Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 15 Similarity

Exercise 15.1

Answer 1.
Answer:
[Diagram: Triangle ABC with point D on side AB and point E on side AC. DE is parallel to BC. Given measurements show AD = 2, DB = 7, and AC = 5.6]

Given: \( \frac{AD}{DB} = \frac{2}{7} \), AC = 5.6
To find: AE = x
Solution: In ΔABC, DE || BC
By BPT: \( \frac{AD}{DB} = \frac{AE}{EC} \)
\( \frac{2}{7} = \frac{x}{5.6 - x} \)
\( \implies 11.2 - 2x = 7x \)
\( \implies 11.2 = 9x \)
\( \implies x = 1.24 \)

In simple words: When a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio. We used this rule to find the unknown length.

📝 Teacher's Note: Draw a triangle on the board and show how parallel lines create equal ratios. This helps students see the pattern clearly.

🎯 Exam Tip: Always write "By BPT" (Basic Proportionality Theorem) when using this rule. Show the cross multiplication step clearly to get full marks.

Answer 2.
Answer:
[Diagram: Triangle PQR with point M on side PQ and point N on side PR. MN is parallel to QR. Given measurements show sides with expressions (x-2), (x+2), (x-1), and x]

Solution: In ΔPQR, MN || QR
By BPT: \( \frac{PM}{MQ} = \frac{PN}{NR} \)
\( \frac{x-2}{x} = \frac{x+2}{x-1} \)
\( \implies (x-2)(x-1) = x(x+2) \)
\( \implies x^2 - x = x^2 - 4 \)
\( \implies -x = -4 \)
\( \implies x = 4 \)

In simple words: We set up equal ratios using the parallel line rule. Then we solved the equation by cross multiplication to find the value of x.

📝 Teacher's Note: Show students how to expand brackets carefully. The middle terms often cancel out, making the solution easier than it looks.

🎯 Exam Tip: Write each step of cross multiplication clearly. Check your answer by substituting back into the original ratios.

Answer 3.
Answer:
[Diagram: Two similar triangles ABC and PQR with corresponding sides labeled. Triangle ABC has sides 6, 9, and x. Triangle PQR has sides 9, y, and 10.5]

Given: ΔABC ~ ΔPQR
To find: AC and QR
Solution: ΔABC ~ ΔPQR
\( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \) (Similar sides of similar triangles)
\( \frac{6}{9} = \frac{9}{y} = \frac{x}{10.5} \)

From \( \frac{6}{9} = \frac{9}{y} \): \( \frac{6}{9} = \frac{x}{10.5} \)
\( \implies 6y = 81 \) \( \implies 63 = 9x \)
\( \implies y = \frac{81}{6} \) \( \implies x = 7 \)
\( \implies y = \frac{27}{2} \) \( \implies AC = 7cm \)
\( \implies QR = 13.5cm \)

In simple words: Similar triangles have all their sides in the same ratio. We used this fact to find the missing sides by setting up equal fractions.

📝 Teacher's Note: Use cutout triangles of different sizes to show similarity. Students can measure and verify that ratios are equal.

🎯 Exam Tip: Write the similarity statement first (ΔABC ~ ΔPQR). Then write the ratio of corresponding sides. This shows you understand the concept.

Answer 4.
Answer:
[Diagram: Two similar quadrilaterals ABCD and PQRS with corresponding sides labeled. Quadrilateral ABCD has sides 12, 10, 15, and x. Quadrilateral PQRS has sides 8, n, m, and 5]

Given: quadrilateral ABCD ~ quadrilateral PQRS
To find: x, m and n
Solution: quadrilateral ABCD ~ quadrilateral PQRS
\( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{DC}{SR} = \frac{AD}{SR} \)
\( \frac{12}{8} = \frac{x}{5} = \frac{15}{m} = \frac{10}{n} \)
\( \frac{12}{8} = \frac{x}{5} \), \( \frac{12}{8} = \frac{15}{m} \), \( \frac{12}{8} = \frac{10}{n} \)
\( 60 = 8x \), \( 4m = 40 \), \( 3n = 20 \)
\( x = \frac{60}{8} \), \( m = 10cm \), \( n = \frac{20}{3} \)
\( x = \frac{15}{2} \), \( m = 10cm \), \( n = 6.66... \)
\( x = 7.5cm, m = 10cm, n = 6.67cm \)

In simple words: Similar shapes have all corresponding sides in the same ratio. We found this common ratio and used it to calculate the unknown sides.

