Question

Abc and def are similar triangles. If area of abc is 100 and def are 49 ..Altitude of abc is 10 then find altitude of def?​

Answer 1

$\underline{ \mathfrak{ \: Given:- \: }}$

• ΔABC and ΔDEF are similar triangles.

• Altitude of ΔABC = 10

$\underline{ \mathfrak{ \: To\:Finf:- \: }}$

• Altitude of ΔDEF

$\underline{ \mathfrak{ \: Solution :- \: }}$

⠀⠀⠀⠀

Let the altitude of ΔDEF be x.

Given that

• ΔABC and ΔDEF are similar triangles.

• Altitude of ΔABC = 10

We know that -

The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Then:-

$\sf \frac{ \triangle ABC}{\triangle DE F } = \frac{(Altitude \: of \: ABC) {}^{2} }{(Altitude \: of \: DE F) {}^{2} } \\ \\ \sf \leadsto \frac{100}{49} = \frac{(10)^2 }{x^2} \\ \\ \leadsto \sf 100x^2 = 49 × 100 \\ \\ \leadsto \sf x^2 = \frac{4900}{100} \\ \\\leadsto \sf x^2 = 49\\\\ \sf \leadsto x = \sqrt{49} \\\\ \leadsto {\boxed{\frak{\purple{x= 7}}}}$

Hence,the altitude of DEF is 7 unit.