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10 days ago

A circle has diameter of 11cm

Mathematics

1 answer:

Zina [9.1K]10 days ago

5 0

Since the diagonal's length ( $Diagonal= 9.9cm$ ) is shorter than the circle's diameter (11 cm), the square can be accommodated within the circle without making contact with its boundary.

Detailed explanation:

Consider that a circle has a diameter of 11 cm and a square with a side length of 7 cm. We need to apply Pythagoras’ Theorem to demonstrate that the square will fit within the circle without touching its boundary. Let's analyze:

One knows that to allow a square to fit within a circle without touching its edge, the diagonal measurement of the square must remain smaller than the circle's diameter. To derive the diagonal's length, use Pythagoras' theorem:

$Hypotenuse ^2 = Perpendicular^2+Base^2$

For a square, $Perpendicular = base = side$

⇒ $Diagonal^2 = 2(side)^2$

⇒ $Diagonal= \sqrt{2}(side)$

⇒ $Diagonal= \sqrt{2}(7)$

⇒ $Diagonal= 9.9cm$

Since the diagonal's length ( $Diagonal= 9.9cm$ ) is shorter than the circle's diameter (11 cm), it can be confirmed that the square fits inside the circle without touching the edge of the circle.

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