Question

25. In the given fig. O is the centre of the circle with AC = 24cm , AB = 7cm
and ZBOD = 90°. Find the area of the shaded region. (Use n = 3.14]​

Answer 1

$\huge{\underline{\bf\orange{Answer :-}}}$

Given:

O is the centre of the circle with AC = 24cm , AB = 7cm

To Find:

the area of the shaded region.

Solution:

In ∆ABC ∠CAB=90° (angle in a semicircle)

So, ∆ABC is right-angled at A

$By \: Pythagoras \: theorem$

$BC {}^{2} = AC {}^{2} + AB$

$BC {}^{2} = 24 {}^{2} + 7 {}^{2}$

$BC {}^{2} = 625$

$BC = \sqrt{625}$

$BC = 25cm$

Therefore, Diameter of circle will be 25 CM.

$And \: Radius = \frac{25}{2} = 12.5cm$

$Area \: of \: Circle = \pi \: r {}^{2}$

$Area \: of \: circle = 3.14 \times 12.5 \times 12.5 = 490.625 cm {}^{2} </p><p>$

$Area \: of \: ∆ABC = \frac{1}{2} \times base \times height$

$= \frac{1}{2} \times 7 \times 24 = 84cm {}^{2}$

$Area \: of \: quadrant = \frac{1}{4} \times \pi \: r {}^{2}$

$\frac{1}{4} \times 3.14 \times 12.5 \times 12.5 = 122.65625cm {}^{2}$

Area of the shaded region = area of the circle −area of the ∆ABC−area of the quadrant COD

$</p><p>Area \: of \: the \: shaded \: region = 490.625 - 84 - 122.65625 = 283.96875cm {}^{2}$

$Hence, \: the \: are \: of \: Shaded \: region \: is \: 283.96875cm {}^{2}$