{CR} \)
\( \frac{DP}{PR} = \frac{DC}{CR} \) (AD = DC, as ABCD is rhombus)
Hence, DP × CR = DC × PR

📝 Teacher's Note: In a rhombus, all sides are equal. Use properties of parallel lines to find alternate angles. Vertically opposite angles are always equal.

🎯 Exam Tip: State that all sides of rhombus are equal. Identify alternate angles and vertically opposite angles clearly. Cross multiply the similarity ratio to get the required product relationship.

Question 17. Given: FB = FD, AE ⊥ FD and FC ⊥ AD. Prove : \( \frac{FB}{AD} = \frac{BC}{ED} \)
Answer:
Given, FB = FD
∴ ∠FDB = ∠FBD ... (1)
In △AED and △FCB,
∠AED = ∠FCB = 90°
∠ADE = ∠FBC [Using (1)]
△AED ∼ △FCB [By AA similarity]
∴ \( \frac{AD}{FB} = \frac{ED}{BC} \)
∴ \( \frac{FB}{AD} = \frac{BC}{ED} \)
In simple words: We use the fact that two equal sides make equal angles. Then we find two similar triangles and use their matching sides to prove the required ratio.

📝 Teacher's Note: Draw the diagram clearly first. Mark all given angles and equal sides. Students often forget to write "By AA similarity" which is needed for marks.

🎯 Exam Tip: Always write the similarity condition (AA, SAS, or SSS) when proving triangles are similar. Write the ratio conclusion clearly at the end.

Question 18. In △ PQR, ∠ Q = 90° and QM is perpendicular to PR, Prove that:
(i) PQ² = PM × PR
(ii) QR² = PR × MR
(iii) PQ² + QR² = PR²
Answer:

[Diagram: A right triangle PQR with right angle at Q, and QM drawn perpendicular to hypotenuse PR, creating point M on PR]

(i) In △ PQM and △ PQR,
∠ PMQ = ∠ PQR = 90°
∠ QPM = ∠ RPQ (Common)
∴ △PQM ∼ △PRQ (By AA similarity)
∴ \( \frac{PQ}{PR} = \frac{PM}{PQ} \)
⇒ PQ² = PM × PR

(ii) In △ QMR and △ PQR,
∠ QMR = ∠ PQR = 90°
∠ QRM = ∠ QRP (Common)
∴ △QMR ∼ △PQR (By AA similarity)
∴ \( \frac{QR}{PR} = \frac{MR}{QR} \)
⇒ QR² = PR × MR

(iii) Adding the relations obtained in (i) and (ii), we get,
PQ² + QR² = PM × PR + PR × MR
= PR(PM + MR)
= PR × PR
= PR²
In simple words: When we drop a line from the right angle to the hypotenuse, it creates similar triangles. We use these similar triangles to prove the three relationships. The third one is just Pythagoras theorem.

📝 Teacher's Note: This is a very important theorem. Practice drawing the altitude to hypotenuse many times. Show students how all three parts connect to prove Pythagoras theorem.

🎯 Exam Tip: Always identify the similar triangles first. Write "By AA similarity" clearly. For part (iii), show that PM + MR = PR to get full marks.

Question 19. In △ ABC, ∠ B = 90° and BD ⊥ AC.
(i) If CD = 10 cm and BD = 8 cm; find AD.
(ii) If AC = 18 cm and AD = 6 cm; find BD.
(iii) If AC = 9 cm, AB = 7 cm; find AD.
Answer:

[Diagram: A right triangle ABC with right angle at B, and BD drawn perpendicular to hypotenuse AC, creating point D on AC]

(i) In △ CDB,
∠ 1 + ∠ 2 + ∠ 3 = 180°
∠ 1 + ∠ 3 = 90° ..... (1)(Since, ∠ 2 = 90°)
∠ 3 + ∠ 4 = 90° .....(2) (Since, ∠ ABC = 90°)
From (1) and (2),
∠ 1 + ∠ 3 = ∠ 3 + ∠ 4
∠ 1 = ∠ 4
Also, ∠ 2 = ∠ 5 = 90°
∴ △CDB ∼ △BDA (By AA similarity)
∴ \( \frac{CD}{BD} = \frac{BD}{AD} \)
⇒ BD² = AD × CD
⇒ (8)² = AD × 10
⇒ AD = 6.4
Hence, AD = 6.4 cm

(ii) Also, by similarity, we have :
\( \frac{BD}{DA} = \frac{CD}{BD} \)
BD² = 6 × (18 - 6)
BD² = 72
Hence, BD = 8.5 cm

(iii)
Clearly, △ADB ∼ △ABC
∴ \( \frac{AD}{AB} = \frac{AB}{AC} \)
AD = \( \frac{7 × 7}{9} = \frac{49}{9} = 5\frac{4}{9} \)
Hence, AD = \( 5\frac{4}{9} \) cm
In simple words: We use the altitude to hypotenuse theorem. When we drop a perpendicular from the right angle to the hypotenuse, we get the formula BD² = AD × CD.

📝 Teacher's Note: This theorem is very useful for finding missing sides. Make students remember the formula BD² = AD × CD. Also show them that AD × AC = AB².

🎯 Exam Tip: Always write the similarity statement first. Use the altitude theorem BD² = AD × CD. Show all calculation steps clearly with proper units.

Question 20. In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the lengths of PN and RM.
Answer:

[Diagram: A parallelogram PQRS with diagonal PR, point L on PR, and line QL extended to meet RS at M and PS extended at N]

In △RLQ and △PLN,
∠RLQ = ∠PLN (Vertically opposite angles)
∠LRQ = ∠LPN (Alternate angles)
△RLQ ∼ △PLN (AA similarity)
∴ \( \frac{RL}{LP} = \frac{RQ}{PN} \)
\( \frac{2}{3} = \frac{10}{PN} \)
PN = 15 cm

In △RLM and △PLQ,
∠RLM = ∠PLQ (Vertically opposite angles)
∠LRM = ∠LPQ (Alternate angles)
△RLM ∼ △PLQ (AA similarity)
∴ \( \frac{RM}{PQ} = \frac{RL}{LP} \)
\( \frac{RM}{16} = \frac{2}{3} \)
RM = \( \frac{32}{3} = 10\frac{2}{3} \) cm
In simple words: We use similar triangles created by the transversal line QL cutting the parallel sides of the parallelogram. The ratio RL:LP helps us find the other segments.

📝 Teacher's Note: Draw the parallelogram carefully and extend the lines clearly. Students should identify the alternate angles formed by parallel lines and transversals.

🎯 Exam Tip: Mark all equal angles in the diagram. Write "alternate angles" and "vertically opposite angles" to show how you got the similarity. Convert mixed fractions to improper fractions in final answers.

Question 21. In quadrilateral ABCD, diagonals AC and BD intersect at point E. Such that AE : EC = BE : ED. Show that ABCD is a parallelogram
Answer:
Given, AE : EC = BE : ED
Draw EF || AB

[Diagram: A quadrilateral ABCD with diagonals AC and BD intersecting at E, and line EF drawn parallel to AB]

In △ ABD, EF || AB
Using Basic Proportionality theorem,
\( \frac{DF}{FA} = \frac{DE}{EB} \)
But, given \( \frac{DE}{EB} = \frac{CE}{EA} \)
∴ \( \frac{DF}{FA} = \frac{CE}{EA} \)
Thus, in △ DCA, E and F are points on CA and DA respectively such that \( \frac{DF}{FA} = \frac{CE}{EA} \)
Thus, by converse of Basic proportionality theorem, FE || DC.
But, FE || AB.
Hence, AB || DC.
Thus, ABCD is a trapezium.
In simple words: When the diagonals cut each other in the same ratio, we can prove that opposite sides are parallel. This makes the quadrilateral a parallelogram.

📝 Teacher's Note: This is a tricky proof. Students need to understand the Basic Proportionality Theorem and its converse. Draw extra construction line EF parallel to AB.

🎯 Exam Tip: Always draw the construction line clearly. Write "by Basic Proportionality Theorem" and "by converse of Basic Proportionality Theorem". Show that both pairs of opposite sides are parallel.

Question 22. In △ ABC, AD is perpendicular to side BC and AD² = BD × DC. Show that angle BAC = 90°
Answer:

[Diagram: A triangle ABC with AD perpendicular to BC, creating point D on BC]

Given: AD ⊥ BC and AD² = BD × DC
To prove: ∠BAC = 90°

In right triangle ABD,
AB² = AD² + BD² (By Pythagoras theorem)
AB² = BD × DC + BD² (Given AD² = BD × DC)
AB² = BD(DC + BD)
AB² = BD × BC

In right triangle ACD,
AC² = AD² + DC² (By Pythagoras theorem)
AC² = BD × DC + DC² (Given AD² = BD × DC)
AC² = DC(BD + DC)
AC² = DC × BC

Adding both equations:
AB² + AC² = BD × BC + DC × BC
AB² + AC² = BC(BD + DC)
AB² + AC² = BC × BC
AB² + AC² = BC²

By the converse of Pythagoras theorem, ∠BAC = 90°
In simple words: We use the given condition AD² = BD × DC to show that AB² + AC² = BC². This means angle A is a right angle by the converse of Pythagoras theorem.

📝 Teacher's Note: This is a beautiful proof connecting the altitude to hypotenuse theorem with Pythagoras theorem. Show students how the given condition leads to the Pythagoras relationship.

🎯 Exam Tip: Use Pythagoras theorem for the two right triangles ABD and ACD. Substitute the given condition carefully. End with "by converse of Pythagoras theorem" for full marks.

Question 23. In the given figure AB // EF // DC; AB = 67.5 cm, DC = 40.5 cm and AE = 52.5 cm.
(i) Name the three pairs of similar triangles.
(ii) Find the lengths of EC and EF.
Answer:

[Diagram: Triangle ABC with point E on side AC, point F on side BC, and line EF parallel to AB and DC. Shows the parallel line configuration forming similar triangles.]

(i) The three pairs of similar triangles are:
△BEF and △BDC
△CEF and △CAB
△ABE and △CDE

(ii) Finding the lengths:
Since △ABE and △CDE are similar:
\( \frac{AB}{CD} = \frac{AE}{CE} \)
\( \frac{67.5}{40.5} = \frac{52.5}{CE} \)
CE = 31.5 cm

Since △CEF and △CAB are similar:
\( \frac{CE}{CA} = \frac{EF}{AB} \)
\( \frac{31.5}{52.5 + 31.5} = \frac{EF}{67.5} \)
\( \frac{31.5}{84} = \frac{EF}{67.5} \)
EF = \( \frac{2126.25}{84} = \frac{405}{16} = 25\frac{5}{16} \) cm

In simple words: When lines are parallel, they make similar triangles. We use the rule that matching sides have the same ratio. This helps us find the missing lengths.

📝 Teacher's Note: Draw three parallel lines on the board. Show how they cut any triangle into similar smaller triangles. Students understand ratios better when they see the pattern.

🎯 Exam Tip: Always write "triangles are similar" first. Then write the ratio clearly. Show all calculation steps to get full marks.

Question 24. In the given figure, QR is parallel to AB and DR is parallel to QB. Prove that PQ² = PD × PA.
Answer:

[Diagram: Triangle PAB with points Q, D, R forming intersecting lines where QR is parallel to AB and DR is parallel to QB.]

Given: QR is parallel to AB. Using Basic Proportionality Theorem:
\( \Rightarrow \frac{PQ}{PA} = \frac{PR}{PB} \) ... (1)

Also, DR is parallel to QB. Using Basic Proportionality Theorem:
\( \Rightarrow \frac{PD}{PQ} = \frac{PR}{PB} \) ... (2)

From (1) and (2), we get:
\( \frac{PQ}{PA} = \frac{PD}{PQ} \)

Cross multiplying:
PQ² = PD × PA

In simple words: When we have two parallel lines in a triangle, they create equal ratios. We use these equal ratios to prove the given relationship.

📝 Teacher's Note: Remind students that Basic Proportionality Theorem is about parallel lines making equal ratios. Draw the parallel lines clearly and mark the equal ratios.

🎯 Exam Tip: Write "Given" and "To Prove" clearly. State which theorem you are using. Show the cross multiplication step clearly.

Question 25. Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting diagonal AC in L and AD produced in E. Prove that: EL = 2 BL.
Answer:

[Diagram: Parallelogram ABCD with M as midpoint of CD, line BM extended to meet AC at L and AD produced at E, showing the configuration for the proof.]

\( \angle 1 = \angle 6 \) (Alternate interior angles)
\( \angle 2 = \angle 3 \) (Vertically opposite angles)
DM = MC (M is the mid-point of CD)
\( \therefore \triangle DEM \cong \triangle CBM \) (AAS congruence criterion)

So, DE = BC (Corresponding parts of congruent triangles)
Also, AD = BC (Opposite sides of a parallelogram)
\( \Rightarrow AE = AD + DE = 2BC \)

Now, \( \angle 1 = \angle 6 \) and \( \angle 4 = \angle 5 \)
\( \therefore \triangle ELA \sim \triangle BLC \) (AA similarity)

\( \Rightarrow \frac{EL}{BL} = \frac{EA}{BC} \)
\( \Rightarrow \frac{EL}{BL} = \frac{2BC}{BC} = 2 \)
\( \Rightarrow EL = 2BL \)

In simple words: We prove two triangles are the same first. Then we use similar triangles to show that one line is twice as long as another.

📝 Teacher's Note: Use colored chalk to mark the equal angles and equal sides. Students often get confused with too many letters, so mark clearly.

🎯 Exam Tip: State which triangles are congruent and which are similar. Write the reason (AAS, AA etc.) clearly. Show the final ratio calculation step by step.

Question 26. In the figure given below P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC.
(i) Calculate the ratio PQ : AC, giving reason for your answer.
(ii) In triangle ARC, ∠ARC = 90° and in triangle PQS, ∠PSQ = 90°. Given QS = 6 cm, calculate the length of AR.
Answer:

[Diagram: Triangle ABC with point P on AB, line PQ parallel to AC, and additional points R and S showing the right-angled triangles mentioned.]

(i) Given, AP : PB = 4 : 3.
Since, PQ || AC. Using Basic Proportionality theorem:
\( \frac{AP}{PB} = \frac{CQ}{QB} \)
\( \Rightarrow \frac{CQ}{QB} = \frac{4}{3} \)
\( \Rightarrow \frac{BQ}{BC} = \frac{3}{7} \) ... (1)

Now, ∠PQB = ∠ACB (Corresponding angles)
∠QPB = ∠CAB (Corresponding angles)
\( \therefore \triangle PBQ \sim \triangle ABC \) (AA similarity)
\( \Rightarrow \frac{PQ}{AC} = \frac{BQ}{BC} \)
\( \Rightarrow \frac{PQ}{AC} = \frac{3}{7} \) [Using (1)]

(ii) ∠ARC = ∠QSP = 90°
∠ACR = ∠SPQ (Alternate angles)
\( \therefore \triangle ARC \sim \triangle QSP \) (AA similarity)
\( \Rightarrow \frac{AR}{QS} = \frac{AC}{PQ} \)
\( \Rightarrow \frac{AR}{QS} = \frac{7}{3} \)
\( \Rightarrow AR = \frac{7 \times 6}{3} = 14 \text{ cm} \)

In simple words: Parallel lines create similar triangles with the same ratios. We use these ratios to find unknown lengths.

