CBSE Class 10 Mathematics Triangles Assignment Set 08

Exercise 1: Similar Figures

Question 1. Fill in the blanks using the correct word given in the brackets:
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
Answer:
(i) Similar. (ii) Similar. (iii) Equilateral. (iv) (a) equal, (b) proportional.
In simple words: Same shape means similar; same shape and same size means congruent.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 2. Give two different examples of a pair of
(i) similar figures   (ii) non-similar figures
Answer:
(i) Any two equilateral triangles are similar; any two squares (or any two circles) are similar.
(ii) A triangle and a rhombus are not similar; a rectangle and a trapezium are not similar.
In simple words: Use the definition of similar figures.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Question 3. State whether the following quadrilaterals are similar or not: the rhombus PQRS with all sides 1.5 cm and the square ABCD with all sides 3 cm.

Answer:
Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are proportional.
The sides are proportional: \( \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CD}{RS}=\frac{DA}{SP}=\frac{3}{1.5}=2 \).
But the angles of the rhombus PQRS are not all 90° (∠P ≠ ∠A, ∠Q ≠ ∠B), so the corresponding angles are not equal.
Hence the quadrilaterals are not similar.
In simple words: Proportional sides alone do not make polygons similar; the angles must also match.

Teacher's Note:
1. Name the corresponding vertices in the same order when you write a similarity statement.
2. State the criterion (AA, SAS or SSS) every time you claim that two triangles are similar.

Exercise 2: Basic Proportionality Theorem

Question 4. In the given figures, DE ∥ BC. Find EC in (i) and AD in (ii).
(i) AD = 1.5 cm, DB = 3 cm, AE = 1 cm.   (ii) AD = ?, DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm.
(i) AD = 1.5 cm, DB = 3 cm, AE = 1 cm.   (ii) DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm.
Answer:
(i) By the Basic Proportionality Theorem \( \frac{AD}{DB}=\frac{AE}{EC}\Rightarrow \frac{1.5}{3}=\frac{1}{EC} \) ⇒ EC = 2 cm.
(ii) \( \frac{AD}{DB}=\frac{AE}{EC}\Rightarrow \frac{AD}{7.2}=\frac{1.8}{5.4} \) ⇒ AD = 7.2 × 1/3 = 2.4 cm.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 5. E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF ∥ QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm   (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm   (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Answer:
By the converse of the Basic Proportionality Theorem, EF ∥ QR if and only if \( \frac{PE}{EQ}=\frac{PF}{FR} \).
(i) \( \frac{PE}{EQ}=\frac{3.9}{3}=1.3 \) and \( \frac{PF}{FR}=\frac{3.6}{2.4}=1.5 \). The ratios are different, so EF is not parallel to QR.
(ii) \( \frac{PE}{QE}=\frac{4}{4.5}=\frac{8}{9} \) and \( \frac{PF}{RF}=\frac{8}{9} \). The ratios are equal, so EF ∥ QR.
(iii) EQ = 1.28 − 0.18 = 1.10 and FR = 2.56 − 0.36 = 2.20, so \( \frac{PE}{EQ}=\frac{0.18}{1.10}=\frac{9}{55} \) and \( \frac{PF}{FR}=\frac{0.36}{2.20}=\frac{9}{55} \). The ratios are equal, so EF ∥ QR.
In simple words: Compare the ratios PE : EQ and PF : FR.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 6. In the figure, if LM ∥ CB and LN ∥ CD, prove that \( \frac{AM}{AB}=\frac{AN}{AD} \).

Answer:
In ΔABC, LM ∥ CB, so \( \frac{AM}{AB}=\frac{AL}{AC} \) ...(i)
In ΔACD, LN ∥ CD, so \( \frac{AN}{AD}=\frac{AL}{AC} \) ...(ii)
From (i) and (ii), \( \frac{AM}{AB}=\frac{AN}{AD} \). Hence proved.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 7. In the figure, DE ∥ AC and DF ∥ AE. Prove that \( \frac{BF}{FE}=\frac{BE}{EC} \).

