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

Answer:

Answer:

$\large{\underline{\boxed{\sf Altitude \: of \: \triangle D EF = 7}}}$

Step-by-step explanation:

Given that -

ΔABC and ΔDEF are similar triangles.

⇒ ΔABC ~ΔDEF

We know that -

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

Also, it is given that -

Altitude of ΔABC = 10

Let the altitude of ΔDEF be x.

$\sf \frac{ \triangle ABC}{\triangle DE F } = \frac{(Altitude \: of \: ABC) {}^{2} }{(Altitude \: of \: DE F) {}^{2} } \\ \\ \sf \implies \frac{100}{49} = \frac{(10)^2 }{x^2} \\ \\ \bf On \: cross \: multiplying : \\ \\ \implies \sf 100x^2 = 49 * 100 \\ \\ \implies \sf x^2 = \frac{4900}{100} \\ \\ \implies \sf x^2 = 49\\\\ \sf \implies x = \sqrt{49} \\\\ \sf \implies x = 7$

Hence, the altitude of DEF is 7 unit.

Answer 2

Answer:

altitude is 13

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