Question

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

Answer 1

Calculate ratio of area of triangles

Given, ratio of perimeter of the triangles is $9:16$.

Let, $a$, $b$, and $c$ be the sides of the the smaller triangle, and $x$, $y$, and $z$ be the corresponding sides of the bigger triangle. Then,
$\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=k$ [Since they are similar triangles]
where $k$ is the proportionality constant

We are given that,
$\begin{array}{ccc}& \frac{a+b+c}{x+y+z}=\frac{9}{16}& \\ \Rightarrow & \frac{xk+yk+zk}{x+y+z}=\frac{9}{16}& \left[\because \frac{a}{x}=k,\frac{b}{y}=k,\frac{c}{z}=k\right]\\ \Rightarrow & \frac{k\left(x+y+z\right)}{x+y+z}=\frac{9}{16}& \\ \Rightarrow & k=\frac{9}{16}& \end{array}$

The area of similar triangles is directly proportional to the square of the ratio of the corresponding sides. Thus,
$\begin{array}{c}\frac{△\text{Area of smaller triangle}}{△\text{Area of bigger triangle}}={\left(\frac{a}{x}\right)}^2\\ ={\left(\frac{xk}{x}\right)}^2\\ ={\left(\frac{9}{16}\right)}^2\\ \therefore \frac{△\text{Area of smaller triangle}}{△\text{Area of bigger triangle}}=\frac{81}{256}\end{array}$

Therefore, the ratio of the area of the given similar triangles is $81:256$.