Exercise 8.3 — Areas of Similar Triangles

Ratio of areas of similar triangles.

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Exercise 8.3 — Areas of Similar Triangles

Exercise 8.3 belongs to Chapter 8, Similar Triangles, in Class 10 Mathematics for students following the CBSE, Telangana, and Andhra Pradesh board syllabi. This exercise is built entirely around one powerful theorem that connects the shape similarity of two triangles to their area ratio: the ratio of the areas of two similar triangles equals the square of the ratio of their corresponding sides.

This is one of the most frequently tested theorems in board exams, and Exercise 8.3 gives you seven different problem types — from finding area ratios using midpoints, to using medians, altitudes, and parallel-line divisions — all built on the same core idea: squares of ratios.

Area Ratio Theorem Midpoint Triangle Parallel Line Division Medians & Altitudes Basic Proportionality Theorem

CORE THEOREM

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides: ar(△ABC) / ar(△PQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)²

💡 Why this matters: This single theorem is the engine behind every question in Exercise 8.3. Whether the problem gives you sides, medians, or altitudes, the moment you know two triangles are similar, you can square any pair of corresponding linear measurements to get the area ratio.

How the Theorem Is Proved

To prove this theorem, we draw altitudes AM ⊥ BC and PN ⊥ QR in two similar triangles ABC and PQR. Since the area of a triangle is half the base times the height, the ratio of areas becomes (BC × AM) / (QR × PN). Using AA similarity in triangles ABM and PQN (both have a right angle, and ∠B = ∠Q from the original similarity), we get AM/PN = AB/PQ. Combining this with the known ratio AB/PQ = BC/QR = AC/PR from triangle similarity, the area ratio simplifies neatly to the square of the side ratio.

Question 1 — Area Ratio of the Midpoint Triangle DEF and △ABC

D, E, F are the midpoints of sides BC, CA, and AB of triangle ABC. We need to find the ratio of the area of triangle DEF (called the medial triangle) to the area of triangle ABC.

Triangle DEF Inside Triangle ABC

D, E, F are midpoints of BC, CA, AB

Question 1 — Solution

Find ar(△DEF) : ar(△ABC)

Step 1 — Use the midpoint property: F is midpoint of AB ⟹ AF = FB ⟹ AF/FB = 1 E is midpoint of AC ⟹ AE = EC ⟹ AE/EC = 1 Step 2 — Apply converse of Basic Proportionality Theorem: Since AF/FB = AE/EC, we get FE ∥ BC Similarly, ED ∥ AB ⟹ BDEF is a parallelogram Step 3 — Find side ratios: FE = BD = (1/2)BC ⟹ FE/BC = 1/2 Similarly: DE/AB = 1/2 and DF/AC = 1/2 Step 4 — Establish similarity: Since FE/BC = DE/AB = DF/AC, by SSS similarity: △DEF ~ △ABC Step 5 — Apply the area ratio theorem: ar(△DEF)/ar(△ABC) = (FE/BC)² = (1/2)² = 1/4 ∴ Ratio of areas of △DEF and △ABC = 1 : 4

✅ Key takeaway: The triangle formed by joining the midpoints of any triangle always has exactly one-quarter the area of the original triangle. This is a useful fact worth memorizing for quick board exam answers.

Question 2 — Finding AX/XB When XY Divides the Triangle Into Equal Areas

In triangle ABC, line XY is drawn parallel to AC, with X on AB and Y on BC, such that XY divides the triangle into two regions of equal area. We need to find the ratio AX/XB.

Question 2 — Solution

Find AX/XB when ar(△BXY) = ar(quad. AXYC)

Given: ar(△BXY) = ar(quad. AXYC) = (1/2) ar(△ABC) Step 1 — Establish similarity using AA: ∠BXY = ∠BAC (XY ∥ AC, corresponding angles) ∠BYX = ∠BCA (XY ∥ AC, corresponding angles) ∠XBY = ∠ABC (common angle) ⟹ △XBY ~ △ABC (by AAA similarity) Step 2 — Apply the area ratio theorem: ar(△ABC) / ar(△XBY) = (AB/XB)² Since ar(△XBY) = (1/2) ar(△ABC): ar(△ABC) / ar(△XBY) = 2/1 ⟹ (AB/XB)² = 2 ⟹ AB/XB = √2 Step 3 — Solve for AX/XB: (AX + XB)/XB = √2 AX/XB + 1 = √2 AX/XB = √2 − 1 ∴ AX/XB = √2 − 1

📌 Algebraic trick to remember: Whenever a ratio like AB/XB appears, and AB = AX + XB, you can always split the fraction as (AX/XB) + (XB/XB) = (AX/XB) + 1. This lets you isolate AX/XB cleanly in one step.

