Triangles — Practice

Question 1

If △ABC ~ △PQR and AB/PQ = 3/5, what is the ratio of their perimeters?

Accepted Answer

Step 1 — In similar triangles all sides are in the same ratio, so perimeters share that ratio.

✦ Answer: 3 : 5.

Question 2

In △ABC, DE ∥ BC with AD = 2 cm, DB = 3 cm, AE = 4 cm. Find EC.

Accepted Answer

Step 1 — By BPT, AD/DB = AE/EC ⇒ 2/3 = 4/EC.

Step 2 — EC = 6 cm.

✦ Answer: EC = 6 cm.

Question 3

Two similar triangles have corresponding sides in ratio 4:9. Find the ratio of their areas.

Accepted Answer

Step 1 — ar ratio = (4/9)² = 16/81.

✦ Answer: 16 : 81.

Question 4

A vertical pole 6 m high casts a shadow 4 m long; at the same time a tower casts a 28 m shadow. Find the tower's height.

Accepted Answer

Step 1 — Equal sun angles ⇒ triangles similar: height/28 = 6/4.

Step 2 — height = 28 × 6/4 = 42 m.

✦ Answer: 42 m.

Question 5

In △ABC, DE ∥ BC. If AD = x, DB = x−2, AE = x+2, EC = x−1, find x.

Accepted Answer

Step 1 — By BPT, x/(x−2) = (x+2)/(x−1).

Step 2 — x(x−1) = (x+2)(x−2) ⇒ x² − x = x² − 4 ⇒ −x = −4 ⇒ x = 4.

✦ Answer: x = 4.

Question 6

Sides of a triangle are 7 cm, 24 cm, 25 cm. Is it right-angled?

Accepted Answer

Step 1 — Check largest side: 25² = 625.

Step 2 — 7² + 24² = 49 + 576 = 625 = 25².

Step 3 — By the converse of Pythagoras, the triangle is right-angled (right angle opposite the 25 cm side).

✦ Answer: Yes, right-angled.

Question 7

The areas of two similar triangles are 81 cm² and 49 cm². If a side of the larger is 6.3 cm, find the corresponding side of the smaller.

Accepted Answer

Step 1 — (side ratio)² = 81/49 ⇒ side ratio = 9/7.

Step 2 — 6.3 / x = 9/7 ⇒ x = 6.3 × 7/9 = 4.9 cm.

✦ Answer: 4.9 cm.

Question 8

State and prove the Basic Proportionality Theorem.

Accepted Answer

Statement — A line parallel to one side of a triangle divides the other two sides in the same ratio.

Step 1 — In △ABC, let DE ∥ BC (D on AB, E on AC). Join BE and CD; draw EM ⊥ AB and DN ⊥ AC.

Step 2 — ar(ADE)/ar(BDE) = AD/DB (same height EM, bases AD, DB). Similarly ar(ADE)/ar(CDE) = AE/EC.

Step 3 — △BDE and △CDE are on the same base DE and between the same parallels DE, BC, so ar(BDE) = ar(CDE).

Step 4 — Hence AD/DB = AE/EC. ∎

✦ Answer: proof by equal-area triangles as above.

Question 9

Prove that the ratio of areas of two similar triangles equals the ratio of the squares of their corresponding sides.

Accepted Answer

Step 1 — Let △ABC ~ △DEF; draw altitudes AP ⊥ BC and DQ ⊥ EF.

Step 2 — ar(ABC)/ar(DEF) = (½·BC·AP)/(½·EF·DQ) = (BC/EF)(AP/DQ).

Step 3 — In △ABP and △DEQ: ∠B = ∠E (similar) and ∠P = ∠Q = 90° ⇒ △ABP ~ △DEQ (AA) ⇒ AP/DQ = AB/DE = BC/EF.

Step 4 — Substitute: ar(ABC)/ar(DEF) = (BC/EF)(BC/EF) = (BC/EF)². ∎

✦ Answer: area ratio = (corresponding side ratio)².

Question 10

In △ABC, D and E are points on AB and AC such that DE ∥ BC and AD:DB = 3:2. Find ar(△ADE) : ar(△ABC).

Accepted Answer

Step 1 — DE ∥ BC ⇒ △ADE ~ △ABC (AA: common angle A and equal corresponding angles).

Step 2 — AD/AB = 3/(3+2) = 3/5.

Step 3 — ar(ADE)/ar(ABC) = (3/5)² = 9/25.

✦ Answer: 9 : 25.

Practice — Triangles

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