Question

The ratio of corresponding sides of similar triangles is 3.5; then find
of their areas.
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Answer 1

Solution :-

Suppose that, △ABC ~△PQR in AB and PQ are the corresponding sides whose ratio is 3 : 5.

To find : Ratio of the areas of the similar triangles.

According to the question,

△ABC /△PQR = AB² / PQ² [ By theorem]

=> △ABC /△PQR = 3² / 5²

=> △ABC /△PQR = 9/25

Hence,

Ratio of the areas of the two similar triangle is 9 : 25

Answer 2

Question:

The ratio of corresponding sides of similar triangles is 3:5. Then, find the ratio of areas of two similar triangle.

Solution:

Assume a ∆ABC and ∆DEF.

And ∆ABC is similar to ∆DEF and it's corresponding sides i.e AB and DE are in the ratio 3:5.

Such that;

AB = 3 and DE = 5

Now..

$\dfrac{ \triangle \: ABC}{ \triangle \: DEF} \: = \: \dfrac{ {(AB)}^{2} }{ {(DE)}^{2} }$

→ $\dfrac{ \triangle \: ABC}{ \triangle \: DEF} \: = \: \dfrac{ {(3)}^{2} }{ {(5)}^{2} }$

→ $\dfrac{ \triangle \: ABC}{ \triangle \: DEF} \: = \: \dfrac{9}{25}$

→ ∆ABC : ∆DEF = 9 : 25

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9 : 25 is the ratio of areas of the two similar triangle.

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