Question

What is the area of square inscribed in a circle of diameter 11cm ?​

Answer 1

Question:

What is the area of square inscribed in a circle of diameter 11cm?

Solution:

Let ABCD be the square inscribed in a circle of diameter 11cm.

$\implies BD=11cm$

Let BC = CD = x.

$In\: \triangle\:BCD \:by\: Pythagoras\: Theorem, \\ BC^2+CD^=BD^2\\ {x}^{2} + {x}^{2} = {(11)}^{2} \\ 2 {x}^{2} = 121 \\ {x}^{2} = \frac{121}{2} \\ x = \sqrt{ \frac{121}{2} } \\ x = \frac{11}{ \sqrt{2} }cm$

Therefore, the side of the square is 11/√2 cm.

$Area\:of\:Square=(side)^2 \\ Area\:of\:Square = {( \frac{11}{ \sqrt{2} }) }^{2} \\ Area\:of\:Square = \frac{121}{2} \\ Area\:of\:Square = 60.5c {m}^{2}$

Hence, the area of the square inscribed is 60.5sq.cm.

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Answer 2

ANSWER:-

Given:

A circle of diameter is 11cm.

To find:

The area of square inscribed in a circle.

Explanation:

We know that area of square: [side]²

&

• The diagonal of square are diameter of the circle.
• The diagonal of square formed a right angle Δ.

Therefore,

Using pythagoras theorem:

[Hypotenuse]² = [Base]² + [Perpendicular]²

In right angle Δ two sides are equal to Hypotenuse.

• Let Base of right angle Δ= R cm
• Let perpendicular of right angle Δ= R cm
• Hypotenuse of right angle Δ, H= 11cm

Now, right angle Δ, we get:

⇒ (11)²= (R)² +(R)²

⇒ 121= 2R²

⇒ $\frac{121}{2} = R^{2}$

⇒ $\sqrt{\frac{121}{2} } =R$

⇒ $\frac{11}{\sqrt{2} } = R$

Therefore,

The side of the square is R= $\frac{11}{\sqrt{2} }cm$

Now,

Area of square= [side]²

Area of square= $(\frac{11}{\sqrt{2} } )^{2}$

Area of square= $(\frac{121}{2} )cm^{2}$

Area of square= 60.5cm².