📝 Teacher's Note: Remind students that in similar figures, we must match corresponding sides correctly. Draw arrows to show which sides correspond to which.

🎯 Exam Tip: Always simplify the common ratio first (like 12/8 = 3/2). This makes calculations easier and reduces mistakes.

Answer 5.
Answer:
[Diagram: Two triangles AOB and COD with point O at the intersection. Lines show measurements including 25-y, 20-x, 6, 9, x, and y]

To find: AO, BO, CO, DO
In ΔAOB and ΔCOD
∠OAB = ∠ODC (90° each)
∠AOB = ∠DOC (vertically opposite angles)
\( \therefore \) ΔAOB ~ ΔDOC (AA corollary)
\( \frac{AO}{DO} = \frac{OB}{OC} = \frac{AB}{DC} \)
\( \frac{x}{20-x} = \frac{2}{3} \), \( \frac{y}{25-y} = \frac{2}{3} \)
\( 3x = 40 - 2x \), \( 3y = 50 - 2y \)
\( 5x = 40 \), \( 5y = 50 \)
\( x = 8 \), \( y = 10 \)
\( AO = 8cm \), \( OB = 10cm \)
\( OD = 20 - 8 = 12cm \), \( OC = 25 - 10 = 15cm \)

In simple words: Two triangles with two equal angles are similar. We used the equal ratios of corresponding sides to find the unknown lengths.

📝 Teacher's Note: Show students how vertically opposite angles are always equal. This is a key step in proving triangles are similar.

🎯 Exam Tip: Write "AA corollary" when two angles are equal. This proves similarity. Then set up the ratio equation to solve for unknowns.

Answer 6.
Answer:
[Diagram: Triangle ABC with point D on side AB and point E on side AC. DE is parallel to BC. Given measurements show DE = 6cm, AE = 3.6cm, AD/DB = 2/3, and DE || BC]

Given: DE = 6cm, AE = 3.6cm, \( \frac{AD}{DB} = \frac{2}{3} \), DE || BC
To find: BC and AC
Solution: In ΔABC, DE || BC
By BPT: \( \frac{AD}{DB} = \frac{AE}{EC} \)
\( \frac{2}{3} = \frac{3.6}{x} \)
\( x = \frac{3.6 \times 3}{2} \)
\( = \frac{1 \times 3}{2} \)
\( x = 5.4 = EC \)
\( \therefore AC = 3.6 + 5.4 = 9cm \)
\( AC = 9cm \)

In ΔADE and ΔABC
∠ADE = ∠ABC
Similarly ∠AED = ∠ACB (corresponding angles)
\( \therefore \) ΔADE ~ ΔABC (AA corollary)
\( \frac{AE}{AC} = \frac{DE}{BC} \) (similar sides of angles)
\( \frac{3.6}{9} = \frac{6}{y} \)
\( y = \frac{9 \times 6}{3.6} \)
\( y = 15 \)
\( BC = 15cm \)

In simple words: We used the parallel line rule to find one part, then used similar triangles to find the other part. Both methods work because parallel lines create similar triangles.

📝 Teacher's Note: Explain that when a line is parallel to one side, it creates two similar triangles. Students can use either BPT or similarity - both give the same answer.

🎯 Exam Tip: Use BPT for ratios of segments. Use similarity for ratios of whole sides. Choose the method that makes calculation easier.

Answer 7.
Answer:
[Diagram: Triangle ABC with point D on side AB and point E on side AC. DE is parallel to BC. Given measurements show various lengths]

To prove: DE || BC
Solution: AB = 5.6cm, AC = 7.2cm
AD = 1.4cm, AE = 1.8cm
DB = 4.2cm, EC = 5.4cm
\( \frac{AD}{DB} = \frac{1.4}{4.2} = \frac{1}{3} \) ---- (1)
\( \frac{AE}{EC} = \frac{1.8}{5.4} = \frac{1}{3} \) ---- (2)
From (1) and (2)
\( \frac{AD}{DB} = \frac{AE}{EC} \)
\( \therefore \) DE || BC (By converse of BPT)

In simple words: We proved that the ratios of the divided sides are equal. The converse of BPT says if ratios are equal, then the line must be parallel.

