The areas of two similar triangles are $100 c m^{2}$ and $49 c m^{2}$ respectively. if the altitude of the ingle is 5cm, find the corresponding altitude of the other.

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Solution

$f o r space t w o space s i m i l a r space t r i a n g l e s space fraction numerator A r e a 1 over denominator A r e a 2 end fraction space equals space open parentheses fraction numerator a l t i t u d e 1 over denominator a l t i t u d e 2 end fraction close parentheses squared h e n c e 100 over 49 space equals space open parentheses 5 over x close parentheses squared h e n c e space x squared space equals space fraction numerator 49 space cross times 25 over denominator 100 end fraction x space equals space 35 over 10 equals space 3.5 space c m$
hence the other altitude is = 3.5cm

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Δ A B C \sim Δ D E F

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49 {c m}^{2}

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