Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In the given figure , AC is the diameter of a circle with radius 5cm. If AB = BC and CD = 8 cm, what is the area of the shaded region to the nearest whole number?

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

28 c m^{2}

29 c m^{2}

30 c m^{2}

45 c m^{2}

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Given that, AC is the diameter of a circle with radius

5 cm

.
And

AB = BC, CD = 8 cm

Diameter =

2 \times Radius

\Rightarrow

AC = 2(5)

\Rightarrow AC = 10 cm

Area of the circle =

π r^{2}

\Rightarrow

Area of the circle

= π (5)^{2}

= 25 π c m^{2}

.
We are aware that any angle that the diameter of a circle subtends is a right angle.
Hence,

△ ABC

is a right angled triangle with

∠ B = 90^{°}

.
Pythagoras theorem states that the square of the hypotenuse side in a right-angled triangle equals the sum of the squares of the other two sides.

∴ A C^{2} = A B^{2} + B C^{2}

[Pythagoras theorem] .

\Rightarrow 100 = 2 A B^{2}

\Rightarrow A B^{2} = 50

\Rightarrow AB = \sqrt{50}

\Rightarrow AB = 5 \sqrt{2}

∴ AB = BC = 5 \sqrt{2}

Area of the triangle =

\frac{1}{2} (Base) (Height)

\Rightarrow

Area of

△ ABC

= \frac{1}{2} \times AB \times BC

\Rightarrow

Area of

△ ABC

\frac{1}{2} \times 5 \sqrt{2} \times 5 \sqrt{2}

\Rightarrow

Area of

△ ABC

= 25 c m^{2}

Similarly,

△ ADC

is a right angled triangle with

∠ D = 90^{°}

∴ A C^{2} = A D^{2} + D C^{2}

\Rightarrow 10^{2} = A D^{2} + 8^{2}

\Rightarrow 100 = A D^{2} + 64

\Rightarrow A D^{2} = 36

\Rightarrow AD = 6

Area of

△ ADC

= \frac{1}{2} \times AD \times DC

\Rightarrow

Area of

△ ADC

\frac{1}{2} \times 6 \times 8

= 24 c m^{2}

Area of shaded region = Area of the circle - Area of

△ ABC

- Area of

△ ADC

\Rightarrow

Area of shaded region =

25 π - 49 = 29.5 c m^{2}

\Rightarrow

Area of shaded region

\sim

30 c m^{2}

So, option 3 is correct.

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th