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# If the ratio of the perimeter of two similar triangles is 9:16, then what is the ratio of the area?

We can use the area theorem of the triangle to solve the above question.

## Answer: The ratio of the area will be 81:256

Let's proceed step by step.

**Explanation:**

Let us consider ∆ABC whose sides are a, b and c respectively and ∆PQR whose sides are p, q and r respectively.

Let ∆ABC and ∆PQR be two similar triangles.

Assume k as a constant.

So, a/p = b/q = c/r = k

Hence , a = pk , b = qk and c = rk.

Given that:

(a + b + c) / (p + q + r) = 9/16

Substituting values of a, b and c: a = pk , b = qk and c = rk, we get

k.(p + q + r) / (p + q + r) = 9/16. [a + b + c = k ( p+q+r)]

Cancelling (p+q+r) from numerator and denominator gives k = 9/16

Area of ∆ABC / Area of ∆PQR = (a/p)^{2 }= (b/q)^{2 }= (c/r)^{2 }= k^{2}.

= (9/16)^{2}

= 81 / 256

= 81 : 256

### So, the ratio of the area is 81 : 256

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