Question

A circle has a diameter 9 cm.

Answer 1

Answer:

Step-by-step explanation:

Since we are to show by Pythagoras that  the square will fit inside the circle without touching the edge of the circle , this means that we must find the value of the length of the diagonal of the square using Pythagoras.

For the square to fit in without touching the edge of the circle , this means that the diagonal of the square must be less than the diameter of the circle.

Length of the square is 5cm.

Recall Pythagoras theorem: it states that

the square on the hypotenuse of a right-angled triangle is equal in area to the sum of the squares on the other two sides

$D^{2}$ = $5^{2}$ + $5^{2}$

$D^{2}$ = 25 + 25

$D^{2}$ = 50

Take the square root of both sides , that is

D = $\sqrt{50}$

D = $\sqrt{25}$ X $\sqrt{2}$ , since 50 = 25 x 2

D = 5$\sqrt{2}$

D ≈ 7.07

Note that the diameter of the circle given is 9 and D < 9 , we can therefore conclude that the square will fit inside the circle without touching the edge of the circle since the diagonal of the square is less than the diameter of the circle