Question

Find the area of a figure formed by a square of side 8 cm and an isosceles triangle with base as one side of the square and perimeter as 18 cm.

Answer 1

Let x cm be one of the equal sides of an isosceles triangle.

Given that the perimeter of the isosceles triangle = 18 cm
Then,
x + x + 8 = 18
⇒ 2x = (18 − 8) cm = 10 cm
⇒ x = 5 cm
Area of the figure formed = Area of the square + Area of the isosceles triangle

= (Side of square)² + $\frac{1}{2}\times \mathrm{Base}\times \sqrt{{\left(\mathrm{Equal}\mathrm{side}\right)}^2-\frac{1}{4}\times {\left(\mathrm{Base}\right)}^2}$
= (8)² + $\frac{1}{2}\times 8\times \sqrt{{\left(5\right)}^2-\frac{1}{4}{\left(8\right)}^2}$
= $64+4\times \sqrt{25-16}$
= $64+4\times \sqrt{9}$
= $64+4\times 3$
= 64 + 12 = 76 cm²