📝 Teacher's Note: The converse of BPT is the opposite direction. If ratios are equal, then lines are parallel. This is how we prove lines are parallel.

🎯 Exam Tip: Calculate both ratios and show they are equal. Then write "By converse of BPT" to complete the proof.

Answer 8.
Answer:
[Diagram: Triangle ABC with point D on side AB and point E on side AC. DE is parallel to BC. Given measurements show AD = 4, DB = 4.5, AE = 6.4, EC = 7.2]

Solution: \( \frac{AD}{DB} = \frac{4}{4.5} = \frac{8}{9} \) ---- (1)
\( \frac{AE}{EC} = \frac{6.4}{7.2} = \frac{8}{9} \) ---- (2)
From (1) and (2)
\( \frac{AD}{DB} = \frac{AE}{EC} \)
\( \therefore \) DE || BC (By converse of BPT)

In simple words: Again we showed that both ratios are equal (8/9). This proves the line DE is parallel to side BC using the converse rule.

📝 Teacher's Note: Give students practice with different numbers. The concept stays the same - equal ratios mean parallel lines.

🎯 Exam Tip: Simplify fractions to their lowest form to make comparison easier. Write the converse statement clearly to show you understand the theorem.

Answer 9.
Answer:
Given: Two triangles ABC and PQR with sides AB = 1.8 cm and PQ = 2.1 cm
To find: \( \frac{Ar.∆ABC}{Ar.∆PQR} \)

Step 1: Use the theorem for ratio of areas of similar triangles.
\( \frac{Ar.∆ABC}{Ar.∆PQR} = \frac{AB^2}{PQ^2} \)

Step 2: Substitute the given values.
\( = \frac{(1.8)^2}{(2.1)^2} \)

Step 3: Simplify the fraction.
\( = \frac{(\frac{9}{5})^2}{1} = (\frac{9}{7})^2 \)

Step 4: Calculate the final ratio.
\( = \frac{81}{49} \)

Required ratio = 36 : 49
In simple words: When two triangles are similar, the ratio of their areas equals the square of the ratio of their matching sides. We squared both side lengths and made a fraction.

[Diagram: Two similar triangles ABC and PQR are shown side by side with their corresponding sides labeled.]

📝 Teacher's Note: Remind students that for similar triangles, area ratio equals side ratio squared. Common mistake is forgetting to square the sides.

🎯 Exam Tip: Always write the theorem first: "Ratio of areas = square of ratio of corresponding sides." Then substitute values step by step.

Answer 10.
Answer:
Given: AD : PS = 4 : 9 and ∆ABC ~ ∆PQR
To find: \( \frac{Ar.∆ABC}{Ar.∆PQR} \)

Step 1: Write the theorem for similar triangles.
\( \frac{Ar.∆ABC}{Ar.∆PQR} = \frac{AB^2}{PQ^2} \)

Step 2: Use the given information about triangles ∆BAD and ∆QPS.
In ∆BAD and ∆QPS:
∠B = ∠Q (∆ABC ~ ∆PQR)
∠AOB = ∠PSQ (90° each)
∆BAD ~ ∆QPS (AA corollary)

Step 3: Find the ratio of corresponding sides.
\( \frac{AB}{PQ} = \frac{AD}{PS} = \frac{4}{9} \) (Similar sides of similar triangles)

Step 4: Calculate the area ratio.
Using the ratios from steps 1 and 3:
\( \frac{Ar.∆ABC}{Ar.∆PQR} = \frac{AD^2}{PS^2} = (\frac{4}{9})^2 = \frac{16}{81} \)

Required ratio is 16 : 81
In simple words: We used the fact that if triangles are similar, their areas are in the ratio of squares of their matching sides. Since AD:PS = 4:9, the area ratio is 16:81.

[Diagram: Two similar triangles with perpendicular lines drawn from vertices to opposite sides, creating right triangles within them.]

📝 Teacher's Note: Show students how perpendicular lines in similar triangles create smaller similar triangles. This connects similarity with area ratios clearly.

🎯 Exam Tip: When given a ratio of corresponding parts, square it to get the area ratio. Write each step clearly to avoid calculation errors.

