Chapter 11 · Question 5

Find the area of the segment $AYB$ of a circle with radius $21\text{ cm}$ if $\angle AOB = 120^\circ$ . (Use $\pi = \frac{22}{7}$ )

Answer Revealed

Direct Answer:

Sector area:

\text{Area of sector } OAYB = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{3} \times \frac{22}{7} \times 441 = 462\text{ cm}^2

. Triangle area: Draw

OM \perp AB

. Since

OA = OB

M

bisects

AB

and

\angle AOM = \angle BOM = 60^\circ

. In right

\triangle OMA

\cos 60^\circ = \frac{OM}{21}

, so

OM = \frac{21}{2}\text{ cm}

\sin 60^\circ = \frac{AM}{21}

, so

AM = \frac{21\sqrt{3}}{2}\text{ cm}

and

AB = 21\sqrt{3}\text{ cm}

. Area of

\triangle OAB = \frac{1}{2} \times 21\sqrt{3} \times \frac{21}{2} = \frac{441\sqrt{3}}{4}\text{ cm}^2

. Segment area:

\text{Area of segment } AYB = 462 - \frac{441\sqrt{3}}{4} = \frac{21}{4}(88 - 21\sqrt{3})\text{ cm}^2

Simple Explanation

The sector (pizza slice for

120^\circ

) area is

462\text{ cm}^2

. The triangle inside has area

\frac{441\sqrt{3}}{4} \approx 191\text{ cm}^2

. The segment (the curved cap) area is the difference: approximately

271\text{ cm}^2

Exam-Ready Structure

Given: Radius

r = 21\text{ cm}

\angle AOB = \theta = 120^\circ

\pi = \frac{22}{7}

. Step 1 — Area of sector $OAYB$ :

\text{Area} = \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{1}{3} \times \frac{22}{7} \times 441 = \frac{1}{3} \times 22 \times 63 = 22 \times 21 = 462\text{ cm}^2

. Step 2 — Area of $\triangle OAB$ : Since

OA = OB

\triangle OAB

is isosceles. Drop perpendicular

OM

AB

. By RHS congruence of

\triangle OMA

and

\triangle OMB

M

is the midpoint of

AB

and

\angle AOM = \angle BOM = 60^\circ

. In right

\triangle OMA

\cos 60^\circ = \frac{OM}{21} \implies OM = 21 \times \frac{1}{2} = \frac{21}{2}\text{ cm}

\sin 60^\circ = \frac{AM}{21} \implies AM = 21 \times \frac{\sqrt{3}}{2} = \frac{21\sqrt{3}}{2}\text{ cm}

. So

AB = 2 \times AM = 21\sqrt{3}\text{ cm}

. Area of

\triangle OAB = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 21\sqrt{3} \times \frac{21}{2} = \frac{441\sqrt{3}}{4}\text{ cm}^2

. Step 3 — Area of segment $AYB$ : Area of segment = Area of sector − Area of triangle =

462 - \frac{441\sqrt{3}}{4} = \frac{21}{4}(88 - 21\sqrt{3})\text{ cm}^2

Key Points

Sector area for 120°: (1/3) × (22/7) × 441 = 462 cm²
Split isosceles triangle OAB into two right triangles with perpendicular OM
OM = 21/2 cm, AB = 21√3 cm
Triangle area = (1/2) × 21√3 × (21/2) = (441√3)/4 cm²
Segment area = 462 − (441√3)/4 = (21/4)(88 − 21√3) cm²

Find the area of the segment $AYB$ of a circle with radius $21\text{ cm}$ if $\angle AOB = 120^\circ$ . (Use $\pi = \frac{22}{7}$ )