Question

The perimeter of an isosceles triangle is 42 cm and its base is $\left(\frac{3}{2}\right)$ times each of the equal sides. Find the length of each side and area of the triangle.

Answer 1

Let a, b and c be the sides of the triangle. Here, a = b and $c\left(\mathrm{base}\right)=\frac{3}{2}a$.
Since the perimeter of the triangle is 42 cm, so
$$\begin{gathered} a+b+c=42 \\ \Rightarrow a+a+\frac{3}{2}a=42 \\ \Rightarrow a=12\mathrm{cm} \\ \Rightarrow c=\frac{3}{2}\times a=\frac{3}{2}\times 12=18\mathrm{cm} \end{gathered}$$
Therefore, the lengths of the sides of the triangle are 12 cm, 12 cm and 18 cm.
Now
$$\begin{gathered} s=\frac{a+b+c}{2}=\frac{12+12+18}{2}=21\mathrm{cm} \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{21\left(21-12\right)\left(21-12\right)\left(21-18\right)} \\ =\sqrt{21\left(9\right)\left(9\right)\left(3\right)}=27\sqrt{7}{\mathrm{cm}}^2 \end{gathered}$$
Hence, the area of the triangle is $27\sqrt{7}{\mathrm{cm}}^2$.