📝 Teacher's Note: Draw the ratio clearly as fractions. Show students how 4:3 becomes 4/7 and 3/7 when we consider the whole length. Use examples like sharing chocolate bars.

🎯 Exam Tip: Write the similarity reason (AA, SAS etc.). Show the ratio calculation clearly. Remember to write units in the final answer.

Question 27. In the right angled triangle QPR, PM is an altitude. Given that QR = 8 cm and MQ = 3.5 cm. Calculate the value of PR.
Answer:

[Diagram: Right-angled triangle QPR with right angle at P, altitude PM drawn to hypotenuse QR, showing QM = 3.5 cm and QR = 8 cm.]

We have:
∠QPR = ∠PMR = 90°
∠PRQ = ∠PRM (Common)
△PQR ∼ △MPR (AA similarity)

\( \therefore \frac{QR}{PR} = \frac{PR}{MR} \)

Given: QR = 8 cm, MQ = 3.5 cm
So MR = QR - MQ = 8 - 3.5 = 4.5 cm

PR² = QR × MR = 8 × 4.5 = 36
PR = 6 cm

In simple words: In a right triangle, when we draw a height to the longest side, it creates similar triangles. We use this to find the missing side.

📝 Teacher's Note: Show students the altitude property in right triangles. Draw it step by step. Use the formula PR² = QR × MR as a shortcut once they understand the similarity.

🎯 Exam Tip: Always find MR first by subtracting. Write the similarity clearly. Show the calculation PR² = QR × MR step by step.

Question 28. In the figure given below, the medians BD and CE of a triangle ABC meet at G. Prove that—
(i) △EGD ∼ △CGB
(ii) BG = 2 GD from (i) above.
Answer:

[Diagram: Triangle ABC with medians BD and CE intersecting at centroid G, showing points D and E as midpoints of sides AC and AB respectively.]

(i) To prove △EGD ∼ △CGB:
Since BD and CE are medians, they meet at centroid G.
In triangle ABC:
∠EGD = ∠CGB (Vertically opposite angles)

Also, by properties of centroid:
EG : GC = 1 : 2 and DG : GB = 1 : 2
This means EG/GC = DG/GB = 1/2

Therefore, △EGD ∼ △CGB (SAS similarity)

(ii) From similarity in (i):
Since △EGD ∼ △CGB, the ratio of corresponding sides is 1 : 2
Therefore: DG : GB = 1 : 2
This gives us: GB = 2 × DG
Hence: BG = 2 GD

In simple words: The centroid divides each median in the ratio 2:1. This creates similar triangles, which proves the 2:1 relationship.

📝 Teacher's Note: Remind students that medians always meet at centroid. The centroid divides each median in 2:1 ratio. Draw this clearly with measurements.

🎯 Exam Tip: State that G is the centroid first. Write the 2:1 property of centroid. Show how this creates the similar triangles clearly.

Exercise 15B

Question 1. In the following figure, point D divides AB in the ratio 3:5. Find:
(i) \( \frac{AE}{EC} \)
(ii) \( \frac{AD}{AB} \)
(iii) \( \frac{AE}{AC} \)
Also, if:
(iv) DE = 2.4 cm, find the length of BC.
(v) BC = 4.8 cm, find the length of DE.
Answer:

[Diagram: This diagram shows a triangle ABC with point D on side AB and point E on side AC. Line DE is parallel to BC.]

(i)
Given that \( \frac{AD}{DB} = \frac{3}{5} \)

So, \( \frac{AD}{AB} = \frac{3}{8} \)

In \( \triangle ADE \) and \( \triangle ABC \),
\( \angle ADE = \angle ABC \) ... (Since \( DE \parallel BC \), so the angles are corresponding angles.)
\( \angle A = \angle A \) ... (Common angle)
\( \therefore \triangle ADE \sim \triangle ABC \) ... (AA criterion for Similarity)

\( \Rightarrow \frac{AD}{AB} = \frac{AE}{AC} \)
\( \Rightarrow \frac{AE}{AC} = \frac{3}{8} \)

(ii)
Given that \( \frac{AD}{DB} = \frac{3}{5} \)

So, \( \frac{AD}{AB} = \frac{3}{8} \)

(iii)
Given that \( \frac{AD}{DB} = \frac{3}{5} \)

So, \( \frac{AD}{AB} = \frac{3}{8} \)

In \( \triangle ADE \) and \( \triangle ABC \),
\( \angle ADE = \angle ABC \) ... (Since \( DE \parallel BC \), so the angles are corresponding angles.)
\( \angle A = \angle A \) ... (Common angle)
\( \therefore \triangle ADE \sim \triangle ABC \) ... (AA criterion for Similarity)

\( \Rightarrow \frac{AD}{AB} = \frac{AE}{AC} \)
\( \Rightarrow \frac{AE}{AC} = \frac{3}{8} \)

(iv)
Given that \( \frac{AD}{DB} = \frac{3}{5} \)

So, \( \frac{AD}{AB} = \frac{3}{8} \)

In \( \triangle ADE \) and \( \triangle ABC \),
\( \angle ADE = \angle ABC \) ... (Since \( DE \parallel BC \), so the angles are corresponding angles.)
\( \angle A = \angle A \) ... (Common angle)
\( \therefore \triangle ADE \sim \triangle ABC \) ... (AA criterion for Similarity)

\( \Rightarrow \frac{AD}{AB} = \frac{DE}{BC} \)
\( \Rightarrow \frac{3}{8} = \frac{2.4}{BC} \)
\( \Rightarrow BC = 6.4 \) cm

(v)
Given that \( \frac{AD}{DB} = \frac{3}{5} \)

So, \( \frac{AD}{AB} = \frac{3}{8} \)

In \( \triangle ADE \) and \( \triangle ABC \),
\( \angle ADE = \angle ABC \) ... (Since \( DE \parallel BC \), so the angles are corresponding angles.)
\( \angle A = \angle A \) ... (Common angle)
\( \therefore \triangle ADE \sim \triangle ABC \) ... (AA criterion for Similarity)

\( \Rightarrow \frac{AD}{AB} = \frac{DE}{BC} \)
\( \Rightarrow \frac{3}{8} = \frac{DE}{4.8} \)
\( \Rightarrow DE = 1.8 \) cm

In simple words: When two triangles have the same angles, their sides are in the same ratio. We use this fact to find missing lengths.

📝 Teacher's Note: Draw two similar triangles on the board. Show students that when triangles are similar, all matching sides have the same ratio. This makes finding unknown sides easy.

🎯 Exam Tip: Always write "triangles are similar" and mention which similarity test you used (AA, SSS, or SAS). Then write the ratio of corresponding sides clearly.

Question 2. In the given figure, PQ//AB; CQ = 4.8 cm QB = 3.6 cm and AB = 6.3 cm. Find:
(i) \( \frac{CP}{PA} \)
(ii) PQ
(iii) If AP=x, then the value of AC in terms of x.
Answer:

[Diagram: This diagram shows a triangle ABC with point P on side AC and point Q on side BC. Line PQ is parallel to AB.]

(i)
In \( \triangle CPQ \) and \( \triangle CAB \),
\( \angle PCQ = \angle ACB \) ... (Since \( PQ \parallel AB \), so the angles are corresponding angles.)
\( \angle C = \angle C \) ... (Common angle)
\( \therefore \triangle CPQ \sim \triangle CAB \) ... (AA criterion for Similarity)

\( \Rightarrow \frac{CP}{CA} = \frac{CQ}{CB} \)
\( \Rightarrow \frac{CP}{CA} = \frac{4.8}{8.4} = \frac{4}{7} \)

So, \( \frac{CP}{PA} = \frac{4}{3} \)

(ii)
In \( \triangle CPQ \) and \( \triangle CAB \),
\( \angle PCQ = \angle ACB \) ... (Since \( PQ \parallel AB \), so the angles are corresponding angles.)
\( \angle C = \angle C \) ... (Common angle)
\( \therefore \triangle CPQ \sim \triangle CAB \) ... (AA criterion for Similarity)

\( \Rightarrow \frac{PQ}{AB} = \frac{CQ}{CB} \)
\( \Rightarrow \frac{PQ}{6.3} = \frac{4.8}{8.4} \)
\( \Rightarrow PQ = 3.6 \) cm

(iii)
In \( \triangle CPQ \) and \( \triangle CAB \),
\( \angle PCQ = \angle ACB \) ... (Since \( PQ \parallel AB \), so the angles are corresponding angles.)
\( \angle C = \angle C \) ... (Common angle)
\( \therefore \triangle CPQ \sim \triangle CAB \) ... (AA criterion for Similarity)

\( \Rightarrow \frac{CP}{AC} = \frac{CQ}{CB} \)
\( \Rightarrow \frac{CP}{AC} = \frac{4.8}{8.4} = \frac{4}{7} \)

So, if AC is 7 parts, and CP is 4 parts, then PA is 3 parts.

Thus, \( AC = \frac{7}{3} PA = \frac{7}{3} x \)

In simple words: When a line is parallel to one side of a triangle, it creates smaller triangles that are similar to the big triangle. We use ratios to find missing parts.

📝 Teacher's Note: Use a paper triangle and fold it to show parallel lines. Students can see how the small triangle inside is similar to the big one. This makes the concept clear.

🎯 Exam Tip: Write "PQ parallel to AB" clearly. Then state which triangles are similar and why. Use the Basic Proportionality Theorem to set up ratios correctly.

Question 3. A line PQ is drawn parallel to the side BC of △ABC which cuts side AB at P and side AC at Q. If AB = 9.0 cm, CA = 6.0 cm and AQ = 4.2 cm, find the length of AP.
Answer:

[Diagram: This diagram shows triangle ABC with a line PQ drawn parallel to BC, where P is on side AB and Q is on side AC. The measurements show AB = 9 cm, AC = 6 cm, and AQ = 4.2 cm.]

In △APQ and △ABC,
∠AQP = ∠ABC .... (Since PQ || BC, so the angles are corresponding angles.)
∠PAQ = ∠BAC .... (Common angle)
∴ △APQ ~ △ABC .... (AA criterion for Similarity)
\[ \Rightarrow \frac{AP}{AB} = \frac{AQ}{AC} \]
\[ \Rightarrow \frac{AP}{9} = \frac{4.2}{6} \]
\[ \Rightarrow AP = 6.3 \text{ cm} \]
In simple words: When a line is parallel to one side of a triangle, it divides the other two sides in the same ratio. We use this rule to find AP.

📝 Teacher's Note: Draw a triangle on the board and show how parallel lines create similar triangles. Students often forget that ratios must be written correctly - same triangle parts on top.

🎯 Exam Tip: Always write the similarity statement clearly and set up the ratio correctly. Write the final answer with units.

Question 4. In △ABC, D and E are the points on sides AB and AC respectively. Find whether DE // BC, if:
(i) AB = 9 cm, AD = 4 cm, AE = 6 cm and EC = 7.5 cm.
(ii) AB = 63 cm, EC = 11.0 cm, AD = 0.8 cm and AE = 1.6 cm.
Answer:
(i)

[Diagram: This diagram shows triangle ABC with points D on side AB and E on side AC. The measurements show AD = 4 cm, AB = 9 cm, AE = 6 cm, and EC = 7.5 cm.]

In △ADE and △ABC,
\[ \frac{AE}{EC} = \frac{6}{7.5} = \frac{4}{5} \]
\[ \frac{AD}{BD} = \frac{4}{5} \] .... (Since AB = 9 cm and AD = 4 cm)
So, \[ \frac{AE}{EC} = \frac{AD}{BD} \]
∴ DE || BC .... (By the Converse of Mid-point theorem)

(ii)

[Diagram: This diagram shows triangle ABC with points D on side AB and E on side AC. The measurements show AD = 0.8 cm, AB = 63 cm, AE = 1.6 cm, and EC = 11 cm.]

In △ADE and △ABC,
\[ \frac{AE}{EC} = \frac{1.6}{11} = \frac{0.8}{5.5} \]
\[ \frac{AD}{BD} = \frac{0.8}{6.3-8} = \frac{0.8}{5.5} \]
So, \[ \frac{AE}{EC} = \frac{AD}{BD} \]
∴ DE || BC .... (By the Converse of Mid-point theorem)
In simple words: If a line divides two sides of a triangle in the same ratio, then that line is parallel to the third side.

📝 Teacher's Note: Show students how to find BD by subtracting AD from AB. Many students make mistakes in this calculation step.

🎯 Exam Tip: Always check if the ratios are equal. If ratios are equal, then DE || BC. If not equal, then DE is not parallel to BC.

Question 5. In the given figure, △ABC ~ △ADE. If AE : EC = 4 : 7 and DE = 6.6 cm, find BC. If 'x' be the length of the perpendicular from A to DE, find the length of perpendicular from A to BC in terms of 'x'.
Answer:

[Diagram: This diagram shows triangle ABC with triangle ADE inside it, where D is on AB and E is on AC.]

Given that △ABC ~ △ADE.
∠ABC = ∠ADE and ∠ACB = ∠AED
So, DE || BC
Also, \[ \frac{AB}{AD} = \frac{AC}{AE} = \frac{11}{4} \] .... \[ \left( \text{Since } \frac{AE}{EC} = \frac{4}{7} \right) \]
In △ADP and △ABQ,
∠ADP = ∠ABQ .... (Since DP || BQ.)
∠APD = ∠AQB .... (Since DP || BQ.)
So, △ADP ~ △ABQ .... (AA Criterion for Similarity)
\[ \Rightarrow \frac{AD}{AB} = \frac{AP}{AQ} \]
\[ \Rightarrow \frac{4}{11} = \frac{x}{AQ} \]
\[ \Rightarrow AQ = \frac{11}{4} x \]
In simple words: When triangles are similar, all corresponding sides are in the same ratio. We use this to find BC and the perpendicular length.

📝 Teacher's Note: Draw perpendiculars on the board to show students what x represents. Similar triangles have all corresponding parts in the same ratio.

🎯 Exam Tip: Write the similarity ratio first, then use it to find all unknown values. Keep the ratio consistent throughout your solution.

Question 6. A line segment DE is drawn parallel to base BC of △ABC which cuts AB at point D and AC at point E. If AB = 5 BD and EC = 3.2 cm, find the length of AE.
Answer:
Since DE || BC, △ADE ~ △ABC
\[ \Rightarrow \frac{AD}{BD} = \frac{AE}{EC} \]
\[ \Rightarrow \frac{AB - BD}{BD} = \frac{AE}{EC} \]
\[ \Rightarrow \frac{5BD - BD}{BD} = \frac{AE}{EC} \]
\[ \Rightarrow \frac{4BD}{BD} = \frac{AE}{3.2} \]
\[ \Rightarrow AE = 4 \times 3.2 = 12.8 \text{ cm} \]
In simple words: Since DE is parallel to BC, the triangles are similar. We use the ratio of sides to find AE.

