pls help me anyone...
Answers
Answer:
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:
128 cm²
Step-by-step explanation:
Given, radius of circle, r = OC = 8cm. ∴ Diameter of the circle = AC = 2 x OC = 2 x 8= 16 cm which is equal to the diagonal of a square. Let side of square be x