Answer 11.
Answer:
Given: ∆ABC ~ ∆DEF
To find: Area of ∆DEF

Step 1: Use the theorem for ratio of areas of similar triangles.
\( \frac{Ar.∆ABC}{Ar.∆DEF} = \frac{BC^2}{EF^2} \)

Step 2: Substitute the known values.
\( \frac{54}{Ar.∆DEF} = (\frac{3}{4})^2 \)

Step 3: Simplify and solve for area of ∆DEF.
\( \frac{54}{Ar.∆DEF} = \frac{9}{16} \)

\( Ar.∆DEF = \frac{54 × 16}{9} = \frac{864}{9} = 96 \text{ cm}^2 \)

In simple words: We used the rule that area ratio equals side ratio squared. Since BC:EF = 3:4, we squared this to get 9:16 for areas. Then we solved to find the unknown area.

[Diagram: Two similar triangles ABC and DEF are shown side by side.]

📝 Teacher's Note: Emphasize cross multiplication as the easiest way to solve proportion problems. Students often get confused with fractions otherwise.

🎯 Exam Tip: Always check your answer by substituting back. If area of DEF is 96 cm², then 54:96 should equal 9:16 when simplified.

Answer 12.
Answer:
Given: ∆ABC ~ ∆XYZ
To find: YZ

Step 1: Use the theorem for ratio of areas of similar triangles.
\( \frac{Ar.∆ABC}{Ar.∆XYZ} = \frac{BC^2}{YZ^2} \)

Step 2: Substitute the known values.
\( \frac{9}{16} = \frac{(2.1)^2}{YZ^2} \)

Step 3: Take square root of both sides.
\( \frac{3}{4} = \frac{2.1}{YZ} \)

Step 4: Solve for YZ.
\( YZ = \frac{2.1 × 4}{3} = \frac{8.4}{3} = 2.8 \text{ cm} \)

In simple words: We found that the area ratio is 9:16, so the side ratio is 3:4 (taking square root). Since BC = 2.1 cm, we calculated YZ using this ratio.

[Diagram: Two similar triangles ABC and XYZ are shown side by side.]

📝 Teacher's Note: Remind students to take square root when going from area ratio to side ratio. Many students forget this step.

🎯 Exam Tip: Write "Taking square root both sides" clearly in your solution. This shows the examiner you understand the relationship between areas and sides.

Answer 13.
Answer:
Given: ∆BCE ~ ∆ACF
To find: Ratio of areas

Step 1: Use Pythagoras theorem in right triangle ABC.
In right triangle ABC: \( AB^2 + BC^2 = AC^2 \)
Since \( 2BC^2 = AC^2 \), we get \( AB = BC \)

Step 2: Apply the theorem for similar triangles.
\( \frac{Ar.∆BCE}{Ar.∆ACF} = \frac{BC^2}{AC^2} \)

Step 3: Substitute the relationship.
\( = \frac{BC^2}{AC^2} = \frac{BC^2}{2BC^2} = \frac{1}{2} \)

Required ratio is 1 : 2
In simple words: We found that AC is bigger than BC by a specific amount. Using the square relationship for similar triangles, the area ratio became 1:2.

[Diagram: A rectangle ABCD with triangles BCE and ACF marked inside, along with point E outside the rectangle.]

📝 Teacher's Note: Draw the figure step by step. Show students how Pythagoras theorem helps find the relationship between sides first.

🎯 Exam Tip: Always establish the side relationship first using given conditions. Then apply the area theorem. This logical sequence gets full marks.

Answer 14.
Answer:
(a) If AN : AC = 5 : 8, find ar(∆AMN) : ar(∆ABC)

Given: \( \frac{AN}{AC} = \frac{5}{8} \)

To Find: \( \frac{Ar.∆AMN}{Ar.∆ABC} \)

In ∆AMN and ∆ABC:
∠AMN = ∠ACB (corresponding angles)
∠ABC = ∠ABC (common)
∴ ∆AMN ~ ∆ABC (AA corollary)

\( \frac{Ar.∆AMN}{Ar.∆ABC} = \frac{AN^2}{AC^2} = (\frac{5}{8})^2 = \frac{25}{64} \)

Required ratio is 25 : 64

(b) If \( \frac{AB}{AM} = \frac{9}{4} \), find \( \frac{Ar.(trapezium BGNM)}{Ar.(∆ABC)} \)

∆AMN ~ ∆ABC [proved above]

\( \frac{Ar.∆AMN}{Ar.∆ABC} = \frac{AM^2}{AB^2} = (\frac{4}{9})^2 = \frac{16}{81} \)

In simple words: We proved the triangles are similar using angle-angle rule. Then we used the area theorem to find ratios by squaring the side ratios.

