Question

This triangle and square have the same perimeter.

Show that the square has an area 50% greater than the triangle.

Show all your working out.

Answer 1

Let $a=$ the side of the rectangle, $b=$ the side of the equilateral triangle.

Given that triangle and square have the same perimeter

$$3b=4a$$

The area of the equilateral triangle

$$S_1=\dfrac{\sqrt{3}}{4}b^2$$

The area of the square

$$S_2=a^2$$

$$\dfrac{S_2}{S_1}=\dfrac{a^2}{\dfrac{\sqrt{3}}{4}b^2}=\dfrac{4\sqrt{3}}{3}(\dfrac{3}{4})^2=\dfrac{3\sqrt{3}}{4}$$

The square has an area approximately 30% greater than the equilateral triangle.