Question

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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:

Right answer is 7.

Follow me for explanation..

Answer 2

Answer:

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

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.

$⟹49100=x2(10)2 \\ On \: cross \: multiplying: \\ ⟹100x2=49∗100 \\ ⟹x2=1004900 \\ ⟹x2=49 \\ ⟹x=49 \\ ⟹x=7 \\ </p><p>Hence, the altitude of DEF is 7 unit.</p><p>$