AI Mentor

Check Your IQ

Free Expert Demo

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

Calculate the area of a segment of a circle with a central angle of $120 °$ and a radius of $21$ unit. Express answer to nearest integer.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

\frac{21}{4} (88 - 21 \sqrt{3)} {unit}^{2}

\frac{21}{4} (88 - 25 \sqrt{3)} {unit}^{2}

\frac{21}{4} (87 - 21 \sqrt{3)} {unit}^{2}

\frac{21}{4} (80 - 21 \sqrt{3)} {unit}^{2}

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

It is given a segment of a circle with a central angle of

120 °

and radius of 21 units. Then,
Question Image

From figure,
Radius of Circle

(OA = OB)

= 21 units

and

∠ AOB = 120^{0}

.
Area of Sector =

\frac{θ}{360^{0}} \times π r^{2}

Then,
Area sector OAB

= \frac{120^{0}}{360^{0}} \times \frac{22}{7} \times 21 \times 21

= 462

sq. units
Now, draw

OP ⊥ AB

, which results in two right angled triangles.
Question Image

Then,

∠ AOP = ∠ BOP = 60 °

.
And

∠ APO = 90 °

. Then,

∠ OAP = ∠ OBP = 30 °

. We know that,

\sin θ = \frac{Perpendicular}{Hypotenuse}

Then, in

ΔOPA

\sin 30^{0} = \frac{OP}{OA} = \frac{OP}{21}

\Rightarrow \frac{1}{2} = \frac{OP}{21}

\{\sin 30^{0} = \frac{1}{2}\}

\Rightarrow OP = \frac{21}{2} unit

And,

\sin 60^{0} = \frac{AP}{OA} = \frac{AP}{21}

\Rightarrow \frac{\sqrt{3}}{2} = \frac{AP}{21}

\{\sin 60^{0} = \frac{\sqrt{3}}{2}\}

\Rightarrow AP = \frac{21 \sqrt{3}}{2} unit

Since,
Area of triangle

= \frac{1}{2} \times Base \times Height

.
So,

ar (ΔOPA) = \frac{1}{2} \times AP \times OP

\Rightarrow ar (ΔOPA) = \frac{1}{2} \times \frac{21 \sqrt{3}}{2} \times \frac{21}{2}

\Rightarrow ar (ΔOPA) = \frac{441}{8} \sqrt{3}

sq. units
And,

ar (ΔOAB) = 2 \times ar (ΔOPA)

\Rightarrow ar (ΔOAB) = 2 \times \frac{441}{8} \sqrt{3}

\Rightarrow ar (ΔOAB) = \frac{441}{4} \sqrt{3}

From figure,
Area of shaded part = Area of sector OAB -

ar (ΔOAB)

Area = 462 - \frac{441}{4} \sqrt{3}

= \frac{21}{4} (88 - 21 \sqrt{3)}

sq. units
Therefore, area of a segment of a circle is

\frac{21}{4} (88 - 21 \sqrt{3)}

sq. units.
Hence, option (1) is correct.

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!