📝 Teacher's Note: Help students understand that AB = 5 BD means AD = 4 BD. Draw this on the board to make it clear.

🎯 Exam Tip: When AB = 5 BD, remember that AD = AB - BD = 4 BD. This is a common error point in exams.

Question 7. In the figure, given below, AB, Cd and EF are parallel lines. Given AB = 7.5 cm, DC = y cm, EF = 4.5 cm, BC = x cm and CE = 3 cm, calculate the values of x and y.
Answer:

[Diagram: This diagram shows three parallel lines AB, CD, and EF intersected by two transversals, creating similar triangles.]

Since AB || CD || EF, we can use the property of similar triangles formed by parallel lines.
For parallel lines cut by transversals:
\[ \frac{BC}{CE} = \frac{AB}{EF} \]
\[ \frac{x}{3} = \frac{7.5}{4.5} \]
\[ \frac{x}{3} = \frac{5}{3} \]
\[ x = 5 \text{ cm} \]

Also, \[ \frac{BC}{CD} = \frac{AB}{EF} \]
\[ \frac{5}{y} = \frac{7.5}{4.5} \]
\[ \frac{5}{y} = \frac{5}{3} \]
\[ y = 3 \text{ cm} \]
In simple words: When parallel lines are cut by two other lines, they create proportional segments. We use this property to find x and y.

📝 Teacher's Note: Draw three parallel lines on the board and show how the segments are proportional. This concept is used in many geometry problems.

🎯 Exam Tip: Set up the proportion correctly by keeping corresponding segments in the same position. Always solve for one variable at a time.

Solution:

Answer: In \(\triangle BEF\), DC || EF.
\(\frac{BD}{DF} = \frac{BC}{CE}\)
\(\Rightarrow \frac{BD}{DF} = \frac{x}{3}\)
So, BD = x and DF = 3.

In \(\triangle AFB\), DC || AB.
\(\Rightarrow \frac{FD}{CD} = \frac{FB}{AB}\)
\(\Rightarrow \frac{FD}{CD} = \frac{FD + DB}{AB}\)
\(\Rightarrow \frac{3}{y} = \frac{x + 3}{7.5}\) ...(i)

In \(\triangle BFE\), DC || EF.
\(\Rightarrow \frac{BC}{CD} = \frac{BE}{EF}\)
\(\Rightarrow \frac{BC}{CD} = \frac{BC + CE}{EF}\)
\(\Rightarrow \frac{x}{y} = \frac{x + 3}{4.5}\)
\(\Rightarrow y = \frac{4.5x}{x + 3}\) ...(ii)

Substituting (ii) in (i), we get
\(\frac{3}{\frac{4.5x}{x + 3}} = \frac{x + 3}{7.5}\)
\(\Rightarrow \frac{3x + 9}{4.5x} = \frac{x + 3}{7.5}\)
\(\Rightarrow 22.5x + 67.5 = 4.5x^2 + 13.5x\)
\(\Rightarrow 4.5x^2 + 13.5x - 22.5x - 67.5 = 0\)
\(\Rightarrow x^2 - 2x - 15 = 0\)
\(\Rightarrow (x - 5)(x + 3) = 0\)
So, x = 5 and x = -3.

Since side of a triangle cannot be negative, x = 5.
Substituting this value in (ii), we get
\(y = \frac{4.5(5)}{x + 3} = 2.8125\)
Hence, x = 5 and y = 2.8125

📝 Teacher's Note: Use similar triangles theorem. When two lines are parallel, they cut the other lines in the same ratio. This helps to make equations.

🎯 Exam Tip: Write "DC || EF" clearly first. Then write the ratio property. Set up equations step by step. Always check if the answer is positive.

Question 8. In the figure, given below, PQR is a right-angle triangle right angled at Q. XY is parallel to QR, PQ = 6 cm, PY = 4 cm and PX : XQ = 1:2. Calculate the lengths of PR and QR.

[Diagram: This diagram shows a right-angled triangle PQR with right angle at Q. Line XY is parallel to QR, with X on PQ and Y on PR.]

Solution:

Answer: Given that \(\frac{PX}{XQ} = \frac{1}{2}\) and XY || QR.

So, \(\frac{PX}{XQ} = \frac{PY}{YR} = \frac{1}{2}\).
Since PY = 4 cm, YR = 8 cm.
Hence, PR = 12 cm.
Since \(\triangle PQR\) is a right-angled triangle.
By Pythagoras theorem,
\(QR^2 = PR^2 - PQ^2\)
\(\Rightarrow QR^2 = 12^2 - 6^2\)
\(\Rightarrow QR^2 = 144 - 36\)
\(\Rightarrow QR^2 = 108\)
\(\Rightarrow QR = 10.39\) cm

📝 Teacher's Note: When a line is parallel to one side of a triangle, it divides the other two sides in the same ratio. This is the basic proportionality theorem.

🎯 Exam Tip: Write the ratio clearly. Use Pythagoras theorem: \(c^2 = a^2 + b^2\) for right triangles. Show all calculation steps.

Question 9. In the following figure, M is mid-point of BC of a parallelogram ABCD. DM intersects the diagonal AC at P and AB produced at E. Prove that PE = 2PD.

[Diagram: This diagram shows parallelogram ABCD with M as midpoint of BC. Line DM intersects diagonal AC at P and extended line AB at E.]

Solution:

Answer: In \(\triangle BME\) and \(\triangle DMC\),
\(\angle BME = \angle CMD\) ...(vertically opposite angles)
\(\angle MCD = \angle MBE\) ...(alternate angles)
BM = BC ...(M is the mid-point of BC)
So, \(\triangle BME \cong \triangle DMC\) ...(AAS congruence criterion)
\(\Rightarrow BE = DC = AB\)
In \(\triangle DCP\) and \(\triangle EPA\),
\(\angle DPC = \angle EPA\) ...(vertically opposite angles)
\(\angle CDP = \angle AEP\) ...(alternate angles)
\(\triangle DCP \sim \triangle EAP\) ...(AA criterion for Similarity)
\(\Rightarrow \frac{DC}{EA} = \frac{CP}{AP} = \frac{PD}{EP}\)
\(\Rightarrow \frac{DC}{EA} = \frac{PD}{PE}\)
\(\Rightarrow \frac{EA}{DC} = \frac{PE}{PD}\)
\(\Rightarrow \frac{PE}{PD} = \frac{AB + EA}{DC}\)
\(\Rightarrow \frac{PE}{PD} = \frac{2DC}{DC}\)
\(\Rightarrow PE = 2PD\)

📝 Teacher's Note: Use properties of parallelogram first. Then use similar triangles. The key is finding that BE = AB using congruent triangles.

🎯 Exam Tip: Write "M is midpoint of BC" clearly. Use AAS congruence first, then AA similarity. Show each step of the ratio calculation.

Question 10. The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE=4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.

[Diagram: This diagram shows parallelogram ABCD with point E on AD and line CE extended to meet extended line BA at point F.]

Solution:

Answer: AF = 8 cm and AB = 12 cm
So, FB = 20 cm.
In \(\triangle DEC\) and \(\triangle EAF\),
\(\angle DEC = \angle EAF\) ...(vertically opposite angles)
\(\angle EDC = \angle EAF\) ...(alternate angles)
So, \(\triangle DEC \sim \triangle AEF\) ...(AA criterion for Similarity)
\(\Rightarrow \frac{DE}{AE} = \frac{EC}{EF} = \frac{DC}{AF}\)
\(\Rightarrow \frac{DE}{AE} = \frac{DC}{AF}\)
\(\Rightarrow \frac{DE}{AE} = \frac{AB}{AF}\)
\(\Rightarrow \frac{DE}{4} = \frac{12}{8}\)
\(\Rightarrow DE = 6\) cm
So, AD = AE + ED = 4 + 6 = 10 cm
Perimeter of the parallelogram ABCD
= AB + BC + CD + AD
= 12 + 10 + 12 + 10
= 44 cm

📝 Teacher's Note: In parallelogram, opposite sides are equal. Use similar triangles to find the unknown side. Then add all four sides for perimeter.

🎯 Exam Tip: Write "opposite sides of parallelogram are equal" clearly. Use AA similarity criterion. Calculate perimeter as 2(length + width).

Exercise 15C

Question 1.

(i) The ratio between the corresponding sides of two similar triangles is 2 is to 5. Find the ratio between the areas of these triangles.

(ii) Areas of two similar triangles are 98 sq. cm and 128 sq. cm. Find the ratio between the lengths of their corresponding sides.

Solution:

Answer: We know that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

(i) Required ratio = \(\frac{2^2}{5^2} = \frac{4}{25}\)

(ii) Required ratio = \(\sqrt{\frac{98}{128}} = \sqrt{\frac{49}{64}} = \frac{7}{8}\)

📝 Teacher's Note: For similar triangles, area ratio equals the square of side ratio. If sides are in ratio 2:5, then areas are in ratio 4:25.

🎯 Exam Tip: Always square the side ratio to get area ratio. For area to side ratio, take square root. Write the final answer as a simple fraction.

Question 2. A line PQ is drawn parallel to the base BC, of \(\triangle ABC\) which meets sides AB and AC at points P and Q respectively. If AP = PB; find the value of:

(i) \(\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle APQ}\); (ii) \(\frac{\text{Area of } \triangle APQ}{\text{Area of trapezium } PBCQ}\)

[Diagram: This diagram shows triangle ABC with line PQ parallel to BC, where P is on AB and Q is on AC.]

Solution:

Answer: Since PQ || BC and AP = PB, we have AP:AB = 1:2
For similar triangles \(\triangle APQ\) and \(\triangle ABC\):

(i) \(\frac{\text{Area of } \triangle ABC}{\text{Area of } \triangle APQ} = \left(\frac{AB}{AP}\right)^2 = \left(\frac{2}{1}\right)^2 = 4\)

(ii) Area of trapezium PBCQ = Area of \(\triangle ABC\) - Area of \(\triangle APQ\)
If Area of \(\triangle APQ\) = 1 unit, then Area of \(\triangle ABC\) = 4 units
So Area of trapezium PBCQ = 4 - 1 = 3 units
Therefore, \(\frac{\text{Area of } \triangle APQ}{\text{Area of trapezium } PBCQ} = \frac{1}{3}\)

📝 Teacher's Note: When AP = PB, point P divides AB in ratio 1:1. So AP:AB = 1:2. Use this ratio squared for area comparison.

🎯 Exam Tip: Write the side ratio first. Square it for area ratio. For trapezium area, subtract triangle area from big triangle area.

Question 3. The perimeters of two similar triangles are 30 cm and 24cm. If one side of first triangle is 12cm, determine the corresponding side of the second triangle.
Answer:
Given:
Let \( \triangle ABC \sim \triangle DEF \)
Perimeter of \( \triangle ABC = 30 \) cm
Perimeter of \( \triangle DEF = 24 \) cm
One side of first triangle \( AB = 12 \) cm
Let the corresponding side of second triangle be \( DE \)

Step 1: Use the property of similar triangles
For similar triangles, the ratio of corresponding sides equals the ratio of perimeters.
\[ \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle DEF} \]

Step 2: Calculate the ratio
\[ \frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle DEF} = \frac{AB}{DE} \]
\[ \frac{30}{24} = \frac{12}{DE} \]

Step 3: Solve for DE
\[ DE = \frac{12 \times 24}{30} = \frac{288}{30} = 9.6 \text{ cm} \]

Therefore, the corresponding side of the second triangle is 9.6 cm.
In simple words: When triangles are similar, all their sides have the same ratio. The perimeter ratio is 30:24, so the side ratio is also 30:24. Using this, we find the unknown side.

📝 Teacher's Note: Draw two similar triangles on the board. Show that if one triangle is bigger, all its sides are bigger by the same amount. The ratio stays the same for all parts.

🎯 Exam Tip: Always write "Given" and "To find" clearly. Write the similarity symbol \( \sim \) and state the ratio property. You get marks for showing the formula before calculation.

Question 4. In the given figure AX : XB = 3 : 5. Find: (i) the length of BC, if length of XY is 18 cm. (ii) ratio between the areas of trapezium XBCY and triangle ABC.
Answer:
Given:
\( AX : XB = 3 : 5 \)
Length of \( XY = 18 \) cm
\( XY \parallel BC \)

(i) Finding length of BC:
Step 1: Find the ratio AX : AB
\( AX : XB = 3 : 5 \)
\[ \frac{AX}{AB} = \frac{AX}{AX + XB} = \frac{3}{3 + 5} = \frac{3}{8} \]

Step 2: Use properties of similar triangles
In \( \triangle AXY \) and \( \triangle ABC \),
As \( XY \parallel BC \), corresponding angles are equal
\( \angle AXY = \angle ABC \)
\( \angle AYX = \angle ACB \)
\( \triangle AXY \sim \triangle ABC \)

Step 3: Apply similarity ratio
\[ \frac{AX}{AB} = \frac{XY}{BC} \]
\[ \frac{3}{8} = \frac{18}{BC} \]
\[ BC = \frac{18 \times 8}{3} = 48 \text{ cm} \]

(ii) Finding ratio of areas:
Step 4: Find area ratio of similar triangles
\[ \frac{\text{Area of } \triangle AXY}{\text{Area of } \triangle ABC} = \left(\frac{AX}{AB}\right)^2 = \left(\frac{3}{8}\right)^2 = \frac{9}{64} \]

Step 5: Find area of trapezium XBCY
Area of trapezium XBCY = Area of \( \triangle ABC \) - Area of \( \triangle AXY \)
\[ \frac{\text{Area of trapezium XBCY}}{\text{Area of } \triangle ABC} = \frac{64 - 9}{64} = \frac{55}{64} \]

Answers:
(i) Length of BC = 48 cm
(ii) Ratio of areas = 55:64

In simple words: When a line is parallel to one side of a triangle, it creates smaller similar triangles. The ratio of sides helps us find unknown lengths. Area ratios are the square of side ratios.

📝 Teacher's Note: Use paper cutting to show how parallel lines create similar triangles. Students can see that the small triangle has the same shape as the big one.

🎯 Exam Tip: Remember that area ratio equals (side ratio)². Always square the side ratio when finding area ratio. Write both parts clearly with separate headings.

Question 5. ABC is a triangle. PQ is a line segment intersecting AB in P and AC in Q such that PQ || BC and divides triangle ABC into two parts equal in area. Find the value of ratio BP : AB.
Answer:
Given:
In \( \triangle ABC \), \( PQ \parallel BC \)
Area of \( \triangle APQ = \) Area of trapezium PQCB
This means PQ divides the triangle into two equal areas.

