Question

Q.33. In the given figure. O is the centre of the circle with AC = 24 cm, AB = 7 cm and <BOD=90°. Find the
area of the shaded region.
​

Answer 1

Answer:

$\huge\colorbox{yellow}{Given\:-}$

AC = 24 cm

AB = 7 cm

BOD = 90°

$\huge\colorbox{yellow}{To\:Find\:-}$

The Area of shaded region

$\huge\colorbox{yellow}{Solution\:-}$

Area of shaded region = Area of circle - Area of the triangle ABC - Area of the quadrant COD

$\small\underbrace\mathrm\red{By\:Pythagoras\:Theorem}$

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

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

${BC}^{2} = 625$

$BC = 25 cm$

The diameter of circle = 25 cm

Radius = 25/2 = 12.5 cm

$then \: area \: of \: circle = {\pi \: r}^{2} = 490.625 \: {cm}^{2}$

$\small\underbrace\mathrm\red{Area\:of\:Δ\:ABC}$

$\frac{1}{2} \times b \times h$

$= 84 \: {cm}^{2}$

Area of quadrant =

$\frac{1}{4} \times \pi \: r^{2}$

$= 122.65625 \: {cm}^{2}$

Now,

Area of Shaded region

$490.625 - 84 - 122.65625 = 283.96875 \: {cm}^{2}$

$\huge\colorbox{yellow}{Thank\:You}$

Answer 2

Given

• ∠BOD = 90°

• AB = 7 cm

• AC = 24 cm

To find

• Area of shaded region

Formulae used:-

• Area of semicircle

       $\frac{1}{2}$πr²

• Area of quadrant

        $\frac{1}{4}$πr²

• Area of triangle

        $\frac{1}{2}$base×height

Solution

∠CAB = 90°(Angle subtended by diameter)

In right ΔCAB,

By pythagoras theorem,

AC² + AB² = BC²

By substituting the values we have in the equation

⤜ 24² + 7² = BC²

⤜ 576 + 49 = BC²

⤜ 625 = BC²

⤜ BC = $\sqrt{625}$

⤜ BC = 25

∴ Diameter is 25 cm.

∴Radius is 12.5 cm or $\frac{25}{2}$

Area of shaded region= Area of semicircle + Area of quadrant-Area ofΔACB

⤜ $\frac{1}{2}$πr² + $\frac{1}{4}$πr² - $\frac{1}{2}$ ×7×24

⤜ $\frac{3}{4}$πr²- $\frac{1}{2}$×7×24

⤜ $\frac{3}{4}$ × $\frac{22}{7}$ × $\frac{625}{4}$ - 7 × 12

⤜ 368.3035 - 84

⤜ 284.3035

⤜ 284.3 cm²

∴ The area of the shaded region is 284.3035 cm².