📝 Teacher's Note: Make sure students identify corresponding angles correctly. Common mistake is mixing up which angles correspond to which.

🎯 Exam Tip: Always prove similarity first before using the area theorem. Write "AA corollary" or "SAS similarity" clearly to show your reasoning.

Answer 15.
Answer:
Given: \( \frac{DE}{BC} = \frac{2}{7} \)
To find: Ratio of areas (Similar sides of similar triangles)

In ∆FDE and ∆FCB:
∠FDE = ∠FCB
∠FED = ∠FBC (Alternate interior angles)
∆FDE ~ ∆FCB (AA corollary)

\( \frac{Ar.∆FDE}{Ar.∆FCB} = \frac{DE^2}{BC^2} = (\frac{2}{7})^2 = \frac{4}{49} \)

In simple words: The line DE is parallel to BC, creating similar triangles. The area ratio is the square of the side ratio, so 2:7 becomes 4:49.

[Diagram: Triangle ABC with line DE drawn parallel to BC, creating smaller similar triangle ADE.]

📝 Teacher's Note: When a line is drawn parallel to one side of a triangle, it always creates similar triangles. This is a very common exam pattern.

🎯 Exam Tip: For parallel line questions, first prove similarity using alternate interior angles. Then apply the area theorem. This is the standard approach.

Answer 16.
Answer:
Given: \( \frac{PT}{TR} = \frac{5}{3} \)
To find: \( \frac{Ar.(∆MTS)}{Ar.(∆MQR)} \)

Step 1: Establish similarity in ∆PST and ∆PRQ.
∠PST = ∠PQR
∠PTS = ∠PRQ (corresponding angles)
∴ ∆PST ~ ∆PRQ (AA corollary)

Step 2: Find the side ratios.
\( \frac{PT}{PR} = \frac{ST}{QR} = \frac{5}{8} \) (Similar sides of similar triangles)

Step 3: Establish similarity in ∆MTS and ∆MQR.
∠MTS = ∠MQR (Alternate interior angles)
∠MST = ∠MRQ
∴ ∆MTS ~ ∆MQR (AA corollary)

Step 4: Calculate the area ratio.
\( \frac{Ar.(∆MTS)}{Ar.(∆MQR)} = \frac{TS^2}{QR^2} = (\frac{5}{8})^2 = \frac{25}{64} \)

i.e. 25 : 64

In simple words: We used the given ratio to find other ratios. Then we applied the area theorem to two similar triangles formed by the parallel lines.

📝 Teacher's Note: This problem has multiple similar triangles. Help students identify each pair separately to avoid confusion.

🎯 Exam Tip: Draw clear diagrams and mark equal angles. This helps you identify all similar triangles correctly and avoid missing any steps.

Answer 17.
Answer:
Given: \( \frac{KL}{KT} = \frac{9}{5} \)
To find: \( \frac{Ar.∆KLM}{Ar.∆KTP} \)

In ∆KLM and ∆KTP:
∠KLM = ∠KTP (Given)
∠LKM = ∠TKP (Common)
∆KLM ~ ∆KTP (AA corollary)

\( \frac{Ar.∆KLM}{Ar.∆KTP} = (\frac{KL}{KT})^2 = (\frac{9}{5})^2 = \frac{81}{25} \)

i.e., 81 : 25

In simple words: We proved the triangles are similar using two equal angles. Then we squared the given side ratio to get the area ratio.

📝 Teacher's Note: Point out that when one angle and the included angle are given, we can use AA similarity. Students sometimes overlook the common angle.

🎯 Exam Tip: Always identify the common angle clearly. Write "common angle" next to it. This shows you understand angle relationships.

Answer 18.
Answer:
In ∆DEF and ∆GHF:
∠DEF = ∠GHF (90° each)
∠DFE = ∠GFH (Common)
∆DEF ~ ∆GHF (AA corollary)

\( \frac{Ar.(∆DEF)}{Ar.(∆GHF)} = \frac{EF^2}{HF^2} \) ---(1)

Step 1: Use Pythagoras theorem in right ∆DEF.
DE² + EF² = DF²
EF² = 10² - 8²
EF² = 100 - 64 = 36
EF = 6

Step 2: Calculate the area ratio.
From (1):
\( \frac{Ar.(∆DEF)}{Ar.(∆GHF)} = (\frac{6}{4})^2 = \frac{36}{16} = \frac{9}{4} \)

i.e., 9 : 4

In simple words: We used Pythagoras theorem to find the missing side. Then we applied the area theorem for similar right triangles.

