Question

Area of square that can be inscribed in a circle of radius 8 cm

Answer 1

Step-by-step explanation:

Given:

Radius = 8 cm

Step-by-step explanation:

Please find the figure containing a circle of radius 8cm.

ABCD is a square inscribed in the circle.

(OA = OB = OC = OD = 8)

ABC is a right angled triangle, as OA = 8, OB = 8

AB = 8 + 8 = 16

According to Pythagoras theorem,

Square of hypotenuse = Sum of squares of other two sides.

A C^{2}=A B^{2}+B C^{2}AC

2

=AB

2

+BC

2

As ABCD is a square all the sides are equal, AB = BC

A C^{2}=2 A B^{2}AC

2

=2AB

2

A C=\sqrt{2} A BAC=

2

AB

16=\sqrt{2} A B16=

2

AB

8 \times 2=\sqrt{2} A B8×2=

2

AB

A B=8 \sqrt{2}AB=8^2

Therefore, side of the square is 8 \sqrt{2}8^2

text{Area of square} = a^{2}Area of square=a^2

\text{Area of a square} = (8 \sqrt{2})^{2}=128 \ \mathrm{cm}^{2}Area of a square=(8^2)2

=128 cm^2

Answer 2

$\bf\huge\underline{Question}$

Area of square that can be inscribed in a circle of radius 8 cm

$\bf\huge\underline{Answer}$

Given, radius of circle, r = OC = 8 cm.

Therefore, Diameter of the circle

=> AC = 2 × OC

=> 2 × 8 = 16 cm,

which is equal to the diagonal of a square.

Let a side of square be x.

In right angled ∆ABC,

AC = AB² + BC²

[by Pythagoras theorem]

=> (16)² = x² + x²

=> 256 = 2x²

=> x² = $\dfrac{256}{2}$ = 128

Therefore, Area of the square

=> x² =128cm²