Question 3 — Proving the Area Ratio Using Corresponding Medians

This question asks for a complete proof that the ratio of areas of two similar triangles equals the square of the ratio of their corresponding medians — not just their sides. The strategy is to create a pair of smaller similar triangles using the medians and apply the main area theorem again.

Question 3 — Proof

Prove ar(△ABC)/ar(△PQR) = (CM/RN)², where CM and RN are corresponding medians

Given: △ABC ~ △PQR, with CM a median to AB and RN a median to PQ Step 1 — From triangle similarity: ar(△ABC)/ar(△PQR) = (AC/PR)² …(1) Step 2 — Use median definitions: AM = BM = (1/2)AB (CM is median in △ABC) PN = QN = (1/2)PQ (RN is median in △PQR) Step 3 — Compare triangles AMC and PNR: ∠A = ∠P (given △ABC ~ △PQR) AC/PR = AB/PQ = [(1/2)AB] / [(1/2)PQ] = AM/PN ⟹ AC/PR = AM/PN Step 4 — Apply SAS similarity: Since ∠A = ∠P and AC/PR = AM/PN, by SAS similarity: △AMC ~ △PNR ⟹ AC/PR = CM/RN …(2) Step 5 — Combine (1) and (2): ar(△ABC)/ar(△PQR) = (AC/PR)² = (CM/RN)² ∴ Ratio of areas of similar triangles = square of ratio of corresponding medians ✓

✅ Pattern to notice: This proof technique — finding a smaller pair of similar triangles using a special line segment (median, altitude, angle bisector) and then linking it back to the main side ratio — is reused throughout this chapter. Questions about altitudes (like Q6) use exactly the same structure.

Question 4 — Finding the Area of △DEF Given Side Lengths and One Area

Given that △ABC ~ △DEF, with BC = 3 cm, EF = 4 cm, and the area of △ABC = 54 sq. cm, we are asked to find the area of △DEF. This is the most direct application of the area ratio theorem in the entire exercise.

Question 4 — Solution

Find ar(△DEF), given BC = 3 cm, EF = 4 cm, ar(△ABC) = 54 cm²

Since △ABC ~ △DEF: ar(△ABC) / ar(△DEF) = (BC/EF)² 54 / ar(△DEF) = (3/4)² 54 / ar(△DEF) = 9/16 ar(△DEF) = (54 × 16) / 9 ar(△DEF) = 6 × 16 ar(△DEF) = 96 cm²

ar(△ABC)

54 cm²

Given

ar(△DEF)

96 cm²

Calculated

Question 5 — Proving Area of △APQ Is 1/16 of Area of △ABC

Line PQ meets AB at P and AC at Q, with AP = 1 cm, BP = 3 cm, AQ = 1.5 cm, and CQ = 4.5 cm. We must prove that ar(△APQ) = (1/16) ar(△ABC).

Question 5 — Proof

Prove ar(△APQ) = (1/16) ar(△ABC)

Step 1 — Check the ratio condition: AP/PB = 1/3 AQ/QC = 1.5/4.5 = 1/3 Since AP/PB = AQ/QC, by the converse of the Basic Proportionality Theorem: PQ ∥ BC Step 2 — Establish similarity: ∠A = ∠A (common angle) ∠P = ∠B (corresponding angles, PQ ∥ BC) ∠Q = ∠C (corresponding angles, PQ ∥ BC) ⟹ △APQ ~ △ABC (by AAA similarity) Step 3 — Apply the area ratio theorem: ar(△APQ)/ar(△ABC) = (AP/AB)² AB = AP + PB = 1 + 3 = 4 ar(△APQ)/ar(△ABC) = (1/4)² = 1/16 ∴ ar(△APQ) = (1/16) ar(△ABC) ✓

💡 Recognizing the pattern: Whenever a question gives you two separate ratios like AP/PB and AQ/QC and they turn out equal, that's your signal to immediately apply the converse of the Basic Proportionality Theorem (BPT) to prove the line is parallel to the third side — the very first move in this type of problem.

Question 6 — Finding the Corresponding Altitude Using the Area Ratio

The areas of two similar triangles are 81 cm² and 49 cm². If the altitude of the bigger triangle is 4.5 cm, we need to find the corresponding altitude of the smaller triangle. Just like medians, altitudes of similar triangles also follow the squared-ratio rule for areas.

