Question

The areas of two similar triangles are 72dm2 and 50dm2. The sum of their perimeters is 226dm. What is the perimeter of each of these triangles?

Answer 1

Answer:

Hence, the perimeter of the triangles are:

P=123.2727 dm

P'=102.7272 dm

Step-by-step explanation:

In two similar triangles:

The ratio of the areas of two triangle is equal to the square of their perimeters.

Let A and A' represents the area of two triangles and P and P' represents their perimeter.

Then they are related as:

$\dfrac{A}{A'}=\dfrac{P^2}{P'^2}$

We are given:

A=72 dm^2  , A'=50 dm^2

and P+P'=226 dm.-----------(1)

i.e. $\dfrac{72}{50}=\dfrac{P^2}{P'^2}\\\\\dfrac{36}{25}=\dfrac{P^2}{P'^2}$

on taking square root on both the side we get:

$\dfrac{P}{P'}=\dfrac{6}{5}\\\\P=\dfrac{6}{5}P'$

Now putting the value of P in equation (1) we obtain:

$\dfrac{6}{5}P'+P'=226\\\\\dfrac{6P'+5\times P'}{5}=226\\\\\dfrac{6P'+5P'}{5}=226\\\\11P'=226\times 5\\\\11P'=1130\\\\P'=\dfrac{1130}{11}=102.7272$

Hence,

P=226-102.7272=123.2727

Hence, the perimeter of the triangles are:

P=123.2727 dm

P'=102.7272 dm