Question

The perimeter of an isosceles triangle is $32$cm. The ratio of the equal side to its base is $3:2$ and the area of the triangle is $32\sqrt{m}$ $c{m}^2$. The value of $m$ is

Answer 1

The perimeter of an isosceles $△=32$cm .

Given that the ratio of the equal side to its base is $3:2$.

So the ratio of all three sides of the isosceles $△=3:3:2$

Sum of the ratios = $3+3+2=8$

Length of equal side = $\frac{3}{8}\times 32=3\times 4=12$

Length of the base = $\frac{2}{8}\times 32=2\times 4=8$

According to Heron's formula, Area of an isosceles $△$ with equal sides $a$ & base $b$

= $\frac{1}{4}\times b\times \sqrt{(4a^2-b^2)}$

$\therefore$ Area of the given isosceles triangle, where $a=12$ & $b=8$

= $\frac{1}{4}\times 8\times \sqrt{(4\times {12}^2)-8^2}$

=$2\times \sqrt{(4\times 144)-64}$ = $2\times \sqrt{576-64}$

= $2\times \sqrt{512}=2\times \sqrt{2\times 16\times 16}$

=$2\times 16\times \sqrt{2}$ = $32\sqrt{2}$

$\therefore m=2$