Question

In the square, AO = 8. What is the perimeter of the square, rounded to the nearest whole number?

Answer 1

If OA = 8

Let each side of the square be x

If OA = 8, then AC = 8 x 2 = 16

Using pythagora's theorem

Hypotenus^2 = opposite^2 + adjacent^2

16^2 = x^2 + x^2

16^2 = 2x^2

256 = 2x^2

Divide both sides by 2

256/2 = 2x^2/2

x^2 = 128

Find the squareroot of 128

`\begin{gathered} x\text{ = }\sqrt[]{128} \\ x\text{ = 11.31} \end{gathered}`

The perimeter of a square = 4S

where S is the sides of the square

Perimeter = 4 x 11.31

Perimeter = 45.25

Approximately 45

The answer is 45