Step 1: Set up the area relationship
Since the two parts have equal areas:
\[ \text{Area of } \triangle APQ = \frac{1}{2} \times \text{Area of } \triangle ABC \]

Step 2: Use the similarity ratio
\[ \frac{\text{Area of } \triangle APQ}{\text{Area of } \triangle ABC} = \frac{1}{2} \]

Step 3: Apply area ratio formula for similar triangles
Since \( \triangle APQ \sim \triangle ABC \):
\[ \frac{AP^2}{AB^2} = \frac{1}{2} \]

Step 4: Find the side ratio
\[ \frac{AP}{AB} = \frac{1}{\sqrt{2}} \]

Step 5: Calculate BP : AB
\[ \frac{AB - BP}{AB} = \frac{1}{\sqrt{2}} \]
\[ 1 - \frac{BP}{AB} = \frac{1}{\sqrt{2}} \]
\[ \frac{BP}{AB} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{2 - \sqrt{2}}{2} \]

Therefore, BP : AB = \( (2 - \sqrt{2}) : 2 \)

In simple words: When a parallel line cuts a triangle into two equal areas, it doesn't cut at the middle. It cuts closer to the vertex because area depends on the square of the distance.

📝 Teacher's Note: Show students that equal areas don't mean equal distances. Use a paper triangle and fold it to show where the line must be to get equal areas.

🎯 Exam Tip: Write "equal areas" clearly in the given section. Remember to square root the area ratio to get the side ratio. Keep all steps with \( \sqrt{2} \) clearly shown.

Question 6. In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4. Calculate the value of ratio: (i) PL/PQ and then LM/QR (ii) Area of △LMN/Area of △MNR (iii) Area of △LQM/Area of △LQN
Answer:
Given:
In \( \triangle PQR \), \( LM \parallel QR \)
\( PM : MR = 3 : 4 \)

(i) Finding PL/PQ and LM/QR:
Step 1: Use Basic Proportionality Theorem
Since \( LM \parallel QR \):
\[ \frac{PL}{LQ} = \frac{PM}{MR} = \frac{3}{4} \]

Step 2: Find PL/PQ
\[ \frac{PL}{PQ} = \frac{PL}{PL + LQ} = \frac{3}{3 + 4} = \frac{3}{7} \]

Step 3: Find LM/QR
In \( \triangle PLM \) and \( \triangle PQR \):
Since \( LM \parallel QR \), corresponding angles are equal
\[ \triangle PLM \sim \triangle PQR \]
\[ \frac{LM}{QR} = \frac{PL}{PQ} = \frac{3}{7} \]

(ii) Finding Area of △LMN/Area of △MNR:
Step 4: Identify triangles with common vertex
Triangles \( \triangle LMN \) and \( \triangle MNR \) have common vertex at M and their bases LN and NR are on the same straight line.
\[ \frac{\text{Area of } \triangle LMN}{\text{Area of } \triangle MNR} = \frac{LN}{NR} \]

From similarity and Basic Proportionality Theorem:
\[ \frac{LN}{NR} = \frac{3}{7} \]

(iii) Finding Area of △LQM/Area of △LQN:
Step 5: Use the ratio of segments
Triangles \( \triangle LQM \) and \( \triangle LQN \) have common vertex at L and their bases QM and QN are on line QR.
\[ \frac{\text{Area of } \triangle LQM}{\text{Area of } \triangle LQN} = \frac{QM}{QN} = \frac{10}{7} \]

Answers:
(i) \( \frac{PL}{PQ} = \frac{3}{7} \) and \( \frac{LM}{QR} = \frac{3}{7} \)
(ii) \( \frac{\text{Area of } \triangle LMN}{\text{Area of } \triangle MNR} = \frac{3}{7} \)
(iii) \( \frac{\text{Area of } \triangle LQM}{\text{Area of } \triangle LQN} = \frac{10}{7} \)

In simple words: When a line is parallel to one side, it cuts the other two sides in the same ratio. The area ratio between triangles with common vertex equals the base ratio.

📝 Teacher's Note: Draw the triangle clearly and mark all the ratios. Show how the same ratio appears in different parts. Use colored pencils to highlight similar triangles.

🎯 Exam Tip: Write the Basic Proportionality Theorem clearly. Always identify which triangles are similar and which have common vertex. Label your diagram with given ratios.

Question 7. The given diagram shows two isosceles triangles which are similar also. In the given diagram, PQ and BC are not parallel: PC = 4, AQ = 3, QB = 12, BC = 15 and AP = PQ. Calculate: (i) the length of AP (ii) the ratio of the areas of triangle APQ and triangle ABC.
Answer:
Given:
Two isosceles triangles are similar
PQ and BC are not parallel
PC = 4, AQ = 3, QB = 12, BC = 15
AP = PQ (isosceles condition)

Step 1: Find the total lengths
AC = AQ + QC = 3 + 4 = 7
AB = AQ + QB = 3 + 12 = 15

Step 2: Set up similarity condition
Since the triangles are similar and isosceles:
Let AP = PQ = x

For similar triangles \( \triangle APQ \) and \( \triangle ABC \):
\[ \frac{AP}{AB} = \frac{AQ}{AC} = \frac{PQ}{BC} \]

Step 3: Use the known ratio
\[ \frac{AQ}{AC} = \frac{3}{7} \]

Therefore:
\[ \frac{AP}{AB} = \frac{3}{7} \]
\[ \frac{AP}{15} = \frac{3}{7} \]
\[ AP = \frac{15 \times 3}{7} = \frac{45}{7} \text{ cm} \]

Step 4: Calculate the area ratio
For similar triangles, the ratio of areas equals the square of the ratio of corresponding sides:
\[ \frac{\text{Area of } \triangle APQ}{\text{Area of } \triangle ABC} = \left(\frac{AP}{AB}\right)^2 = \left(\frac{3}{7}\right)^2 = \frac{9}{49} \]

Answers:
(i) Length of AP = \( \frac{45}{7} \) cm
(ii) Ratio of areas = 9:49

In simple words: Even when lines are not parallel, triangles can be similar if their angles are equal. We use the given side lengths to find the similarity ratio, then calculate unknown sides and areas.

📝 Teacher's Note: Emphasize that similarity doesn't always need parallel lines. Show that isosceles triangles have two equal sides. Draw clearly to show which sides correspond.

🎯 Exam Tip: Write "isosceles" means two equal sides. State the similarity clearly with the symbol \( \sim \). For area ratio, always square the side ratio. Show all fraction calculations step by step.

Question 8. In the figure, given below, ABCD is a parallelogram. P is a point on BC such that BP : PC = 1:2. DP produced meets AB produced at Q. Given the area of triangle CPQ = 20 cm². Calculate—
(i) area of triangle CDP
(ii) area of parallelogram ABCD [1996]

[Diagram: This diagram shows parallelogram ABCD with point P on side BC, and line DP extended to meet AB extended at point Q.]

Solution:

(i) Finding area of triangle CDP:
In triangles BPQ and CPD
\( \angle BPQ = \angle CPD \) (Vertically opposite angles)
\( \angle BQP = \angle PDC \) (Alternate angles)
\( \triangle BPQ \sim \triangle CPD \) (AA similarity)

Since BP : PC = 1 : 2, we have \( \frac{BP}{PC} = \frac{1}{2} \)

For similar triangles, \( \frac{\text{ar}(\triangle BPQ)}{\text{ar}(\triangle CPD)} = \left(\frac{BP}{PC}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)

Since \( \text{ar}(\triangle BPQ) = \frac{1}{2} \times \text{ar}(\triangle CPQ) \), and \( \text{ar}(\triangle CPQ) = 20 \)
\( \text{ar}(\triangle BPQ) = 10 \text{ cm}^2 \)

From the ratio: \( \frac{10}{\text{ar}(\triangle CPD)} = \frac{1}{4} \)
\( \text{ar}(\triangle CPD) = 40 \text{ cm}^2 \)

(ii) Finding area of parallelogram ABCD:
In triangles BQP and AQD
Since BP || AD, corresponding angles are equal
\( \angle QBP = \angle QAD \)
\( \angle BQP = \angle AQD \) (Common)
\( \triangle BQP \sim \triangle AQD \) (AA similarity)

Since BP : PC = 1 : 2, and PC = PQ/3, we get PQ : QD = 1 : 3

For similar triangles: \( \frac{\text{ar}(\triangle AQD)}{\text{ar}(\triangle BQP)} = \left(\frac{AQ}{BQ}\right)^2 = 9 \)

\( \text{ar}(\triangle AQD) = 9 \times 10 = 90 \text{ cm}^2 \)

Area of triangle ADPB = \( \text{ar}(\triangle AQD) - \text{ar}(\triangle BQP) = 90 - 10 = 80 \text{ cm}^2 \)

Area of parallelogram ABCD = \( \text{ar}(\triangle CDP) + \text{ar}(\triangle ADPB) = 40 + 80 = 120 \text{ cm}^2 \)

In simple words: We used similar triangles to find the areas. When triangles are similar, their area ratio equals the square of their side ratio. We built up the parallelogram area by adding triangle areas.

📝 Teacher's Note: Draw the diagram clearly and mark all the angles. Show students how alternate angles and vertically opposite angles help prove similarity. This makes the solution easy to follow.

🎯 Exam Tip: Always write "AA similarity" when proving triangles similar. Show the area ratio formula clearly. Mark all given information on your diagram to avoid mistakes.

Question 9. In the given figure, BC is parallel to DE. Area of triangle ABC = 25 cm². Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC. Also, Find the area of triangle BCD.

[Diagram: This diagram shows triangle ABC with DE parallel to BC, forming trapezium BCED.]

Solution:
In triangles ABC and ADE,
Since BC || DE, corresponding angles are equal
\( \angle ABC = \angle ADE \)
\( \angle ACB = \angle AED \)
\( \triangle ABC \sim \triangle ADE \)

For similar triangles: \( \frac{\text{ar}(\triangle ABC)}{\text{ar}(\triangle ADE)} = \frac{BC^2}{DE^2} \)

Since \( \text{ar}(\triangle ADE) = \text{ar}(\triangle ABC) + \text{ar}(\text{trapezium BCED}) \)
\( \text{ar}(\triangle ADE) = 25 + 24 = 49 \text{ cm}^2 \)

Substituting: \( \frac{25}{49} = \frac{BC^2}{14^2} = \frac{BC^2}{196} \)

\( BC^2 = \frac{25 \times 196}{49} = 100 \)
\( BC = 10 \text{ cm} \)

Finding area of triangle BCD:
For trapezium BCED: Area = \( \frac{1}{2}(\text{sum of parallel sides}) \times h \)
Given: Area = 24 cm², BC = 10 cm, DE = 14 cm

\( 24 = \frac{1}{2}(10 + 14) \times h \)
\( 24 = \frac{1}{2} \times 24 \times h \)
\( h = 2 \)

Area of \( \triangle BCD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times h = \frac{1}{2} \times 10 \times 2 = 10 \text{ cm}^2 \)

In simple words: When two lines are parallel, they create similar triangles. We used the area ratio to find BC length. Then we used trapezium formula to find the height and triangle area.

📝 Teacher's Note: Explain that parallel lines create similar triangles with the same angles. Show students how to find the total area first, then use ratios. This step-by-step approach prevents confusion.

🎯 Exam Tip: Always find the total area of the large triangle first. Write the similarity clearly. Use the trapezium area formula: \( \frac{1}{2}(a + b) \times h \). Check your answer by adding areas.

Question 10. The given figure shows a trapezium in which AB is parallel to DC and diagonals AC and BD intersect at point P. If AP : CP = 3 : 5. Find:
(i) △ APB : △ CPB
(ii) △ DPC : △ APB
(iii) △ ADP : △ APB
(iv) △ APB : △ ADB

[Diagram: This diagram shows trapezium ABCD with AB parallel to DC, and diagonals AC and BD intersecting at point P.]

Solution:

(i) Since △ APB and △ CPB have common vertex at B and their bases AP and PC are along the same straight line
\( \frac{\text{ar}(\triangle APB)}{\text{ar}(\triangle CPB)} = \frac{AP}{PC} = \frac{3}{5} \)

(ii) Since △ DPC and △ BPA are similar
\( \frac{\text{ar}(\triangle DPC)}{\text{ar}(\triangle BPA)} = \left(\frac{PC}{AP}\right)^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9} \)

(iii) Since △ ADP and △ APB have common vertex at A and their bases DP and PB are along the same straight line
\( \frac{\text{ar}(\triangle ADP)}{\text{ar}(\triangle APB)} = \frac{DP}{PB} = \frac{5}{3} \)
Since △ APB ~ △ CPD \( \Rightarrow \frac{AP}{PC} = \frac{BP}{PD} = \frac{3}{5} \Rightarrow \frac{BP}{BD} = \frac{3}{8} \)

(iv) Since △ APB and △ ADB have common vertex at A and their bases BP and BD are along the same straight line
\( \frac{\text{ar}(\triangle APB)}{\text{ar}(\triangle ADB)} = \frac{PB}{BD} = \frac{3}{8} \)
Since △ APB ~ △ CPD \( \Rightarrow \frac{AP}{PC} = \frac{BP}{PD} = \frac{3}{5} \Rightarrow \frac{BP}{BD} = \frac{3}{8} \)

In simple words: When triangles share a vertex and have bases on the same line, their area ratio equals their base ratio. For similar triangles, area ratio is the square of side ratio.

📝 Teacher's Note: Draw the trapezium clearly with all points marked. Show students how triangles with same height have area ratio equal to base ratio. This is a key concept for trapezium problems.

🎯 Exam Tip: Always identify which triangles share a common vertex first. Write the ratios clearly as fractions. For similar triangles, remember to square the side ratio for area ratio.

Question 11. In the given figure, ARC is a triangle. DE is parallel to BC and \( \frac{AD}{DB} = \frac{3}{2} \).
(i) Determine the ratios \( \frac{AD}{AB}, \frac{DE}{BC} \).
(ii) Prove that △DEF is similar to △CBF. Hence, find \( \frac{EF}{FB} \).
(iii) What is the ratio of the areas of △DEF and △BFC?

[Diagram: This diagram shows triangle ABC with DE parallel to BC, and point F where DE and BC intersect when extended.]

Solution:

(i) Given: DE || BC and \( \frac{AD}{DB} = \frac{3}{2} \)

In △ ADE and △ ABC,
\( \angle A = \angle A \) (Corresponding Angles)
\( \angle ADE = \angle ABC \) (Corresponding Angles)
\( \triangle ADE \sim \triangle ABC \) (By AA similarity)

\( \frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC} \) ........(1)

Now, \( \frac{AD}{AB} = \frac{AD}{AD + DB} = \frac{3}{3 + 2} = \frac{3}{5} \)

Using (1), we get \( \frac{AD}{AB} = \frac{3}{5} = \frac{DE}{BC} \) ........(2)

(ii) To prove △DEF ~ △CBF:
In △DEF and △CBF,
\( \angle DFE = \angle CFB \) (Vertically opposite angles)
\( \angle DEF = \angle CBF \) (Alternate angles, since DE || BC)
\( \triangle DEF \sim \triangle CBF \) (AA similarity)

From similarity: \( \frac{DE}{CB} = \frac{EF}{FB} = \frac{DF}{CF} \)
Since \( \frac{DE}{BC} = \frac{3}{5} \), we have \( \frac{EF}{FB} = \frac{3}{5} \)

(iii) Ratio of areas of △DEF and △BFC:
Since △DEF ~ △CBF, \( \frac{\text{ar}(\triangle DEF)}{\text{ar}(\triangle BFC)} = \left(\frac{DE}{BC}\right)^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \)

In simple words: Parallel lines create similar triangles with equal angles. The ratio of corresponding sides is the same. For areas, we square the side ratio.

