Question

9. In Fig. 5.27, AB and CD are two diameters of a circle
(with centre O) perpendicular to each other and OD
is the diameter of the smaller circle. If
OA= 7 cm, find the area of the shaded region.​

Answer 1

Explanation:

so area of the shaded region is 66.5cm2

Answer 2

Solution:

Given:

=> AB = CD = 2 × 7 = 14 cm.

=> OD = OC = OB = OA = 7 cm.

=> Diameter AB ⊥ Diameter CD.

=> Radius = 7/2

To Find:

=> Area of shaded region.

Formula used:

$\sf{\implies Area\;of\;circle=\pi r^{2}}$

$\sf{\implies Area\;of\;semi\;circle=\dfrac{1}{2}\pi r^{2}}$

$\sf{\implies Area\;of\;triangle = \dfrac{1}{2} \times base\times height}$

So,

$\sf{\implies Area\;of\;smaller\;circle=\pi r^{2}}$

$\sf{\implies \dfrac{22}{7}\times \bigg(\dfrac{7}{2}\bigg)^{2}}$

$\sf{\implies \dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}}$

$\sf{\implies 38.5\;cm^{2}}$

$\sf{Now,\;area\;of\;semicircle = \dfrac{1}{2} \pi r^{2}}$

$\sf{\implies \dfrac{1}{2} \times \dfrac{22}{7}\times 7\times 7}$

$\sf{\implies 77\;cm^{2}}$

$\sf{\implies Area\;of\;triangle(ABC)=\dfrac{1}{2}\times Base\times Height}$

$\sf{\implies \dfrac{1}{2} \times 14\times 7}$

$\sf{\implies 49\;cm^{2}}$

Now, Area of shaded region = Area of smaller circle + Area of semicircle - Area of triangle (ABC)

$\sf{\implies 38.5\;cm^{2}+77\;cm^{2}-49\;cm^{2}}$

${\boxed{\boxed{\sf{\implies So,\;Area\;of\;shaded\;region=66.5\;cm^{2}}}}}$