Question

A circle has diameter of 11cm

A square has side length of 7cm
Use pythagoras’ Theorem to show that the square will fit inside the circle without ruching the edge of the circle.

Any suggestions would be greatly appreciated ThankYou!

Answer 1

The square will not crunch the edge of the circle,, Using Pythagoras theorem we found the diagonal of the square (10) to be less that the diameter of the circle (11)

How to show the square will fit into the circle

The largest dimension of the square is the diagonal. Once the diagonal is calculated and it is less than the diameter of the circle, that means the square can fit inside the circle without crunching the edge of the circle.

The diagonal is solved with Pythagoras theorem

diagonal of he square = √(7² + 7²)

diagonal of he square = √(49 + 49)

diagonal of he square = 9.9 approximately 10

10 is less than 11, The square will not crunch the edge of the circle

Answer 2

Since length of diagonal (
`Diagonal= 9.9cm` ) is less than diameter of circle ( 11 cm ) , Therefore , the square will fit inside the circle without touching the edge of the circle.

Explanation:

Here we have , A circle has diameter of 11 cm A square has side length of 7 cm . Use Pythagoras’ Theorem to show that the square will fit inside the circle without touching the edge of the circle . Let's find out:

We know the concept that for any square to fit inside the circle without touching the edge of circle , diagonal of square must be less than diameter of circle . Let's find out length of diagonal by using Pythagoras Theorem :

`Hypotenuse ^2 = Perpendicular^2+Base^2`

For a square ,
`Perpendicular = base = side`

⇒
`Diagonal^2 = 2(side)^2`

⇒
`Diagonal= √(2)(side)`

⇒
`Diagonal= √(2)(7)`

⇒
`Diagonal= 9.9cm`

Since length of diagonal (
`Diagonal= 9.9cm` ) is less than diameter of circle ( 11 cm ) , Therefore , the square will fit inside the circle without ruching the edge of the circle.