Question

The perimeter of an isosceles triangle is $32\mathrm{cm}$.

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

Find the area of the triangle.

Answer 1

Step 1. Find out the sides of the triangle.

Let $PQR$ be an isosceles triangle with a perimeter of $32\mathrm{cm}$.

Since it is given that the ratio of its equal side to its base is $3:2$.

Therefore, the sides of the triangle are $\mathrm{PQ}=\mathrm{PR}=3x,\mathrm{QR}=2x$

Since the perimeter of a triangle is the sum of all three sides.

$$\begin{gathered} \therefore 3x+3x+2x=32 \\ \Rightarrow 8x=32 \\ \Rightarrow x=\frac{32}{8} \\ \Rightarrow x=4\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{PQ}=\mathrm{PR}=3\times 4=12\mathrm{cm} \\ \mathrm{QR}=2x=2\times 4=8\mathrm{cm} \end{gathered}$$

Thus, the sides of a triangle are $a=12\mathrm{cm},b=12\mathrm{cm}\mathrm{and}c=8\mathrm{cm}.$

Step 2. Calculate the area of the triangle.

The Semi-perimeter $\left(s\right)$ of an isosceles triangle is computed as:

$$\begin{gathered} s=\frac{a+b+c}{2} \\ =\frac{12+12+8}{2} \\ =\frac{32}{2} \\ =16\mathrm{cm} \end{gathered}$$

The area of an isosceles triangle is computed by using heron's formula:

$$\begin{gathered} A=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{16(16-12)(16-12)(16-8)} \\ =\sqrt{16\times 4\times 4\times 8} \\ =\sqrt{2048} \\ =32\sqrt{2}{\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of an isosceles triangle is $32\sqrt{2}{\mathrm{cm}}^2$.