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 = k².

= (9/16)²

= 81 / 256

= 81 : 256

So, the ratio of the area is 81 : 256