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A triangle ABC has sides AB = AC = 5cm and BC = 6 cm Triangle A’ B’ C’ is the reflection of the triangle ABC in a line parallel to AB placed at a distance 2 cm from AB, outside the triangle ABC. Triangle A” B” C” is the reflection of the triangle A’ B’ C’ in a line parallel to B’ C’ placed at a distance of 2 cm From B’ C’ outside the triangle A’ B’ C’. Find the distance between A and A”

A triangle ABC has sides AB = AC = 5cm and BC = 6 cm Triangle A’ B’ C’ is the reflection of the triangle ABC in a line parallel to AB placed at a distance 2 cm from AB, outside the triangle ABC. Triangle A” B” C” is the reflection of the triangle A’ B’ C’ in a line parallel to B’ C’ placed at a distance of 2 cm From B’ C’ outside the triangle A’ B’ C’. Find the distance between A and A”

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
6 years ago
Hello Student,
Please find the answer to your question
Let L be the line parallel to side AB of ∆ ABC, at a distance of 2 cm from AB, in which the first reflection ∆A’ B’ C’ is obtained. Let L’ be the second line parallel to B’ C’, at a distance of 2 cm from B’ C’, in which reflection of ∆A’ B’ C’ is taken as ∆A” B” C”.
In figure, size of ∆A” B” C” is same to the size of ∆A’ B’ C’.
236-699_12345.png
From figure AA’ = 4cm and A’A” = 12 cm. So to find AA” it suffices to know ∠AA’A”, clearly
∠AA’A” = 90° + α where sin α = 3/5
⇒ cos ∠AA’A” = cos(90° + α) = sin α = -3/5 and hence
AA” = √(AA’)2 + (A’ A”)2 – 2AA’ x A’ A”. cos(90° + α)
= √16 + 144 + 96 x 3/5
= √1088/5 = 8√17/5 cm.

Thanks
Aditi Chauhan
askIITians Faculty

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