📝 Teacher's Note: Show students how to find the ratio AD:AB from AD:DB. Mark all angles clearly to show similarity. Explain why vertically opposite and alternate angles are equal.

🎯 Exam Tip: Write "AA similarity" clearly when proving similar triangles. Always convert part-to-part ratios to part-to-whole ratios first. For area ratios, square the side ratio.

Question 12. In the given figure, ∠B = ∠E, ∠ACD = ∠BCE, AB=10.4 cm and DE=7.8 cm. Find the ratio between areas of the ΔABC and ΔDEC.
Answer:
Given: ∠ACD = ∠BCE
∠ACD + ∠BCD = ∠BCE + ∠BCD
∠ACB = ∠DCE
Also, given ∠B = ∠E
∴ ΔABC ~ ΔDEC
\[ \frac{\text{ar}(ΔABC)}{\text{ar}(ΔDEC)} = \left(\frac{AB}{DE}\right)^2 = \left(\frac{10.4}{7.8}\right)^2 = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \]
In simple words: When two triangles have the same angles, their areas are in the ratio of squares of their matching sides. Here we square the ratio 10.4:7.8 to get 16:9.

[Diagram: This diagram shows two triangles ABC and DEC with point C common to both triangles, where angles are marked to show the similarity.]

📝 Teacher's Note: Show students that when triangles share a common vertex like C, we can use angle-angle similarity. The key is to identify which sides match which.

🎯 Exam Tip: Always write "triangles are similar" first. Then use the formula: ratio of areas = (ratio of corresponding sides)². Don't forget to square the ratio.

Question 13. Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:
(i) \( \frac{\text{area of ΔADC}}{\text{area of ΔFEB}} \)
(ii) \( \frac{\text{area of ΔFEB}}{\text{area of ΔABC}} \)
Answer:
(i) AB = AC(Given)
∴ ∠FBE = ∠ACD
∠BFE = ∠ADC
ΔEFB ~ ΔADC (AA similarity)
∴ \( \frac{\text{ar}(ΔADC)}{\text{ar}(ΔEFB)} = \left(\frac{AC}{BE}\right)^2 = \left(\frac{AC}{BC + CE}\right)^2 = \left(\frac{13}{18}\right)^2 = \frac{169}{324} \) ... (1)

(ii) Similarly, it can be proved that ΔADB ~ ΔEFB
∴ \( \frac{\text{ar}(ΔADB)}{\text{ar}(ΔEFB)} = \left(\frac{AB}{BE}\right)^2 = \left(\frac{13}{18}\right)^2 = \frac{169}{324} \) ... (2)

From (1) and (2),
\( \frac{\text{ar}(ΔABC)}{\text{ar}(ΔEFB)} = \frac{169 + 169}{324} = \frac{338}{324} = \frac{169}{162} \)
∴ ar(ΔEFB) : ar(ΔABC) = 162 : 169
In simple words: In an isosceles triangle, when we drop perpendiculars, we create similar triangles. We use the property that ratio of areas equals square of ratio of corresponding sides.

[Diagram: This diagram shows isosceles triangle ABC with AD perpendicular to BC, and EF perpendicular to AB, creating similar triangles.]

📝 Teacher's Note: Remind students that in isosceles triangles, the perpendicular from vertex to base creates two equal right triangles. This property helps find similar triangles.

🎯 Exam Tip: For isosceles triangles, always mark equal sides and equal angles first. Write "AA similarity" clearly and show which angles are equal.

Exercise 15D

Question 1. A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A' B' C'. Calculate:
(i) the length of AB, if A' B' = 6 cm.
(ii) the length of C' A' if CA = 4 cm.
Answer:
(i) Given that ABC is a triangle that has been enlarged by scale factor m = 2.5 to the triangle A'B'C'.
A' B' = 6 cm
So, AB(2.5) = A' B'
⇒ AB(2.5) = 6
⇒ AB = 2.4 cm

(ii) Given that ABC is a triangle that has been enlarged by scale factor m = 2.5 to the triangle A'B'C'.
A' B' = 6 cm
So, AB(2.5) = A' B'
⇒ AB(2.5) = 6
⇒ AB = 2.4 cm
If CA = 4 cm.
So, CA(2.5) = C' A'
⇒ (4)(2.5) = C' A'
⇒ C' A' = 10 cm
In simple words: When we enlarge by scale factor 2.5, each side becomes 2.5 times longer. To find the original length, we divide by 2.5.

📝 Teacher's Note: Use a simple example like enlarging a 2cm line to 5cm (scale factor 2.5). Students can see that 2 × 2.5 = 5.

🎯 Exam Tip: Remember: enlarged length = original length × scale factor. To find original length, divide enlarged length by scale factor.

Question 2. A triangle LMN has been reduced by scale factor 0.8 to the triangle L' M' N'. Calculate:
(i) the length of M' N', if MN = 8 cm.
(ii) the length of LM, if L' M' = 5.4 cm.
Answer:
(i) Given that LMN is a triangle that has been reduced by scale factor m = 0.8 to the triangle L'M'N'.
MN = 8 cm
So, MN(0.8) = M' N'
⇒ (8)(0.8) = M' N'
⇒ M' N' = 6.4 cm

(ii) Given that LMN is a triangle that has been reduced by scale factor m = 0.8 to the triangle L'M'N'.
L' M' = 5.4 cm
So, LM(0.8) = L' M'
⇒ LM(0.8) = L' M'
⇒ LM(0.8) = 5.4
⇒ LM = 6.75 cm
In simple words: When we reduce by scale factor 0.8, each side becomes 0.8 times the original (smaller). To find original length from reduced length, we divide by 0.8.

📝 Teacher's Note: Explain that scale factors less than 1 make things smaller. 0.8 means "8 out of 10" or "4 out of 5" of the original size.

🎯 Exam Tip: For scale factors less than 1, the new shape is smaller. Always check if your answer makes sense - reduced should be smaller than original.

Question 3. A triangle ABC is enlarged, about the point O as centre of enlargement, and the scale factor is 3. Find:
(i) A' B', if AB = 4 cm.
(ii) BC, if B' C' = 15 cm.
(iii) OA, if OA'= 6 cm.
(iv) OC', if OC = 21 cm.
Also, state the value of:
(a) \( \frac{OB'}{OB} \)
(b) \( \frac{C'A'}{CA} \)
Answer:
(i) Given that ABC is enlarged and the scale factor m = 3 to the triangle A'B'C'.
AB = 4 cm
So, AB(3) = A' B'
⇒ (4)(3) = A' B'
⇒ A' B' = 12 cm

(ii) Given that ABC is enlarged and the scale factor m = 3 to the triangle A'B'C'.
B' C' = 15 cm
So, BC(3) = B' C'
⇒ BC(3) = 15
⇒ BC = 5 cm

(iii) Given that ABC is enlarged and the scale factor m = 3 to the triangle A'B'C'.
OA' = 6 cm
So, OA(3) = OA'
⇒ OA(3) = 6
⇒ OA = 2 cm

(iv) Given that triangle ABC is enlarged and the scale factor is m = 3 to the triangle A'B'C'.
OC = 21 cm
So, (OC)3 = OC'
i.e. 21 x 3 = OC'
i.e. OC' = 63 cm

The ratio of the lengths of two corresponding sides of two similar triangles.
(a) Given that ABC is enlarged and the scale factor m = 3 to the triangle A'B'C'.
⇒ \( \frac{OB'}{OB} = 3 \)
(b) Given that ABC is enlarged and the scale factor m = 3 to the triangle A'B'C'.
⇒ \( \frac{C'A'}{CA} = 3 \)
In simple words: When we enlarge from a center point O, all distances from O also get multiplied by the scale factor. All sides of the triangle also get multiplied by the same factor.

📝 Teacher's Note: Use a photocopier example - when you enlarge at 300%, everything becomes 3 times bigger, including distances from the center.

🎯 Exam Tip: In enlargement, both sides of triangle AND distances from center get multiplied by scale factor. All ratios equal the scale factor.

Question 4. A model of an aeroplane is made to a scale of 1:400. Calculate:
(i) the length, in cm, of the model; if the length of the aeroplane is 40 m.
(ii) the length, in m, of the aeroplane, if length of its model is 16 cm.
Answer:
The ratio of the lengths of two corresponding sides of two similar triangles.
A model of an aeroplane is made to a scale of 1:400.

So, the length of the model = \( \frac{1}{400} \times 4000 = 10 \) cm

(ii) The ratio of the lengths of two corresponding sides of two similar triangles.
A model of an aeroplane is made to a scale of 1:400.

So, the length of the aeroplane = \( 400 \times \frac{16}{100} = 64 \) m
In simple words: Scale 1:400 means the model is 400 times smaller than the real plane. So real length ÷ 400 = model length.

📝 Teacher's Note: Explain that 1:400 means "1 unit on model = 400 units in real life." Convert 40 m to 4000 cm for easier calculation.

🎯 Exam Tip: For scale 1:400, divide real size by 400 to get model size. Multiply model size by 400 to get real size. Watch your units - convert m to cm when needed.

Question 5. The dimensions of the model of a multistory building are 1.2 m × 75 cm × 2 m. If the scale factor is 1:30; find the actual dimensions of the building.
Answer:
Given:
Model dimensions = 1.2 m × 75 cm × 2 m
Scale factor = 1:30

Step 1: Convert all dimensions to same units.
Model dimensions = 1.2 m × 0.75 m × 2 m

Step 2: Find actual dimensions.
The scale factor is 1:30, so actual dimensions = 30 × model dimensions

Step 3: Calculate each dimension.
Length = \( 30 \times 1.2 = 36 \) m
Width = \( 30 \times 0.75 = 22.5 \) m
Height = \( 30 \times 2 = 60 \) m

Actual dimensions of the building = 36 m × 22.5 m × 60 m
In simple words: The model is 30 times smaller than the real building. So we multiply each side of the model by 30 to get the real building size.

📝 Teacher's Note: Always convert all measurements to the same unit first. Show students that 75 cm = 0.75 m before doing calculations.

🎯 Exam Tip: Write "Given" and list all values clearly. Show the conversion step. Write the final answer with proper units.

Question 6. On a map drawn to a scale of 1: 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°. Calculate:
(i) the actual lengths of AB and BC in km.
(ii) the area of the plot in sq. km.
Answer:
Given:
Scale = 1:2,50,000
Map measurements: AB = 3 cm, BC = 4 cm
Angle ABC = 90°

Part (i): Find actual lengths

For AB:
Map length AB = 3 cm
Actual length AB = \( 3 \times 2,50,000 = 7,50,000 \) cm = 7.5 km

For BC:
Map length BC = 4 cm
Actual length BC = \( 4 \times 2,50,000 = 10,00,000 \) cm = 10 km

Part (ii): Find area of the plot
Since angle ABC = 90°, this is a right triangle.
Area = \( \frac{1}{2} \times AB \times BC \)
Area = \( \frac{1}{2} \times 7.5 \times 10 = 37.5 \) sq. km

In simple words: The map scale means 1 cm on map = 2,50,000 cm in real life. We multiply map distances by this number. Then we find area using the triangle formula.

📝 Teacher's Note: Remind students to convert cm to km by dividing by 100,000. Show them that 7,50,000 cm = 7.5 km step by step.

🎯 Exam Tip: Always convert the final answer to the units asked in the question. Here it asks for km, so convert from cm to km clearly.

Question 7. A model of a ship of made to a scale 1 : 300
(i) The length of the model of ship is 2 m. Calculate the lengths of the ship.
(ii) The area of the deck ship is 180,000 m². Calculate the area of the deck of the model.
(iii) The volume of the model in 6.5 m³. Calculate the volume of the ship. (2016)
Answer:
Given:
Scale factor k = \( \frac{1}{300} \)

Part (i): Length of the ship
Length of model = k × Length of ship
\( 2 = \frac{1}{300} \times \) Length of ship
Length of ship = \( 2 \times 300 = 600 \) m

Part (ii): Area of deck of model
Area of model = k² × Area of ship
Area of model = \( \left(\frac{1}{300}\right)^2 \times 180,000 \)
Area of model = \( \frac{1}{90000} \times 180,000 = 2 \) m²

Part (iii): Volume of ship
Volume of model = k³ × Volume of ship
\( 6.5 = \left(\frac{1}{300}\right)^3 \times \) Volume of ship
Volume of ship = \( 6.5 \times 27000000 = 17,55,00,000 \) m³

In simple words: For length, we multiply by scale factor. For area, we multiply by scale factor squared. For volume, we multiply by scale factor cubed.

📝 Teacher's Note: Show students the pattern: length uses k, area uses k², volume uses k³. This is because area has 2 dimensions and volume has 3 dimensions.

🎯 Exam Tip: Remember the rule - linear measurements use k, area uses k², volume uses k³. Write this formula clearly in your answer.

Question 7(old). A model of ship is made to a scale of 1: 200.
(i) The length of the model is 4 m; calculate the length of the ship.
(ii) The area of the deck of the ship is 160000 m²; find the area of the deck of the model.
(iii) The volume of the model is 200 litres; calculate the volume of the ship in m³.
Answer:
Given:
Scale factor k = \( \frac{1}{200} \)

Part (i): Length of ship
Length of model = k × actual length of ship
Actual length of ship = \( 4 \times 200 = 800 \) m

Part (ii): Area of deck of model
Area of model = k² × area of deck of ship
Area of model = \( \left(\frac{1}{200}\right)^2 \times 160000 = 4 \) m²

Part (iii): Volume of ship
Volume of model = k³ × volume of ship
Volume of ship = \( \frac{1}{k^3} \times 200 \) litres
Volume of ship = \( (200)^3 \times 200 \) litres = 1600000000 litres = 1600000 m³

In simple words: The model is 200 times smaller. So real ship is 200 times longer, 200² times bigger in area, and 200³ times bigger in volume.

📝 Teacher's Note: Help students remember that 1000 litres = 1 m³. This conversion often comes in exams.

🎯 Exam Tip: Always check what units the question asks for in the answer. Convert litres to m³ when needed by dividing by 1000.

Question 8. An aeroplane is 30 in long and its model is 15 cm long. If the total outer surface area of the model is 150 cm², find the cost of painting the outer surface of the aeroplane at the rate of Rs. 120 per sq. m. Given that 50 sq. m of the surface of the aeroplane is left for windows.
Answer:
Step 1: Find the scale factor.
15 cm represents 30 m
1 cm represents \( \frac{30}{15} = 2 \) m
1 cm² represents \( 2 \times 2 = 4 \) m²

Step 2: Find actual surface area.
Surface area of model = 150 cm²
Actual surface area of aeroplane = \( 150 \times 4 = 600 \) m²

Step 3: Find area to be painted.
50 m² is left for windows
Area to be painted = \( 600 - 50 = 550 \) m²

Step 4: Calculate cost.
Cost of painting per m² = Rs. 120
Cost of painting 550 m² = \( 120 \times 550 = \) Rs. 66000

In simple words: First we find how much bigger the real plane is than the model. Then we calculate the real surface area. We subtract window area and find the cost.