📝 Teacher's Note: Remind students that Pythagoras theorem is very useful in similarity problems. Always check if you have a right triangle first.

🎯 Exam Tip: Show the Pythagoras calculation step by step. Even if it seems obvious, the examiner wants to see your working clearly.

Exercise 15.2

Answer 1.
Answer:
Scale = 1 : 500
1cm represents 500cm
\( \frac{500}{100} = 5m \)
1cm represents 5m
Length of model = \( \frac{50}{5} = 10cm \)
Breadth of model = \( \frac{40}{5} = 8cm \)
Height of model = \( \frac{70}{5} = 14cm \)
In simple words: When 1cm on the model equals 5m in real life, we divide all real measurements by 5 to get the model size. Think of it like making a toy house that is much smaller than a real house.

📝 Teacher's Note: Show students how to find what 1cm represents first. Then use division to find all model measurements. This is the same as making a scale drawing on paper.

🎯 Exam Tip: Always write what 1cm represents first. Then show each calculation step clearly. Write all measurements with correct units (cm).

Answer 2.
Answer:
20cm represents 400m
1cm represents \( \frac{400}{20} = 20cm \)
Width of model = \( \frac{100}{20} = 5cm \)
Length of model = 20cm
Surface area of the deck of the model = 5cm × 20cm = 100cm²
In simple words: First we find what 1cm represents. Then we use division to find all model sizes. The deck area is just length times width of the model.

📝 Teacher's Note: Help students see that area calculation uses the model measurements, not the real measurements. It is like finding the area of a rectangle on paper.

🎯 Exam Tip: Find the scale first. Use model measurements to calculate model area. Write the final answer with cm² as the unit.

Answer 3.
Answer:
Scale: 1 : 500
1cm represents 500cm
= \( \frac{500}{100} = 5m \)
1cm represents 5m
(i) Actual length of ship = 60 × 5m = 300m
(ii) 1 cm² represents 5m × 5m = 25m²
Deck area of the ship = 1500000m²
Deck area of the model = \( \frac{1500000}{25} \) cm² = 60000cm²
(iii) 1 cm³ represents 5m × 5m × 5m = 125 m³
Volume of the model = 200 cm³
Volume of the ship = 200 × 125m³ = 25000 m³
In simple words: For length, we multiply model size by scale factor. For area, we divide real area by scale factor squared. For volume, we multiply model volume by scale factor cubed.

📝 Teacher's Note: Teach students that area scale is (length scale)² and volume scale is (length scale)³. Use examples like how 2×2 squares fit in a 4×4 square to show area scaling.

🎯 Exam Tip: Remember - length scale 1:500 means area scale 1:25 and volume scale 1:125. Show these calculations clearly in your working.

Answer 4.
Answer:
15cm represents = 30m
1cm represents \( \frac{30}{15} = 2m \)
1 cm² represents 2m × 2m = 4 m²
Surface area of the model = 150 cm²
Actual surface area of aeroplane = 150 × 2 × 2 m² = 600 m²
50 m² is left out for windows
Area to be painted = 600 - 50 = 550 m²
Cost of painting per m² = Rs. 120
Cost of painting 550 m² = 120 × 550 = Rs. 66000
In simple words: We find the real area of the plane using the scale. Then we subtract the window area. Finally we multiply by the cost per square meter to get total cost.

📝 Teacher's Note: Show students that we use the real area (600 m²) for calculations, not the model area. The windows are measured in real area too.

🎯 Exam Tip: Find real surface area first. Subtract window area. Then multiply by cost per m². Show each step clearly and write Rs. in the final answer.

Answer 5.
Answer:
1cm on map represents 12500m on land
1 cm represents 12.5km on land
Length of river on map = 54cm
Actual length of the river = 54 × 12.5 = 675.00km = 675km
In simple words: The map scale tells us how much real distance each cm on the map represents. We multiply the map distance by this scale to get the real distance.

📝 Teacher's Note: Show students a real map and explain how we can measure distances using the scale. This makes the concept very clear and practical.

