- ABC- DEF and their areas are respectively 100 cm 2 and 49 cm 2- If the altitude of - ABC is 5 cm- find the corresponding altitude of - DEF-

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Standard IX

Mathematics

Properties of Isosceles Triangle

Δ ABC∼Δ DEF a...

Question

$Δ A B C \sim Δ D E F$ and their areas are respectively $100 {c m}^{2}$ and $49 {c m}^{2}$ . If the altitude of $Δ A B C$ is $5 c m$ , find the corresponding altitude of $Δ D E F$ .

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Solution

Given $△ A B C ≅ △ D E F$
$\frac{A r e a_{1}}{A r e a_{2}} = \frac{(S i d e_{1})^{2}}{(S i d e_{2})^{2}}$
Given $A r e a_{1} = 100 c m^{2}, A r e a_{2} = 49 c m^{2}$
$a l t i t u d e_{1} = 5 c m, a l t i t u d e_{2} = x$
So $\frac{100}{49} = \frac{5^{2}}{X^{2}}$
$X^{2} = \frac{25 \times 49}{100}$
$X = \frac{5 \times 7}{100}$
$x = 3.5 c m$

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Given △ABC≅△DEFArea1Area2=(Side1)2(Side2)2Given Area1=100cm2,Area2=49cm2altitude1=5cm,altitude2=xSo 10049=52X2X2=25×49100X=5×7100x=3.5 cm

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