Exercise 11.3 — Triangles

Exercise 11.3 — Areas: Triangles on the Same Base and Parallels

Exercise 11.3 is part of Chapter 11, Areas, of Class 9 Mathematics for CBSE, Telangana, and Andhra Pradesh board students. This is one of the most proof-intensive exercises in the chapter — every question asks you to show or prove an area relationship, making it critical for board exams where 4–5 mark proofs are commonly asked.

The entire exercise rests on two powerful theorems about triangles sharing the same base. Once you deeply understand those two theorems, all nine problems in this exercise follow naturally. The exercise also weaves in key concepts from earlier — medians, parallelograms, midpoint theorem, and trapeziums — making it an excellent revision tool.

The Two Foundation Theorems — Know These Before Attempting Any Problem

Every question in Exercise 11.3 is either a direct application of one of these theorems, or a multi-step proof that builds on them. Study the theorem statements and their converses carefully.

These two theorems are converses of each other. Theorem 1 lets you go from parallel lines → equal areas, while Theorem 2 lets you go from equal areas → parallel lines. Question 5 and Question 9 in this exercise rely specifically on Theorem 2.

Worked Example — The Median Always Creates Two Equal-Area Triangles

Before diving into the exercise problems, the textbook presents this important worked example. The result it proves is used as a ready-made tool in Problems 1, 2, and 4. Understanding it fully saves significant time.

Question 1 — E is the Midpoint of Median AD

In △ABC, AD is the median and E is the midpoint of AD. We need to show two things: (i) ar(△ABE) = ar(△ACE), and (ii) ar(△ABE) = ¼ ar(△ABC). This problem is a beautiful two-layer application of the median property.

Question 2 — Diagonals of a Parallelogram Make 4 Equal Triangles

In parallelogram ABCD, diagonals AC and BD intersect at O. The task is to prove that the four triangles created — △AOB, △BOC, △COD, and △AOD — all have equal areas. This is a direct consequence of the fact that the diagonals of a parallelogram bisect each other.

Question 3 — Segment CD Bisected by AB Implies Equal Triangle Areas

△ABC and △ABD share base AB, and line segment CD is bisected by AB at point O. We must prove that ar(△ABC) = ar(△ABD). The trick here is to identify the hidden medians within smaller triangles and then combine their area equalities.

Question 4 — Midpoints D, E, F of Sides of △ABC: Three Results

This is the most multi-part question in the exercise. D, E, F are the midpoints of BC, CA, and AB respectively in △ABC. We need to prove that (i) BDEF is a parallelogram, (ii) ar(△DEF) = ¼ ar(△ABC), and (iii) ar(BDEF) = ½ ar(△ABC). The key tool here is the Midpoint Theorem: the line joining midpoints of two sides of a triangle is parallel to the third side and half its length.

Question 5 — Equal-Area Triangles on Same Base Imply DE ∥ BC

D and E are points on sides AB and AC of △ABC respectively. Given that ar(△DBC) = ar(△EBC), we need to prove that DE ∥ BC. This uses Theorem 2 (the converse) — equal areas on the same base forces the triangles between the same parallels.

Question 6 — XY ∥ BC Through A, Proving ar(△ABE) = ar(△ACF)

A line XY is drawn through A parallel to BC. Lines BE ∥ CA and CF ∥ BA are extended to meet XY at E and F. We must show ar(△ABE) = ar(△ACF). The key is to spot two parallelograms sharing the same base BC.

Question 7 — Trapezium Diagonals: ar(△AOD) = ar(△BOC)

In trapezium ABCD with AB ∥ DC, the diagonals AC and BD meet at O. Prove that ar(△AOD) = ar(△BOC). The approach is elegant — find two whole triangles with equal areas, then subtract the common part.

Question 8 — Pentagon ABCDE: ar(△ACB) = ar(△ACF) and ar(AEDF) = ar(ABCDE)

ABCDE is a pentagon. A line through B parallel to AC meets DC extended at F. We must prove two things: that triangles ACB and ACF are equal in area, and that the area of quadrilateral AEDF equals the area of the original pentagon.

Question 9 — Proving PQSR and RSBA Are Trapeziums

Given ar(△RAS) = ar(△RBS) and ar(△QRB) = ar(△PAS), prove that both quadrilaterals PQSR and RSBA are trapeziums. This problem requires applying the converse theorem twice — going from equal areas back to parallel lines.

Question 10 — The Land Exchange Problem (Application)

A villager Ramayya owns a quadrilateral plot ABCD and agrees to donate a triangular corner △MCD (where M is the midpoint of BC) to the gram panchayat for a school. In exchange, he should receive an equal area of land adjoining his plot, converting his remaining land into a triangle of the same total area. This is a real-life application of the theorem that two triangles on the same base between the same parallels are equal.

Common Mistakes to Avoid in Exercise 11.3

• Using the wrong theorem direction: Theorem 1 goes from parallel lines → equal areas. Theorem 2 (the converse) goes from equal areas → parallel lines. In Q5 and Q9 you need Theorem 2. Many students apply Theorem 1 in both directions incorrectly.
• Forgetting to identify the shared base: Before applying either theorem, always state clearly which base (or equal bases) the two triangles share. This is the starting condition for both theorems.
• Confusing "same base" with "same vertex": Two triangles can share a base without sharing any vertices (like △DBC and △EBC in Q5). The base is the bottom edge, not the top vertex.
• Skipping equation numbering in proofs: Board examiners expect you to label intermediate results (1), (2), (3)… and refer back to them. This is standard proof format for Telangana and AP exams.
• In Q4, not stating which parallelogram's diagonal is used: There are three parallelograms formed — BDEF, DCEF, and AFDE. Each gives one congruence result. You must use all three to build the complete proof for part (ii).

Quick Reference — All 10 Problems at a Glance

What Exercise 11.3 Prepares You For

The proof techniques and area-comparison strategies you practise here are directly tested in board exams across CBSE, Telangana, and Andhra Pradesh. 4-mark and 5-mark questions from this exercise appear regularly in SA1 and SA2 papers. Q1, Q2, Q4, and Q7 are among the most frequently examined.

The concepts from this exercise also connect forwards to Coordinate Geometry (Chapter 7 / Class 9), where area of a triangle is computed using vertex coordinates, and the same equal-area results can be verified algebraically. The midpoint theorem you used in Q4 is a central result in Triangles proofs as well.

For Class 10, these area ideas build toward similarity of triangles, where the ratio of areas of similar triangles equals the square of the ratio of corresponding sides — a direct extension of the base-height area thinking developed here.