📝 Teacher's Note: Make sure students understand that if length scale is 1:2, then area scale is 1:4. Use a simple square example to show this.

🎯 Exam Tip: Don't forget to subtract the window area before calculating cost. This is a common mistake that students make.

Exercise 15E

Question 1. In the following figure, XY is parallel to BC, AX = 9 cm, XB = 4.5cm and BC = 18 cm. Find:
(i) \( \frac{AY}{YC} \) (ii) \( \frac{YC}{AC} \) (iii) XY
Answer:
[Diagram: Triangle ABC with line XY parallel to BC, where X is on AB and Y is on AC]

Given:
XY || BC, AX = 9 cm, XB = 4.5 cm, BC = 18 cm

Part (i): Find \( \frac{AY}{YC} \)
Since XY || BC, triangles AXY and ABC are similar
\( \frac{AX}{AB} = \frac{AY}{AC} \)
AB = AX + XB = 9 + 4.5 = 13.5 cm
\( \frac{9}{13.5} = \frac{AY}{AC} \)
\( \frac{AY}{YC} = \frac{2}{1} \)

Part (ii): Find \( \frac{YC}{AC} \)
Since XY || BC, triangles AXY and ABC are similar
\( \frac{AX}{AB} = \frac{AY}{AC} \)
\( \frac{9}{13.5} = \frac{AY}{AC} \)
From part (i), AY : YC = 2 : 1
So \( \frac{YC}{AC} = \frac{4.5}{13.5} = \frac{1}{3} \)

Part (iii): Find XY
Since triangles AXY and ABC are similar:
\( \frac{XY}{BC} = \frac{AX}{AB} = \frac{9}{13.5} = \frac{2}{3} \)
XY = \( \frac{2}{3} \times 18 = 12 \) cm

In simple words: When a line is parallel to one side of a triangle, it divides the other two sides in the same ratio. We use this property to find all the answers.

📝 Teacher's Note: Draw the triangle clearly and mark all given measurements. Show students that AX:XB = AY:YC when XY || BC.

🎯 Exam Tip: Always write "XY || BC" first to show you know the parallel line theorem. Then write the ratio property clearly.

Question 2. In the following figure, ABCD to a trapezium with AB//DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP=6 cm and BE = 15 cm. Calculate:
(i) EC (ii) AF (iii) PE
Answer:
[Diagram: Trapezium ABCD with AB parallel to DC, and internal lines forming triangles]

Given:
ABCD is a trapezium with AB || DC
AB = 9 cm, DC = 18 cm, CF = 13.5 cm, AP = 6 cm, BE = 15 cm

Part (i): Find EC
In triangles AEB and FEC,
∠AEB = ∠FEC (vertically opposite angles)
∠BAE = ∠CFE (Since AB || DC)
∆AEB ~ ∆FEC (AA criterion for similarity)
\( \frac{AE}{FE} = \frac{BE}{CE} = \frac{AB}{FC} \)
\( \frac{15}{CE} = \frac{9}{13.5} \)
CE = 22.5 cm

Part (ii): Find AF
In triangles APB and FPD,
∠APB = ∠FPD (vertically opposite angles)
∠BAP = ∠DFP (Since AB || DF)
∆APB ~ ∆FPD (AA criterion for similarity)
\( \frac{AP}{FP} = \frac{AB}{FD} \)
\( \frac{6}{FP} = \frac{9}{31.5} \)
FP = 21 cm
So, AF = AP + PF = 6 + 21 = 27 cm

Part (iii): Find PE
From the given information and using similar triangles properties, PE can be calculated using the intersecting lines in the trapezium.

In simple words: In a trapezium with parallel sides, when diagonals intersect, they create similar triangles. We use the ratios from these similar triangles to find unknown lengths.

📝 Teacher's Note: Show students how to identify similar triangles in a trapezium. Mark the equal angles clearly to help them see the similarity.

🎯 Exam Tip: Always write which triangles are similar and why (AA, SSS, or SAS). Then write the ratio of corresponding sides clearly.

Question 3. In the following figure, AB, CD and EF are perpendicular to the straight line BDF.

[Diagram: Shows three perpendicular lines AB, CD, and EF standing on a straight line BDF. There are triangles formed with intersecting lines connecting the tops of these perpendiculars.]

If AB = x and CD = z unit and EF = y unit, prove that: \( \frac{1}{x} + \frac{1}{y} = \frac{1}{z} \).

Answer:
In \( \triangle FDC \) and \( \triangle FBA \),
\( \angle FDC = \angle FBA \) ...(Since DC||AB)
\( \angle DFC = \angle BFA \) ...(common angle)
\( \triangle FDC \sim \triangle FBA \) ....(AA criterion for Similarity)

\( \Rightarrow \frac{DC}{AB} = \frac{DF}{BF} \)
\( \Rightarrow \frac{z}{x} = \frac{DF}{BF} \) ....(i)

In \( \triangle BDC \) and \( \triangle BFE \),
\( \angle BDC = \angle BFE \) ...(Since DC||FE)
\( \angle DBC = \angle FBE \) ...(common angle)
\( \triangle BDC \sim \triangle BFE \) ....(AA criterion for Similarity)

\( \Rightarrow \frac{BD}{BF} = \frac{DC}{EF} \)
\( \Rightarrow \frac{BD}{BF} = \frac{z}{y} \) ....(ii)

Adding (i) and (ii), we get
\( \frac{BD}{BF} + \frac{DF}{BF} = \frac{z}{y} + \frac{z}{x} \)

\( \Rightarrow 1 = \frac{z}{y} + \frac{z}{x} \)

\( \Rightarrow \frac{1}{z} = \frac{1}{x} + \frac{1}{y} \)

Hence proved.
In simple words: We use similar triangles to show how the three perpendicular lines are related. When we add the ratios, we get the formula that connects all three lengths.

📝 Teacher's Note: Draw three sticks standing up on a table. Show students how the shadows make triangles. This helps them see similar triangles in real life.

🎯 Exam Tip: Always write "AA criterion for Similarity" clearly. Mark the equal angles first. Then write the ratio step by step to get full marks.

Question 4. Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that: \( \frac{AB}{PQ} = \frac{AD}{PM} \).

[Diagram: Shows two similar triangles ABC and PQR with their medians AD and PM drawn from vertices A and P to the opposite sides.]

Answer:
Given that \( \triangle ABC \sim \triangle PQR \).

\( \Rightarrow \frac{AB}{PQ} = \frac{AD}{PM} \)

\( \angle ABC = \angle PQR \), that is, \( \angle ABD = \angle PQM \)
Also, \( \angle ADB = \angle PMQ \) ....(both are right angles)
So, \( \triangle ABD \sim \triangle PQM \) ....(AA criterion for Similarity)

\( \Rightarrow \frac{AB}{PQ} = \frac{AD}{PM} \)
In simple words: When two triangles are similar, their medians are also in the same ratio as their sides. A median is a line from a vertex to the middle of the opposite side.

📝 Teacher's Note: Show that medians are like "half-way lines". When triangles are similar, everything scales the same way - sides, medians, heights, everything.

🎯 Exam Tip: Write "corresponding medians" in your answer. Show that the triangles formed by the medians are also similar using AA criterion.

Question 5. Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, prove that: \( \frac{AB}{PQ} = \frac{AD}{PM} \).

[Diagram: Shows two similar triangles ABC and PQR with their altitudes AD and PM drawn perpendicular to the bases BC and QR respectively.]

Answer:
Given that \( \triangle ABC \sim \triangle PQR \).

\( \angle ABC = \angle PQR \), that is, \( \angle ABD = \angle PQM \)
Also, \( \angle ADB = \angle PMQ \) ....(both are right angles)
So, \( \triangle ABD \sim \triangle PQM \) ....(AA criterion for Similarity)

\( \Rightarrow \frac{AB}{PQ} = \frac{AD}{PM} \)
In simple words: When two triangles are similar, their heights (altitudes) are also in the same ratio as their sides. An altitude is a line from a vertex straight down to the opposite side.

📝 Teacher's Note: Use a ladder against a wall to explain altitude. The ladder is like the altitude - it goes straight up from the base to the top point.

🎯 Exam Tip: Write "both are right angles" when talking about altitudes. This is the key step that makes the triangles similar by AA criterion.

Question 6. Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that: \( \frac{AB}{PQ} = \frac{AD}{PM} \).

[Diagram: Shows two similar triangles ABC and PQR with angle bisectors AD and PM drawn from vertices A and P to the opposite sides.]

Answer:
Given that \( \triangle ABC \sim \triangle PQR \).

\( \Rightarrow \angle BAC = \angle QPR \)

\( \Rightarrow \frac{1}{2}\angle BAC = \frac{1}{2}\angle QPR \)

\( \Rightarrow \angle BAD = \angle QPM \)
Also, \( \angle ABC = \angle PQR \), that is, \( \angle ABD = \angle PQM \)
So, \( \triangle ABD \sim \triangle PQM \) ....(AA criterion for Similarity)

\( \Rightarrow \frac{AB}{PQ} = \frac{AD}{PM} \)
In simple words: When two triangles are similar, their angle bisectors are also in the same ratio as their sides. An angle bisector cuts an angle into two equal parts.

📝 Teacher's Note: Use a paper fold to show angle bisector. When you fold an angle in half, the crease is the angle bisector. Similar triangles have everything in the same ratio.

🎯 Exam Tip: Write that half of equal angles are equal. This gives you the first angle for similarity. Then use the fact that triangles are similar for the second angle.

Question 7. In the following figure, \( \angle AXY = \angle AYX \). If \( \frac{BX}{AX} = \frac{CY}{AY} \), show that triangle ABC is isosceles.

[Diagram: Shows triangle ABC with points X and Y on sides AB and AC respectively, with line XY drawn parallel to BC.]

Answer:
Given that \( \angle AXY = \angle AYX \).
So, AX = AY....(Sides opposite equal angles are equal.)

Also, \( \frac{BX}{AX} = \frac{CY}{AY} \) .....(By the Basic Proportionality theorem)
So, BX = CY
Thus, AX + BX = AY + CY
\( \Rightarrow AB = AC \)
Hence, \( \triangle ABC \) is an isosceles triangle.
In simple words: When two angles in a triangle are equal, the sides opposite them are also equal. This makes the big triangle ABC have two equal sides, so it becomes isosceles.

📝 Teacher's Note: Draw an isosceles triangle on the board. Show students that equal angles make equal sides. This is a basic property they must remember.

🎯 Exam Tip: Write "sides opposite equal angles are equal" as your first step. Then use the given ratio to prove AB = AC. Always end with "hence, triangle is isosceles."

Question 8. In the following diagram, lines l, m and n are parallel to each other. Two transversals p and q intersect the parallel lines at points A, B, C and P, Q, R as shown. Prove that: \( \frac{AB}{BC} = \frac{PQ}{QR} \).

[Diagram: Shows three parallel lines l, m, n intersected by two transversals p and q at points A, B, C and P, Q, R respectively.]

Answer:
This follows directly from the Basic Proportionality Theorem (Thales' theorem).

When parallel lines are cut by two transversals, the segments formed on one transversal are proportional to the corresponding segments on the other transversal.

Since lines l, m, n are parallel and transversals p and q cut them:
\( \frac{AB}{BC} = \frac{PQ}{QR} \)
In simple words: When parallel lines are cut by two slanting lines, they cut both lines in the same ratio. This is like cutting two sticks with the same pattern.

📝 Teacher's Note: Use three parallel rulers and two pencils as transversals. Show students how the distances are always in the same ratio. This makes the theorem very clear.

🎯 Exam Tip: Write "Basic Proportionality Theorem" or "Thales' theorem" in your answer. This theorem is the key to solving such parallel line problems.

Question 9. In the following figure, DE //AC and DC //AP. Prove that: \( \frac{BE}{EC} = \frac{BC}{CP} \)

[Diagram: Triangle ABC with parallel lines creating intersections at points D, E, P]

Answer:

Join AR.

In triangle ACR, BX || CR. By Basic Proportionality theorem,
\( \frac{AB}{BC} = \frac{AX}{XR} \) ... (1)

In triangle APR, XQ || AP. By Basic Proportionality theorem,
\( \frac{PQ}{QR} = \frac{AX}{XR} \) ... (2)

From (1) and (2), we get,
\( \frac{AB}{BC} = \frac{PQ}{QR} \)

Since DE||AC, by Basic Proportionality theorem:
\( \frac{BE}{EC} = \frac{BD}{DA} \) ... (By the Basic Proportionality theorem)

Since DC||AP, by Basic Proportionality theorem:
\( \frac{BC}{CP} = \frac{BD}{DA} \) ... (By the Basic Proportionality theorem)

Hence, \( \frac{BE}{EC} = \frac{BC}{CP} \)

📝 Teacher's Note: Draw the lines clearly and mark the parallel signs. Show students how to identify which triangles to use for Basic Proportionality theorem. The key is finding the same ratio in two different triangles.

🎯 Exam Tip: Always write "By Basic Proportionality theorem" after each ratio. Number your equations (1), (2) etc. and clearly show how you combine them to get the final answer.

Question 10. In the figure given below, AB//EF// CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm. Calculate:
(i) EF
(ii) AC

[Diagram: Triangle with three parallel lines AB, EF, CD creating segments]

Answer:

(i)

In triangles PCD and PEF,
∠CPD = ∠EPF ... (vertically opposite angles)
∠DCE = ∠FEP ... (Since DC||EF.)
△PCD ~ △PEF ... (AA criterion for Similarity)

⟹ \( \frac{27}{EF} = \frac{15}{7.5} \)
⟹ EF = 13.5 cm

(ii)

Since EF||AB, △CEF ~ △CAB.
⟹ \( \frac{EC}{AC} = \frac{EF}{AB} \)

⟹ \( \frac{22.5}{AC} = \frac{13.5}{22.5} \)
⟹ AC = 37.5 cm

📝 Teacher's Note: Make students identify all parallel lines first. Then show which triangles are similar. The ratios of corresponding sides are equal - this is the main idea.

🎯 Exam Tip: Write "Given:" and list all measurements. Show each step of calculation clearly. Always write units (cm) in your final answer.

Question 11. In △ABC, ∠ABC = ∠DAC. AB = 8 cm, AC = 4 cm, AD = 5 cm.
(i) Prove that △ACD is similar to △BCA.
(ii) Find BC and CD.
(iii) Find area of △ACD: area of △ABC. (2014)

Answer:

(i)
In △ACD and △BCA,
∠DAC = ∠ABC ... (given)
∠ACD = ∠BCA ... (common angles)
△ACD ~ △BCA ... (AA criterion for Similarity)

(ii)
In △ACD and △BCA,
∠DAC = ∠ABC ... (given)
∠ACD = ∠BCA ... (common angles)
△ACD ~ △BCA ... (AA criterion for Similarity)

⟹ \( \frac{AC}{BC} = \frac{CD}{CA} = \frac{AD}{AB} \)

⟹ \( \frac{4}{BC} = \frac{CD}{4} = \frac{5}{8} \)

⟹ \( \frac{4}{BC} = \frac{5}{8} \)

⟹ BC = \( \frac{32}{5} \) = 6.4 cm

⟹ \( \frac{CD}{4} = \frac{5}{8} \)

⟹ CD = \( \frac{20}{8} \) = 2.5 cm

(iii)
In △ACD and △BCA,
∠DAC = ∠ABC ... (given)
∠ACD = ∠BCA ... (common angles)
△ACD ~ △BCA ... (AA criterion for Similarity)

⟹ \( \frac{ar(△ACD)}{ar(△ABC)} = \frac{AD^2}{AB^2} \)

⟹ \( \frac{ar(△ACD)}{ar(△ABC)} = \frac{5^2}{8^2} = \frac{25}{64} \)

📝 Teacher's Note: Help students see that angle ABC equals angle DAC means the triangles have the same shape. When triangles are similar, the ratio of areas equals the ratio of squares of corresponding sides.

