Question

Each of equal sides of an isosceles triangle is 4 cm greater than its height. If the base of the triangle is 24 cm, calculate the perimeter and the area of the triangle.

Answer 1

Given that,

Equal sides of an isosceles triangle are $4cm$ greater than its height.

The base of the triangle is $24cm$.

To find out,

The area and perimeter of the triangle.

Let the height of the triangle be $hcm$

Hence, equal sides will be $(h+4)cm$

We know that, area of a triangle $=\frac{1}{2}\times base\times height$

Hence, area of the given triangle $=\frac{1}{2}\times 24\times h$

$=12hc{m}^2................\left(1\right)$

We also know that, according to Heron's formula, area of a triangle is $\sqrt{s(s-a)(s-b)(s-c)}$

where, $a,b,c$ are the three sides and $s$ is the semi-perimeter of the triangle.

Thus, $s=\frac{a+b+c}{2}$

Here, semi-perimeter $s=\frac{h+4+h+4+24}{2}$

$=\frac{2h+32}{2}$

$=(h+16)cm$

And area $=\sqrt{(h+16)(h+16-24)(h+16-(h+4\left)\right)(h+16-(h+4\left)\right)}$

$=\sqrt{(h+16)(h-8)\left(12\right)\left(12\right)}$

$=12\sqrt{(h+16)(h-8)}..............\left(2\right)$

Comparing $\left(1\right)$ and $\left(2\right)$, we get:

$12h=12\sqrt{(h+16)(h-8)}$

$\Rightarrow h=\sqrt{(h+16)(h-8)}$

Squaring both sides, we get:

$h^2=(h+16)(h-8)$

$\Rightarrow h^2=h^2+8h-128$

$\Rightarrow 8h=128$

$\Rightarrow h=\frac{128}{8}$

$\Rightarrow h=16cm$

Hence, area $=12h=12\times 16$

$=192c{m}^2$

Also, perimeter $=\text{Sum of all sides}$

$=(h+4)+(h+4)+24$

$=20+20+24$

$=64cm$

Hence, the area of the given isosceles triangle is $192c{m}^2$ and its perimeter is $64cm$.