Question

please give answer of question no 8
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Answer 1

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Given:

Radius = 8 cm

Step-by-step explanation:

Let there be a circle with center O

ABCD is a square inscribed in the circle.

(OA = OB = OC = OD = 8) (Ref. Image at the bottom)

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.

∴ $AC^{2}  = AB^{2} + BC^{2}$

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

∴ $AC^{2}  = 2AB^{2}$

∴ $AC = \sqrt{2} AB$

∴ $16 = \sqrt{2} AB$

∴ 8 × 2 = $\sqrt{2} AB$

∴ $AB = 8\sqrt{2}$

Therefore, side of the square = $8\sqrt{2}$

Area of a square = $Side^{2}$

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

Answer 2

Given

radius of circle= 8cm

let the side of a square be= a

according to properties of a square it's all find are equal and it's all internal angle are of 90 degree.

by Pythagoras theorem

a sqrt2= 16cm

a = 16/sqrt2

Now area of square = side ^2 = 16*16/2 = 128 cm^2