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Areas Related to Circles
RD Sharma
CBSE
Grade 10
Question:
Four equal circles, each of radius $a$, touch each other externally. Show that the area between them is $\left(4 - \pi\right) a^2$. Use this formula to calculate the area when $a = 7\text{ cm}$. (Take $\pi = 22/7$)
Step-by-Step Solution
Key Concept: Part 1: Centres form a square of side $2a$. Area of square $= (2a)^2 = 4a^2$. 4 corner sectors form a full circle of radius $a \Rightarrow \text{Area} = \pi a^2$. Enclosed Area $= 4a^2 - \pi a^2 = (4 - \pi)a^2$. Part 2: For $a = 7\text{ cm}$, Enclosed Area $= \left(4 - \dfrac{22}{7}\right) \times 49 = \dfrac{6}{7} \times 49 = 42\text{ cm}^2$.
Part 1: Square side $= 2a \Rightarrow \text{Square Area} = 4a^2$. [1.5 Marks] 4 corner sectors of angle $90^\circ$ form a full circle of radius $a \Rightarrow \text{Area} = \pi a^2$. [1.5 Marks] Enclosed Area $= 4a^2 - \pi a^2 = (4 - \pi)a^2$. Proved! [1.0 Mark] Part 2: For $a = 7\text{ cm}$, Enclosed Area $= \left(4 - \dfrac{22}{7}\right) \times 49 = \dfrac{6}{7} \times 49 = 42\text{ cm}^2$. [1.0 Mark]
--- 🎯 Official CBSE Marking Scheme: Finding square area $= 4a^2$: 1.5 Marks Finding 4 sectors area $= \pi a^2$: 1.5 Marks Proving enclosed area formula $(4 - \pi)a^2$: 1.0 Mark Evaluating area for $a=7\text{ cm} \Rightarrow 42\text{ cm}^2$: 1.0 Mark
Correct Answer:
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