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Areas Related to Circles
RD Sharma
CBSE
Grade 10
Question:
In an equilateral triangle $ABC$ of side $12\text{ cm}$, three circular arcs of radius $6\text{ cm}$ are drawn with vertices $A, B, C$ as centres. Find the area of the region enclosed between the three arcs. (Use $\pi = 3.14$ and $\sqrt{3} = 1.732$)
Step-by-Step Solution
Key Concept: Area of equilateral $\Delta ABC = \dfrac{\sqrt{3}}{4} \times 144 = 36\sqrt{3} = 36 \times 1.732 = 62.352\text{ cm}^2$.<br>3 corner sectors of angle $60^\circ$ form a semi-circle of radius $6\text{ cm}$: $\text{Area} = \dfrac{1}{2} \times 3.14 \times 36 = 56.52\text{ cm}^2$.<br>Enclosed Area $= 62.352 - 56.52 = 5.832\text{ cm}^2$.
Area of equilateral $\Delta ABC = \dfrac{\sqrt{3}}{4} \times 12^2 = 36\sqrt{3} = 36 \times 1.732 = 62.352\text{ cm}^2$. [2.0 Marks] Sum of 3 corner sector areas $= 3 \times \left(\dfrac{60}{360} \times 3.14 \times 36\right) = \dfrac{1}{2} \times 3.14 \times 36 = 56.52\text{ cm}^2$. [2.0 Marks] Enclosed Area $= 62.352 - 56.52 = 5.832\text{ cm}^2$. [1.0 Mark]
--- 🎯 Official CBSE Marking Scheme: Calculating triangle area $= 62.352\text{ cm}^2$: 2.0 Marks Calculating 3 sectors area $= 56.52\text{ cm}^2$: 2.0 Marks Evaluating enclosed area $= 5.832\text{ cm}^2$: 1.0 Mark
Correct Answer:
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