Triangles — Andhra Pradesh (SSC) Class 10 Mathematics Solutions (Free)

The Basic Proportionality Theorem (also called Thales' Theorem) says this: take a triangle, and draw a line across it that runs parallel to one of its sides. Wherever that line cuts the other two sides, it splits both of them in identical proportions. So for triangle ABC, with a line parallel to side BC cutting AB at D and AC at E, the theorem guarantees: AD/DB = AE/EC The similarity behind the ratio: a parallel line crossing two sides of a triangle creates a smaller triangle (ADE) at the top that is a scaled copy of the whole triangle (ABC). This happens because the parallel line forces ∠ADE to equal ∠ABC and ∠AED to equal ∠ACB (corresponding angles on parallel lines), and triangle ADE already shares angle A with the full triangle — three matching angles is enough (in fact two is enough, by AA) to establish similarity. Once you know the two triangles are similar, the equal ratio AD/DB = AE/EC is really just restating that both segments were scaled down by the same factor.

Since the triangles are similar, all corresponding sides share the same scale factor. Scale factor = PQ/AB = 12/6 = 2 QR = BC × scale factor = 9 × 2 = 18 cm RP = CA × scale factor = 8 × 2 = 16 cm QR = 18 cm, RP = 16 cm.

By the Basic Proportionality Theorem, since DE ∥ BC: AD/DB = AE/EC Given: AD = 5 cm, DB = 10 cm, AE = 4 cm. Find EC. 5/10 = 4/EC Cross-multiplying: 5 × EC = 10 × 4 5 × EC = 40 EC = 8 cm

For similar triangles, the ratio of areas equals the SQUARE of the ratio of corresponding sides. Given: side ratio = 4:9, area of smaller triangle = 32 cm². Area ratio = (4/9)² = 16/81 Area of smaller / Area of larger = 16/81 32 / Area of larger = 16/81 Area of larger = 32 × 81/16 = 162 cm² The area of the larger triangle is 162 cm².

Since DE ∥ BC, triangle ADE is similar to triangle ABC. Given AD:DB = 4:3, side AB is divided into 4 + 3 = 7 equal parts, with AD representing 4 of those 7 parts. AD/AB = 4/7 Since the triangles are similar: DE/BC = AD/AB DE/21 = 4/7 DE = 21 × 4/7 = 12 cm

Given: triangle ABC, right-angled at B, with BD ⊥ AC, meeting AC at D. To prove: AB² = AD × AC Proof: Step 1: In triangles ABD and ACB: ∠ADB = ∠ABC = 90° (BD ⊥ AC by construction; ABC is right-angled at B, given) ∠A is common to both triangles. Step 2: By AA similarity, △ABD ~ △ACB. Step 3: Since the triangles are similar, corresponding sides are proportional: AB/AC = AD/AB Step 4: Cross-multiplying: AB × AB = AD × AC AB² = AD × AC Hence proved.

For similar triangles, the ratio of perimeters equals the same ratio as their corresponding sides. Given: side ratio = 3:8, perimeter of smaller triangle = 36 cm. Perimeter of smaller / Perimeter of larger = 3/8 36 / Perimeter of larger = 3/8 Perimeter of larger = 36 × 8/3 = 96 cm The perimeter of the larger triangle is 96 cm.

Checking whether all three pairs of corresponding sides share the same ratio (SSS similarity criterion): PQ/AB = 6/10 = 0.6 QR/BC = 9/15 = 0.6 RP/CA = 12/20 = 0.6 All three ratios equal 0.6 (or 3/5). Since all corresponding sides are in the same ratio, the triangles ARE similar by the SSS similarity criterion, with triangle PQR being a smaller version of triangle ABC at a scale factor of 3/5.

Since DE ∥ BC, by the Basic Proportionality Theorem: AD/AB = AE/AC Given: AD = 3 cm, AB = 9 cm, AC = 12 cm. Find AE. 3/9 = AE/12 AE = 12 × 3/9 = 4 cm Now find EC: EC = AC − AE = 12 − 4 = 8 cm AE = 4 cm, EC = 8 cm.