Triangles — Practice

Question 1

In △ABC, D lies on AB and E on AC with DE ∥ BC. If AD = 4 cm, DB = 6 cm and AE = 5 cm, find EC.

Accepted Answer

Step 1 — DE ∥ BC, so the Basic Proportionality Theorem applies.
  AD/DB = AE/EC

Step 2 — Substitute.
  4/6 = 5/EC

Step 3 — Cross-multiply.
  4 × EC = 30, so EC = 7.5

✦ Answer: EC = 7.5 cm.

Question 2

In △PQR, E lies on PQ and F on PR with PE = 2 cm, EQ = 3 cm, PF = 3 cm and FR = 4.5 cm. Is EF ∥ QR?

Accepted Answer

Step 1 — Test the two ratios.
  PE/EQ = 2/3
  PF/FR = 3/4.5 = 2/3

Step 2 — They are equal, so by the converse of BPT the line divides the two sides in the same ratio and must be parallel to the third.

✦ Answer: Yes, EF ∥ QR.

Question 3

In △ABC, AB = 3 cm, AC = 4.5 cm and ∠A = 65°. In △DEF, DE = 4 cm, DF = 6 cm and ∠D = 65°. Are the triangles similar? Name the criterion.

Accepted Answer

Step 1 — Check whether the equal angle is the included one.
  In △ABC, ∠A lies between AB and AC. In △DEF, ∠D lies between DE and DF. It is, in both.

Step 2 — Compare the ratios of those including sides.
  AB/DE = 3/4 = 0.75
  AC/DF = 4.5/6 = 0.75

Step 3 — Equal ratios with the included angle equal is exactly the SAS similarity criterion.

✦ Answer: Yes — △ABC ~ △DEF by SAS similarity.

Question 4

A girl 1.5 m tall stands 3 m away from a lamp-post and casts a shadow 1.8 m long. Find the height of the lamp-post.

Accepted Answer

Step 1 — Set up the two triangles.
  The lamp-post, the girl and their shadows form two triangles sharing the tip of the shadow. Both the post and the girl stand vertical, so both have a right angle at the ground, and the angle at the shadow tip is common.

Step 2 — Two equal angles is enough, so the triangles are similar by AA.

Step 3 — The distance from the shadow tip to the base of the post is 1.8 + 3 = 4.8 m; to the girl's feet it is 1.8 m.
  height of post / height of girl = 4.8 / 1.8
  h / 1.5 = 4.8 / 1.8

Step 4 — Solve.
  h = 1.5 × 4.8 / 1.8 = 4

✦ Answer: The lamp-post is 4 m tall.

Practice — Triangles

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