Question

The area of two similar triangles are 169 $c{m}^2$ and 121 $c{m}^2$ respectively, If the longest side of the largest triangle is 26 cm, find the longest side of the smaller triangle?

Answer 1

Given the area of ${\Delta }_1=169{cm}^2$ and area of ${\Delta }_2=121{cm}^2$

Let longest side of the smaller triangle be $x$.

For two similar triangles, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$⟹\frac{Areaof{\Delta }_1}{Areaof{\Delta }_2}={\left(\frac{Longestsideof{\Delta }_1}{Longestsideof{\Delta }_2}\right)}^2$

Substituting the values, we get

$⟹\frac{169}{121}={\left(\frac{26}{x}\right)}^2$

$⟹\sqrt{\frac{169}{121}}=\frac{26}{x}$

$\frac{13}{11}=\frac{26}{x}$

$x=22cm$

The longest side of smaller triangle is $22cm$