Question

The vertices of a square ABCD lie on the circumference of a circle, radius 8 cm
a) calculate the area of the square.

Answer 1

$\cos {45}^{\circ }=\frac{a/2}{8}$

$\frac{1}{\sqrt{2}}=\frac{9}{16}$

$\therefore a=\frac{16}{\sqrt{2}}=8\sqrt{2}$

$\therefore$ area of square $=a^2=(8\sqrt{2}{)}^2$

$=128c{m}^2$