🎯 Exam Tip: Convert the scale to easy units first (12.5km per cm). Then multiply map distance by scale. Write the final answer in km.

Answer 6.
Answer:
(i) Scale: 1 : 200000
∴ 1cm represents 200000cm
= \( \frac{200000}{1000 × 100} = 2km \)
1cm represents 2km
(ii) 1cm represents 2 km
11² + 16² represents 2 × 2 = 4km²
(iii) 4km² is represented by 1 km²
1km² is represented by \( \frac{1}{4} \) cm²
20km² is represented by \( \frac{1}{4} × 20cm² = 5cm² \)
Area on map that represents the plot of land = 5cm²
In simple words: First we find what distance 1cm represents. Then we find what area 1cm² represents. Finally we calculate how much map area we need for 20km² of real land.

📝 Teacher's Note: Help students understand that area scale is different from length scale. If length scale is 1:200000, then area scale is 1:40000000000 (much bigger number).

🎯 Exam Tip: Find length scale first. Then find area scale by squaring. Use area scale to convert real area to map area. Show all steps clearly.

Answer 7.
Answer:
Actual area = 1872km²
Area on map represents 117 cm²
Let 1cm represents x km
∴ 1cm² represents x × x km²
Actual area = x × x × 117 km²
1872 = x² × 117
x² = \( \frac{1872}{117} \)
x² = 16
x = 4
∴ 1cm represents 4 km
Length of coastline on map = 44cm
Actual length of coastline = 44 × 4 km = 176 km
In simple words: We work backwards from the given areas to find what 1cm represents. Once we know this scale, we can find the real length of the coastline.

📝 Teacher's Note: This is like solving a puzzle. We use the area information to find the scale, then use the scale to find lengths. It is working backwards.

🎯 Exam Tip: Set up the equation using area relationship. Solve for x to find the scale. Then use this scale to find the required length.

Answer 8.
Answer:
Scale: 1 : 25000
∴ 1 cm represents 25000cm
= \( \frac{25000}{1000 × 100} = 0.25km \)
∴ 1cm represents 0.25km

[Diagram: This shows a right triangle ABC where angle B is the right angle. Point A is at the top, B is at bottom left, and C is at bottom right.]

Actual length of AB = 6 × 0.25 = 1.50 km
Area of △ABC = \( \frac{1}{2} × BC × AB \)
= \( \frac{1}{2} × 8 × 6 = 24cm² \)
1cm represents 0.25 km
1 cm² represents 0.25 × 0.25km²
The area of plot = 0.25 × 0.25 × 24 km²
= 0.0625 × 24
= 1.5km²
In simple words: We find the triangle area on the map first. Then we use the area scale to convert this map area to real land area.

📝 Teacher's Note: Show students how to find triangle area using the formula. Then explain that we need to convert this map area to real area using the scale.

🎯 Exam Tip: Find triangle area on map first. Then find what 1cm² represents in real area. Multiply map area by this scale to get real area in km².

Answer 9.
Answer:
Scale: 1 : 25000

1 cm represents 25000 cm

\[ = \frac{25000}{1000 \times 100} = 0.25 \text{ km} \]

1 cm represents 0.25 km
 

[Diagram: A rectangle ABCD with diagonal AC drawn from top-left corner A to bottom-right corner C]

\[ AC^2 = AB^2 + BC^2 \]
\[ = 12^2 + 16^2 \]
\[ = 144 + 256 \]
\[ AC^2 = 400 \]
\[ AC = 20 \text{ cm} \]

Actual length of diagonal = 20 × 0.25
= 5.00
= 5 km

1 cm represents 0.25 km
1 cm² represents 0.25 × 0.25 km²

The area of the rectangle ABCD = AB × BC
= 16 × 12 = 192 cm²

The area of the plot = 0.25 × 0.25 × 192 km²
= 12 km²

In simple words: We used the scale to find real distances. Then we used Pythagoras theorem to find the diagonal. Finally we calculated the real area by converting from map area.

📝 Teacher's Note: Draw a rectangle on the board and show students how to use Pythagoras theorem. Make them understand that scale works for both length and area calculations. Show them that area scale is (length scale)².

🎯 Exam Tip: Always write the scale first. Show all conversion steps clearly. For area, remember to square the scale factor. Write final answers with correct units (km for length, km² for area).