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Areas Related to Circles
NCERT Exemplar Ch 10
CBSE_NCERT_EXEMPLAR_CH10
Grade 10
Question:
Find the area of the segment $AYB$ of a circle of radius $21\text{ cm}$ if central angle is $120^\circ$.
Step-by-Step Solution
Key Concept: Segment area $= \text{Sector area} - \text{Area of } \Delta OAB$. Sector area $= \dfrac{120}{360} \pi (21)^2$. $\text{Area}(\Delta OAB) = \dfrac{1}{2} r^2 \sin 120^\circ$.
Stepwise Solution:
Sector area $= \dfrac{1}{3} \times \dfrac{22}{7} \times 21 \times 21 = 462\text{ cm}^2$. [1.0 Mark]
$\text{Area}(\Delta OAB) = \dfrac{1}{2} (21)^2 \sin 120^\circ = \dfrac{441}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{441\sqrt{3}}{4}\text{ cm}^2$.
Segment area $= \left(462 - \dfrac{441\sqrt{3}}{4}\right)\text{ cm}^2$. [1.0 Mark]
Marking Scheme:
• Finding sector area $= 462\text{ cm}^2$: 1.0 Mark
• Finding triangle area and segment area $= 462 - \dfrac{441\sqrt{3}}{4}\text{ cm}^2$: 1.0 Mark
Sector area $= \dfrac{1}{3} \times \dfrac{22}{7} \times 21 \times 21 = 462\text{ cm}^2$. [1.0 Mark]
$\text{Area}(\Delta OAB) = \dfrac{1}{2} (21)^2 \sin 120^\circ = \dfrac{441}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{441\sqrt{3}}{4}\text{ cm}^2$.
Segment area $= \left(462 - \dfrac{441\sqrt{3}}{4}\right)\text{ cm}^2$. [1.0 Mark]
Marking Scheme:
• Finding sector area $= 462\text{ cm}^2$: 1.0 Mark
• Finding triangle area and segment area $= 462 - \dfrac{441\sqrt{3}}{4}\text{ cm}^2$: 1.0 Mark
Correct Answer:
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