Answer:
In ΔABC, DE ∥ AC, so \( \frac{BE}{EC}=\frac{BD}{DA} \) ...(i)
In ΔBAE, DF ∥ AE, so \( \frac{BF}{FE}=\frac{BD}{DA} \) ...(ii)
From (i) and (ii), \( \frac{BE}{EC}=\frac{BF}{FE} \). Hence proved.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 8. In the figure, DE ∥ OQ and DF ∥ OR. Show that EF ∥ QR.

Answer:
In ΔPQO, DE ∥ OQ, so \( \frac{PD}{DO}=\frac{PE}{EQ} \) ...(i)
In ΔPOR, DF ∥ OR, so \( \frac{PD}{DO}=\frac{PF}{FR} \) ...(ii)
From (i) and (ii), \( \frac{PE}{EQ}=\frac{PF}{FR} \). By the converse of the Basic Proportionality Theorem, EF ∥ QR.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 9. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR. Show that BC ∥ QR.

Answer:
In ΔOPQ, AB ∥ PQ, so \( \frac{OA}{AP}=\frac{OB}{BQ} \) ...(i)
In ΔOPR, AC ∥ PR, so \( \frac{OA}{AP}=\frac{OC}{CR} \) ...(ii)
From (i) and (ii), \( \frac{OB}{BQ}=\frac{OC}{CR} \). By the converse of the Basic Proportionality Theorem, BC ∥ QR.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 10. Using the Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Answer:
Let P be the mid-point of AB in ΔABC and let PQ ∥ BC meet AC at Q. We must prove that Q is the mid-point of AC.
By the Basic Proportionality Theorem \( \frac{AP}{PB}=\frac{AQ}{QC} \). Since P is the mid-point of AB, AP = PB, so \( \frac{AQ}{QC}=1 \) and AQ = QC.
Hence Q is the mid-point of AC. Proved.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 11. Using the converse of the Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Answer:
Let P and Q be the mid-points of AB and AC of ΔABC. Then AP = PB and AQ = QC, so \( \frac{AP}{PB}=1=\frac{AQ}{QC} \).
The line PQ divides the sides AB and AC in the same ratio. By the converse of the Basic Proportionality Theorem, PQ ∥ BC. Proved.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 12. ABCD is a trapezium in which AB ∥ DC and its diagonals intersect each other at the point O. Show that \( \frac{AO}{BO}=\frac{CO}{DO} \).

Answer:
Draw EO ∥ AB meeting AD at E. Since AB ∥ DC, EO ∥ DC as well.
In ΔADC, OE ∥ DC, so \( \frac{AE}{ED}=\frac{AO}{CO} \) ...(i)
In ΔABD, OE ∥ AB, so \( \frac{AE}{ED}=\frac{BO}{DO} \) ...(ii)
From (i) and (ii), \( \frac{AO}{CO}=\frac{BO}{DO} \), that is \( \frac{AO}{BO}=\frac{CO}{DO} \). Hence proved.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.

Question 13. The diagonals of a quadrilateral ABCD intersect each other at the point O such that \( \frac{AO}{BO}=\frac{CO}{DO} \). Show that ABCD is a trapezium.

Answer:
Draw EO ∥ AB meeting AD at E. In ΔDAB, EO ∥ AB, so \( \frac{DE}{EA}=\frac{DO}{OB} \) ...(i)
It is given that \( \frac{AO}{BO}=\frac{CO}{DO} \), that is \( \frac{DO}{OB}=\frac{CO}{AO} \) ...(ii)
From (i) and (ii), \( \frac{DE}{EA}=\frac{CO}{AO} \). Equivalently \( \frac{AE}{ED}=\frac{AO}{OC} \), where E lies on AD and O lies on AC of ΔADC. By the converse of the Basic Proportionality Theorem, EO ∥ DC.
But EO ∥ AB, so AB ∥ DC. Hence ABCD is a trapezium.
In simple words: Apply the Basic Proportionality Theorem.

Teacher's Note:
1. Check which side is parallel to which and write the Basic Proportionality Theorem for the correct triangle.
2. The converse of the theorem is used to prove that two lines are parallel.