🎯 Exam Tip: Always prove similarity first using AA or SAS method. Then use the property that ratios of corresponding sides are equal. For area ratio, remember it equals the square of the side ratio.

Question 12. In the given triangle P, Q and R are the midpoints of sides AB, BC and AC respectively. Prove that triangle PQR is similar to triangle ABC.

[Diagram: Triangle ABC with midpoints P, Q, R creating smaller triangle PQR inside]

Answer:

In triangle ABC, PR || BC. By Basic proportionality theorem,
\( \frac{AP}{PB} = \frac{AR}{RC} \)

Also, in triangle PAR and triangle ABC,
∠PAR = ∠BAC (Common)
∠APR = ∠ABC (Corresponding angles)
△PAR ~ △BAC (AA similarity)

\( \frac{PR}{BC} = \frac{AP}{AB} \)

\( \frac{PR}{BC} = \frac{1}{2} \) (As P is the mid-point of AB.)

PR = \( \frac{1}{2} \)BC

Similarly, PQ = \( \frac{1}{2} \)AC

RQ = \( \frac{1}{2} \)AB

Thus, \( \frac{PR}{BC} = \frac{PQ}{AC} = \frac{RQ}{AB} \)

⟹ △QRP ~ △ABC (SSS similarity)

📝 Teacher's Note: Show students that when you connect midpoints, you get a triangle that is exactly half the size. All sides of the smaller triangle are half of the bigger triangle's sides.

🎯 Exam Tip: Write clearly that P, Q, R are midpoints. Use the midpoint theorem to show each side is half. Then use SSS similarity criterion to prove the triangles are similar.

Question 13. In the following figure, AD and CE are medians of △ ABC. DF is drawn parallel to CE. Prove that:
(i) EF = FB;
(ii) AG : GD = 2 : 1

[Diagram: Triangle ABC with medians AD and CE intersecting at G, with DF parallel to CE]

Answer:

(i) Since DF || CE and D is the midpoint of BC, by the converse of midpoint theorem, F is the midpoint of BE.
Therefore, EF = FB.

(ii) In any triangle, the medians divide each other in the ratio 2:1.
Since G is the point of intersection of medians AD and CE,
AG : GD = 2 : 1

📝 Teacher's Note: Explain that medians are lines from a vertex to the midpoint of the opposite side. When two medians cross, they always divide each other in the ratio 2:1. This is a special property.

🎯 Exam Tip: For part (i), use the midpoint theorem. For part (ii), state the centroid property clearly - "medians divide each other in ratio 2:1". This gets you full marks.

Question 14. The two similar triangles are equal in area. Prove that the triangles are congruent.
Answer:
Let us assume two similar triangles as \( \triangle ABC \) ~ \( \triangle PQR \)
Now \( \frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)} = \left(\frac{AB}{PQ}\right)^2 = \left(\frac{BC}{QR}\right)^2 = \left(\frac{AC}{PR}\right)^2 \)

Since area \( (\triangle ABC) = \) area \( (\triangle PQR) \)
Therefore AB = PQ
BC = QR
AC = PR

So, respective sides of two similar triangles are also of same length
So, \( \triangle ABC \cong \triangle PQR \) (by SSS rule)
In simple words: When two similar triangles have the same area, their matching sides must be equal. This means the triangles are exactly the same size and shape — they are congruent.

📝 Teacher's Note: Show students that similar means same shape but different size. When areas are equal, the sizes must also be equal. Use paper cutouts to show this clearly.

🎯 Exam Tip: Always state that similar triangles with equal areas have equal corresponding sides. Then write "SSS rule" to prove congruence.

Question 15. The ratio between the altitudes of two similar triangles is 3 : 5; write the ratio between their:
(i) medians
(ii) perimeters
(iii) areas
Answer:
(i) The ratio between the medians of two similar triangles is same as the ratio between their sides. Required ratio = 3:5
(ii) The ratio between the perimeters of two similar triangles is same as the ratio between their sides. Required ratio = 3:5
(iii) The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides. Required ratio = \( (3)^2 : (5)^2 = 9 : 25 \)
In simple words: For similar triangles, altitudes, medians and perimeters all have the same ratio as their sides. But area ratio is the square of the side ratio.

📝 Teacher's Note: Make students remember that linear measurements have the same ratio as sides, but area is always the square of that ratio. Use simple examples like 2:3 gives area ratio 4:9.

🎯 Exam Tip: For linear measurements write the same ratio. For area, square the ratio. Always show your calculation for area ratio.

Question 16. The ratio between the areas of two similar triangles is 16 : 25. Find the ratio between their:
(i) perimeters
(ii) altitudes
(iii) medians
Answer:
The ratio between the areas of two similar triangles is same as the square of the ratio between their corresponding sides.
So, the ratio between the sides of the two triangles = 4:5
(i) The ratio between the perimeters of two similar triangles is same as the ratio between their sides. Required ratio = 4:5
(ii) The ratio between the altitudes of two similar triangles is same as the ratio between their sides. Required ratio = 4:5
(iii) The ratio between the medians of two similar triangles is same as the ratio between their sides. Required ratio = 4:5
In simple words: When you know area ratio, take square root to find side ratio. Then all linear measurements like perimeter, altitude and median have this same ratio.

📝 Teacher's Note: Start by finding the square root of area ratio to get side ratio. Show that \( \sqrt{16} : \sqrt{25} = 4 : 5 \). This is the key step.

🎯 Exam Tip: First find the side ratio from area ratio by taking square root. Write this step clearly. Then all linear ratios are the same as side ratio.

Question 17. The following figure shows a triangle PQR in which XY is parallel to QR. If PX: XQ = 1:3 and QR = 9 cm, find the length of XY. Further, if the area of ∆ PXY = x cm²; find in terms of x, the area of :
(i) triangle PQR
(ii) trapezium XQRY
Answer:
In \( \triangle PXY \) and \( \triangle PQR \), XY is parallel to QR, so corresponding angles are equal.
\( \angle PXY = \angle PQR \)
\( \angle PYX = \angle PRQ \)
Hence, \( \triangle PXY \) ~ \( \triangle PQR \) (By AA similarity criterion)

\( \frac{PX}{PQ} = \frac{XY}{QR} \)

\( \Rightarrow \frac{1}{4} = \frac{XY}{QR} \) (PX : XQ = 1 : 3 \( \Rightarrow \) PX : PQ = 1 : 4)

\( \Rightarrow \frac{1}{4} = \frac{XY}{9} \)

\( \Rightarrow XY = 2.25 \) cm

(i) We know that the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
\( \frac{\text{Ar}(\triangle PXY)}{\text{Ar}(\triangle PQR)} = \left(\frac{PX}{PQ}\right)^2 \)

\( \frac{x}{\text{Ar}(\triangle PQR)} = \left(\frac{1}{4}\right)^2 = \frac{1}{16} \)

\( \text{Ar}(\triangle PQR) = 16x \) cm²

(ii) Ar (trapezium XQRY) = Ar \( (\triangle PQR) \) - Ar \( (\triangle PXY) \)
= \( (16x - x) \) cm²
= \( 15x \) cm²
In simple words: When lines are parallel, triangles are similar. We use ratios to find missing lengths. The bigger triangle has 16 times the area because \( 4^2 = 16 \).

📝 Teacher's Note: Draw the diagram clearly. Show that PX:XQ = 1:3 means PX:PQ = 1:4. Students often get confused with this ratio conversion.

🎯 Exam Tip: Always convert the given ratio to the ratio you need. Write "By AA similarity" clearly. Show the area calculation step by step using the square of ratios.

Question 18. On a map, drawn to a scale of 1 : 20000, a rectangular plot of land ABCD has AB = 24 cm, and BC = 32 cm. Calculate :
(i) The diagonal distance of the plot in kilometre
(ii) The area of the plot in sq. km
Answer:
Scale :- 1 : 20000
1 cm represents 20000 cm = \( \frac{20000}{1000 \times 100} = 0.2 \) km

(i) \( AC^2 = AB^2 + BC^2 \)
= \( 24^2 + 32^2 \)
= 576 + 1024 = 1600
AC = 40 cm

Actual length of diagonal = 40 × 0.2 km = 8 km

(ii) 1 cm represents 0.2 km
1 cm² represents 0.2 × 0.2 km²
The area of the rectangle ABCD = AB × BC
= 24 × 32 = 768 cm²
Actual area of the plot = 0.2 × 0.2 × 768 km² = 30.72 km²
In simple words: We first find distances on the map, then multiply by the scale to get real distances. For area, we multiply by the scale twice (once for length, once for width).

📝 Teacher's Note: Show students that 1:20000 means 1 cm on map = 20000 cm in reality. For area, the scale factor is squared. Use simple examples first.

🎯 Exam Tip: Convert the scale to convenient units first. Use Pythagoras theorem for diagonal. Remember that area scale factor is the square of length scale factor.

Question 19. The dimensions of the model of a multi-storeyed building are 1m by 60 cm by 1.20 m. If the scale factor is 1 : 50, find the actual dimensions of the building. Also, find :
(i) the floor area of a room of the building, if the floor area of the corresponding room in the model is 50 sq cm
(ii) the space (volume) inside a room of the model, if the space inside the corresponding room of the building is 90m³
Answer:
The dimensions of the building are calculated as below.
Length = 1 × 50 m = 50 m
Breadth = 0.60 × 50 m = 30 m
Height = 1.20 × 50 m = 60 m
Thus, the actual dimensions of the building are 50 m × 30 m × 60 m.

(i) Floor area of the room of the building = \( 50 \times \left(\frac{50}{1}\right)^2 = 125000 \text{ cm}^2 = \frac{125000}{100 \times 100} = 12.5 \text{ m}^2 \)

(ii) Volume of the model of the building
= \( 90 \times \left(\frac{1}{50}\right)^3 = 90 \times \left(\frac{1}{50}\right) \times \left(\frac{1}{50}\right) \times \left(\frac{1}{50}\right) = 90 \times \left(\frac{100 \times 100 \times 100}{50 \times 50 \times 50}\right) \text{ cm}^3 \)
= 720 cm³
In simple words: For actual size, multiply by 50. For area, multiply by \( 50^2 \). For volume, we work backwards from building to model, so we divide by \( 50^3 \).

📝 Teacher's Note: Show clearly that linear scale is 1:50, area scale is 1:2500, and volume scale is 1:125000. Use blocks or boxes to demonstrate these relationships.

🎯 Exam Tip: Linear dimensions multiply by scale factor. Area multiplies by square of scale factor. Volume multiplies by cube of scale factor. Convert units carefully.

Question 20. In ∆ABC, ∠ACB = 90° and CD ⊥ AB. Prove that : \( \frac{BC^2}{AC^2} = \frac{BD}{AD} \)
Answer:
(i) In \( \triangle PQL \) and \( \triangle RMP \)
\( \angle LPQ = \angle QRP \) (Given)
\( \angle RQP = \angle RPM \) (Given)
\( \triangle PQL \) ~ \( \triangle RMP \) (AA similarity)

(ii) As \( \triangle PQL \) ~ \( \triangle RMP \) (Proved above)
\( \frac{PQ}{RP} = \frac{QL}{PM} = \frac{PL}{RM} \)
\( \Rightarrow QL \times RM = PL \times PM \)

(iii) \( \angle LPQ = \angle QRP \) (Given)
\( \angle Q = \angle Q \) (Common)
\( \triangle PQL \) ~ \( \triangle RQP \) (AA similarity)
\( \Rightarrow \frac{PQ}{RQ} = \frac{QL}{QP} = \frac{PL}{PR} \)
\( \Rightarrow PQ^2 = QR \times QL \)
In simple words: When we have a right triangle with a height drawn to the hypotenuse, we get similar triangles. These give us useful relationships between the sides.

📝 Teacher's Note: Draw the figure clearly and mark all angles. Show students how the altitude creates two smaller triangles that are similar to the original triangle and to each other.

🎯 Exam Tip: Always mark equal angles in the diagram. Write "AA similarity" clearly. State the proportional sides and cross-multiply to get the required result.

Question 21. A triangle ABC with AB = 3 cm, BC = 6 cm and AC = 4 cm is enlarged to ∆DEF such that the longest side of ∆DEF = 9 cm. Find the scale factor and hence, the lengths of the other sides of ∆DEF.
Answer:
Triangle ABC is enlarged to DEF. So, the two triangles will be similar.
\( \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \)
Longest side in ∆ ABC = BC = 6 cm
Corresponding longest side in ∆ DEF = EF = 9 cm
Scale factor = \( \frac{EF}{BC} = \frac{9}{6} = \frac{3}{2} = 1.5 \)
\( \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{2}{3} \)
DE = \( \frac{3}{2} \) AB = \( \frac{9}{2} \) = 4.5 cm
DF = \( \frac{3}{2} \) AC = \( \frac{12}{2} \) = 6 cm
In simple words: When we enlarge a triangle, all sides grow by the same ratio. This ratio is called scale factor. Here, all sides of ∆DEF are 1.5 times bigger than ∆ABC.

📝 Teacher's Note: Show students how to find the longest side first. Then use that to find the scale factor. A common mistake is using the wrong sides for comparison.

🎯 Exam Tip: Always write "scale factor" clearly. Show the calculation step by step. Write all side lengths with units (cm).

Question 22. Two isosceles triangles have equal vertical angles. Show that the triangles are similar. If the ratio between the areas of these two triangles is 16 : 25, find the ratio between their corresponding altitudes.
Answer:
[Diagram: Two isosceles triangles ABC and PQR are shown with vertex angles A and P marked equal]
Let ABC and PQR be two isosceles triangles.
Then, \( \frac{AB}{AC} = \frac{1}{1} \) and \( \frac{PQ}{PR} = \frac{1}{1} \)
Also, ∠ A = ∠ P (Given)
∴ ∆ABC ~ ∆PQR (SAS similarity)

Let AD and PS be the altitude in the respective triangles.
We know that the ratio of areas of two similar triangles is equal to the square of their corresponding altitudes.
\( \frac{Ar(∆ABC)}{Ar(∆PQR)} = \left(\frac{AD}{PS}\right)^2 \)
\( \frac{16}{25} = \left(\frac{AD}{PS}\right)^2 \)
\( \frac{AD}{PS} = \frac{4}{5} \)
In simple words: Two isosceles triangles with the same top angle are similar. The ratio of their heights is found using the square root of the area ratio.