Question 6 — Solution

Find the smaller triangle's altitude, given areas 81 cm², 49 cm² and bigger altitude = 4.5 cm

Setup: ar(△ABC) = 81 cm², ar(△DEF) = 49 cm² Altitude from A in △ABC: AP = 4.5 cm Let altitude from D in △DEF: DQ = x cm Step 1 — Show triangles ABP and DEQ are similar: ∠B = ∠E (given △ABC ~ △DEF) ∠P = ∠Q = 90° (both are altitudes) ⟹ △ABP ~ △DEQ (by AA similarity) ⟹ AB/DE = AP/DQ …(1) Step 2 — Apply area ratio theorem: ar(△ABP)/ar(△DEQ) = (AB/DE)² = (AP/DQ)² (using eq. 1) 81/49 = (4.5/x)² Step 3 — Solve for x: 4.5/x = √(81/49) = 9/7 x = (4.5 × 7) / 9 x = 0.5 × 7 x = 3.5 cm

Bigger △

81 cm²

Altitude = 4.5 cm

Smaller △

49 cm²

Altitude = 3.5 cm

Key Theorems Used in This Exercise

Theorem	Statement	Used In
Area Ratio Theorem	Ratio of areas of similar triangles = square of ratio of corresponding sides	Q1, Q2, Q4, Q5
Converse of BPT	If a line divides two sides of a triangle in the same ratio, it is parallel to the third side	Q1, Q5
AA / AAA Similarity	If two (or three) angles of one triangle equal those of another, the triangles are similar	Q2, Q5, Q6
SAS Similarity	If one angle is equal and the sides including it are proportional, triangles are similar	Q3
SSS Similarity	If all three corresponding sides are proportional, triangles are similar	Q1

Common Mistakes to Avoid

Forgetting to square the ratio: The single most common mistake is using the side ratio directly as the area ratio. Always remember to square the side, median, or altitude ratio before equating it to the area ratio.
Confusing AB with AX or XB: In problems like Q2 and Q5, where a point divides a side into two parts, be careful to identify whether the question wants the ratio of a part to the whole side, or one part to the other part. Misreading this leads to squaring the wrong fraction.
Skipping the similarity proof: Before applying the area ratio theorem, you must first establish that the two triangles are actually similar (using AA, SAS, or SSS). Jumping straight to squaring side ratios without this step loses marks in board exams.
Mixing up medians/altitudes with sides: The area ratio theorem works for any pair of corresponding linear measurements — not just the three main sides. Medians (Q3) and altitudes (Q6) follow the exact same squared-ratio rule, but require an extra similarity proof first.
Sign or root errors in square root steps: When solving equations like (AB/XB)² = 2, remember to take the positive square root only, since lengths cannot be negative.

⛔ Exam trap: In Question 5 type problems, students sometimes use AP/AQ to write the area ratio instead of AP/AB. Always identify the correct pair of corresponding sides between the two similar triangles before squaring.

Quick Reference — All Answers at a Glance

Question	What Was Asked	Answer
Q1	ar(△DEF) : ar(△ABC), D,E,F are midpoints	1 : 4
Q2	AX/XB when XY divides triangle into equal areas	√2 − 1
Q3	Prove area ratio = square of median ratio	Proved: ar(△ABC)/ar(△PQR) = (CM/RN)²
Q4	ar(△DEF), given BC=3, EF=4, ar(△ABC)=54 cm²	96 cm²
Q5	Prove ar(△APQ) = (1/16) ar(△ABC)	Proved using AP/AB = 1/4
Q6	Corresponding altitude of smaller triangle	3.5 cm

What This Exercise Prepares You For

Exercise 8.3 is a high-weightage section in board exams across Telangana SSC, AP SSC, and CBSE Class 10 syllabi, often appearing as 2-mark direct-application questions or 4-mark proof questions. Mastery here is essential before moving on to the Pythagoras Theorem applications later in this chapter, since both rely heavily on triangle similarity techniques.

To strengthen the foundational concepts used throughout this exercise, revisit Introduction to Similar Triangles for the AA, SAS, and SSS similarity criteria, and the Basic Proportionality Theorem, which is used directly in Questions 1 and 5 of this exercise. These concepts also connect to Class 9 Triangles, where the fundamentals of triangle congruence and properties are first introduced.

📐 Board Exam Strategy (Telangana & AP SSC, CBSE): For proof-based questions like Q3 and Q5, always structure your answer in this order: state the given information, prove similarity using the correct criterion (AA/SAS/SSS), then apply the area ratio theorem with squared ratios. This structured approach is what examiners look for when awarding full marks.