📝 Teacher's Note: Remind students that isosceles triangles have two equal sides. The vertex angle is the angle between the two equal sides.

🎯 Exam Tip: Write "SAS similarity" clearly. For area ratios, always use the square of the side ratio. Show the square root calculation.

Question 23. In ∆ABC, AP: PB = 2 :3. PO is parallel to BC and is extended to Q so that CQ is parallel to BA.
Find:
(i) area ∆APO: area ∆ABC.
(ii) area ∆APO: area ∆CQO.
Answer:
[Diagram: Triangle ABC with point P on AB such that AP:PB = 2:3, line PO parallel to BC, and line CQ parallel to BA]
In triangle ABC, PO || BC. Using Basic proportionality theorem,
\( \frac{AP}{PB} = \frac{AO}{OC} \)
\( \Rightarrow \frac{AO}{OC} = \frac{2}{3} \) ... (1)

(i)
∠PAO = ∠BAC (Common)
∠APO = ∠ABC (Corresponding angles)
∆APO ~ ∆ABC (AA similarity)
\( \frac{Ar(∆APO)}{Ar(∆ABC)} = \left(\frac{AO}{AC}\right)^2 = \left(\frac{2}{2+3}\right)^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25} \)

(ii)
∠POA = ∠COQ (Vertically opposite angles)
∠PAO = ∠QCO (Alternate angles)
∆AOP ~ ∆COQ (AA similarity)
\( \frac{Ar(∆AOP)}{Ar(∆COQ)} = \left(\frac{AO}{CO}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \)
In simple words: When lines are parallel, triangles formed have areas in the ratio of squares of their corresponding sides.

📝 Teacher's Note: Draw clear diagrams showing parallel lines. Students often confuse which triangles to compare. Mark the equal angles clearly.

🎯 Exam Tip: Write "AA similarity" for each part. Show the ratio calculation step by step. Always square the side ratios to get area ratios.

Question 24. The following figure shows a triangle ABC in which AD and BE are perpendiculars to BC and AC respectively.
Show that:
(i) ∆ADC ~ ∆BEC
(ii) CA × CE = CB × CD
(iii) ∆ ABC ~ ∆DEC
(iv) CD × AB = CA × DE
Answer:
[Diagram: Triangle ABC with perpendiculars AD to BC and BE to AC, intersecting at point inside the triangle]
(i) ∠ADC = ∠BEC = 90°
∠ACD = ∠BCE (Common)
∆ADC ~ ∆BEC (AA similarity)

(ii) From part (i),
\( \frac{AC}{BC} = \frac{CD}{EC} \) ... (1)
⇒ CA × CE = CB × CD

(iii) In ∆ABC and ∆DEC,
From (1),
\( \frac{AC}{BC} = \frac{CD}{EC} \Rightarrow \frac{AC}{CD} = \frac{BC}{EC} \)
∠DCE = ∠BCA (Common)
∆ABC ~ ∆DEC (SAS similarity)

(iv) From part (iii),
\( \frac{AC}{DC} = \frac{AB}{DE} \)
⇒ CD × AB = CA × DE
In simple words: When two lines are perpendicular to the sides of a triangle, they create similar triangles. This helps us find equal products of sides.

📝 Teacher's Note: Emphasize that perpendicular lines create 90° angles. This is the key to proving similarity. Show students how to identify common angles.

🎯 Exam Tip: Always mark 90° angles in the diagram. Write "AA similarity" or "SAS similarity" clearly. Show cross multiplication for the products.

Question 25. In the given figure, ABC is a triangle with ∠EDB = ∠ACB. Prove that ∆ABC ~ ∆EBD. If BE=6 cm, EC = 4 cm, BD = 5 cm and area of ∆BED = 9 cm². Calculate the (i) length of AB (ii) area of ∆ABC
Answer:
[Diagram: Triangle ABC with point D inside and point E on BC such that BE = 6 cm, EC = 4 cm]
In ∆ ABC and ∆ EBD,
∠ ACB = ∠ EDB (given)
∠ ABC = ∠ EBD (common)
∆ABC ~ ∆EBD (by AA- similarity)

(i) We have, \( \frac{AB}{BE} = \frac{BC}{BD} \Rightarrow AB = \frac{6 \times 10}{5} = 12 \) cm

(ii) \( \frac{\text{Area of } ∆ABC}{\text{Area of } ∆BED} = \left(\frac{AB}{BE}\right)^2 \)
\( \Rightarrow \text{Area of } ∆ABC = \left(\frac{12}{6}\right)^2 \times 9 \text{ cm}^2 \)
= 4 × 9 cm² = 36 cm²
In simple words: When triangles are similar, their areas have a ratio equal to the square of their corresponding side ratios.

📝 Teacher's Note: Help students identify BC = BE + EC = 6 + 4 = 10 cm. This is often missed. Show how to use the similarity ratio correctly.

🎯 Exam Tip: First prove similarity using AA. Then find BC by adding BE and EC. Use the square of side ratios for area calculations.

Question 26. In the given figure, ABC is a right-angled triangle with ∠BAC = 90°. (i) Prove ∆ADB ~ ∆CDA. (ii) If BD = 18 cm, CD = 8 cm, find AD. (iii) Find the ratio of the area of ∆ADB is to area of ∆CDA.
Answer:
[Diagram: Right triangle ABC with right angle at A, and altitude AD drawn to hypotenuse BC]
(i) Let ∠CAD = x
⇒ m∠DAB = 90° - x
⇒ m∠DBA = 180° - (90° + 90° - x) = x
⇒ ∠CAD = ∠DBA ....(1)
In ∆ ADB and ∆ CDA,
∠ADB = ∠CDA ....[Each 90°]
∠ABD = ∠CAD ....[From (1)]
∴ ∆ ADB ~ ∆ CDA ....[By A.A.]

(ii) Since the corresponding sides of similar triangles are proportional, we have
\( \frac{BD}{AD} = \frac{AD}{CD} \)
\( \Rightarrow \frac{18}{AD} = \frac{AD}{8} \)
⇒ AD² = 18 × 8 = 144
⇒ AD = 12 cm

(iii) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
\( \Rightarrow \frac{Ar(∆ ADB)}{Ar(∆ CDA)} = \frac{AD^2}{CD^2} = \frac{12^2}{8^2} = \frac{144}{64} = \frac{9}{4} = 9 : 4 \)
In simple words: When we draw altitude to hypotenuse in a right triangle, it creates two smaller similar triangles. These help us find unknown lengths.

📝 Teacher's Note: This is a very important theorem in geometry. The altitude to hypotenuse creates similar triangles. Show students that AD² = BD × CD always works.

🎯 Exam Tip: Write the similarity proof first using angle relationships. For part (ii), use AD² = BD × CD formula. For ratios, always simplify to lowest terms.

Question 27. In the given figure, AB and DE are perpendicular to BC.
(i) Prove that ∆ABC ~ ∆DEC
(ii) If AB = 6 cm: DE = 4 cm and AC = 15 cm. Calculate CD.
(iii) Find the ratio of the area of ∆ABC : area of ∆DEC.

[Diagram: This diagram shows triangle ABC with point D on side AC and point E on side BC. AB is perpendicular to BC, and DE is perpendicular to BC, creating two similar right triangles.]

Solution:

(i)
In ∆ABC and ∆DEC,
∠ABC = ∠DEC .... (both are right angles)
∠ACB = ∠DCE .... (common angles)
∴ ∆ABC ~ ∆DEC .... (AA criterion for similarity)

(ii)
In ∆ABC and ∆DEC,
∠ABC = ∠DEC .... (both are right angles)
∠ACB = ∠DCE .... (common angles)
∴ ∆ABC ~ ∆DEC .... (AA criterion for similarity)
⇒ \( \frac{AB}{DE} = \frac{AC}{CD} \)
⇒ \( \frac{6}{4} = \frac{15}{CD} \)
⇒ CD = 10 cm

(iii)
In ∆ABC and ∆DEC,
∠ABC = ∠DEC .... (both are right angles)
∠ACB = ∠DCE .... (common angles)
∴ ∆ACB ~ ∆DEC .... (AA criterion for similarity)
\( \frac{ar(∆ABC)}{ar(∆DEC)} = \frac{AB^2}{DE^2} = \frac{6^2}{4^2} = \frac{36}{16} \)
⇒ \( \frac{ar(∆ABC)}{ar(∆DEC)} = \frac{9}{4} \)

In simple words: When two triangles have the same angles, they are similar. Similar triangles have sides in the same ratio. The area ratio equals the square of the side ratio.

📝 Teacher's Note: Show students that perpendicular lines make right angles. When two triangles share an angle and both have right angles, they are similar by AA test.

🎯 Exam Tip: Always write "AA criterion for similarity" to get marks. For area ratio, use the formula: area ratio = (side ratio)².

Question 28. ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that:
(i) ∆ADE ~ ∆ACB.
(ii) If AC = 13 cm, BC = 5 cm and AE=4 cm. Find DE and AD.
(iii) Find, area of ∆ADE : area of quadrilateral BCED. (2015)

[Diagram: This diagram shows right triangle ABC with right angle at B. Point D is on side AB, and DE is perpendicular to AC at point E.]

Solution:

(i)
In ∆ADE and ∆ACB,
∠AED = ∠ABC .... (both are right angles)
∠DAE = ∠CAB .... (common angles)
∴ ∆ADE ~ ∆ACB .... (AA criterion for similarity)

(ii)
In ∆ADE and ∆ACB,
∠AED = ∠ABC .... (both are right angles)
∠DAE = ∠CAB .... (common angles)
∴ ∆ADE ~ ∆ACB .... (AA criterion for similarity)
⇒ \( \frac{AE}{AB} = \frac{DE}{BC} = \frac{AD}{AC} \)
⇒ \( \frac{4}{AB} = \frac{DE}{5} = \frac{AD}{13} \) .... (i)
In right ∆ABC,
⇒ AB² + BC² = AC²
⇒ AB² + 5² = 13²
⇒ AB² = 144
⇒ AB = 12 cm
From (i), we get
\( \frac{4}{12} = \frac{DE}{5} = \frac{AD}{13} \)
So, DE = \( \frac{20}{12} = \frac{5}{3} = 1\frac{2}{3} \) cm
\( \frac{AD}{13} = \frac{4}{12} \) ⇒ AD = \( \frac{52}{12} = 4\frac{1}{3} \) cm

(iii)
We need to find the area of ∆ADE and quadrilateral BCED.
Area of ∆ADE = \( \frac{1}{2} \times AE \times DE = \frac{1}{2} \times 4 \times \frac{5}{3} = \frac{10}{3} \) cm²
Area of quad.BCED = Area of ∆ABC - Area of ∆ADE
= \( \frac{1}{2} \times BC \times AB - \frac{10}{3} \)
= \( \frac{1}{2} \times 5 \times 12 - \frac{10}{3} \)
= \( 30 - \frac{10}{3} \)
= \( \frac{80}{3} \) cm²
Thus ratio of areas of ∆ADE to quadrilateral BCED = \( \frac{\frac{10}{3}}{\frac{80}{3}} = \frac{1}{8} \)

In simple words: We use similar triangles to find unknown sides. Then we calculate areas using the triangle area formula and subtract to find the quadrilateral area.

📝 Teacher's Note: Remind students to use Pythagoras theorem first to find the missing side. Then use similarity ratios. Common mistake is forgetting to convert improper fractions.

🎯 Exam Tip: Show all steps clearly. Write "Given", find AB first using Pythagoras, then use similarity ratios. Always simplify final answers.

Question 29. Given: AB // DE and BC // EF. Prove that:
(i) \( \frac{AD}{DG} = \frac{CF}{FG} \)
(ii) ∆DFG ~ ∆ACG.

[Diagram: This diagram shows triangle ABC with points D, E, F creating parallel lines. AB is parallel to DE, and BC is parallel to EF.]

Solution:

(i) In ∆ AGB, DE || AB, by Basic proportionality theorem,
\( \frac{GD}{DA} = \frac{GE}{EB} \) .... (1)
In ∆ GBC, EF || BC, by Basic proportionality theorem,
\( \frac{GE}{EB} = \frac{GF}{FC} \) .... (2)
From (1) and (2), we get,
\( \frac{GD}{DA} = \frac{GF}{FC} \)
\( \frac{AD}{DG} = \frac{CF}{FG} \)

(ii)
From (i), we have:
\( \frac{AD}{DG} = \frac{CF}{FG} \)
∠DGF = ∠AGC (Common)
∴ ∆DFG ~ ∆ACG (SAS similarity)

In simple words: When lines are parallel, they create proportional segments. We use the basic proportionality theorem (also called Thales theorem) to prove ratios are equal.

📝 Teacher's Note: Draw parallel lines clearly on the board. Show students how parallel lines cut transversals into proportional parts. This is the basic proportionality theorem.

🎯 Exam Tip: Always mention "Basic proportionality theorem" by name. For similarity, write which test you used - AA, SAS, or SSS. Here we used SAS.

Question 30.

i.
In ∆PQR and ∆SPR,
∠PSR = ∠QPR ... given
∠PRQ = ∠PRS ... common angle
⇒ ∆PQR ~ ∆SPR (AA Test)

ii. Find the lengths of QR and PS.
Since ∆PQR ~ ∆SPR ... from (i)
⇒ \( \frac{PQ}{SP} = \frac{QR}{PR} = \frac{PR}{SR} \) .... (a)
\( \frac{QR}{PR} = \frac{PR}{SR} \) .... from (a)
⇒ \( \frac{QR}{6} = \frac{6}{3} \)

In simple words: When triangles are similar, their corresponding sides are in the same ratio. We use this ratio to find unknown sides.

📝 Teacher's Note: Show students how to identify corresponding sides in similar triangles. The order of vertices in the similarity statement tells us which sides correspond.

🎯 Exam Tip: Write the similarity statement with vertices in correct order. This shows which sides and angles correspond. Always write "AA Test" or other similarity test used.

ii.
\[ QR = \frac{6 \times 6}{3} = 12 \text{ cm} \]
\[ \frac{PQ}{SP} = \frac{PR}{SR} \text{ ... from (a)} \]
\[ \frac{8}{SP} = \frac{6}{3} \]
\[ SP = \frac{8 \times 3}{6} = 4 \text{ cm} \]

📝 Teacher's Note: Show students that when triangles are similar, the ratios of matching sides are equal. Use this rule to find unknown sides.

🎯 Exam Tip: Always write the proportion clearly first. Then cross-multiply to find the unknown value. Show all steps.

iii.
\[ \frac{\text{area of } \triangle PQR}{\text{area of } \triangle SPR} = \frac{PQ^2}{SP^2} = \frac{8^2}{4^2} = \frac{64}{16} = 4 \]

📝 Teacher's Note: For similar triangles, the ratio of areas equals the square of the ratio of matching sides. This is a key formula students must remember.

🎯 Exam Tip: Write the formula "ratio of areas = square of ratio of sides" first. Then substitute values. This gets you full marks.