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Q2:
Find the area of a quadrant of a circle whose circumference is 22 cm. (Unless stated otherwise, use $\pi = \frac{22}{7}$)

Solution :

Given: The circumference of a circle is $C = 22 \text{ cm}$.

To find: The area of a quadrant of the circle.

r r 90°

Step 1: Finding the radius of the circle.

The formula for the circumference of a circle is given by:

$C = 2\pi r$

Substituting the given value $C = 22 \text{ cm}$ and $\pi = \frac{22}{7}$:

$22 = 2 \times \left(\frac{22}{7}\right) \times r$

To isolate $r$, multiply both sides by $7$ and divide by $(2 \times 22)$:

$r = \frac{22 \times 7}{2 \times 22}$

$r = \frac{7}{2} \text{ cm}$

$r = 3.5 \text{ cm}$

Step 2: Defining the area of a quadrant.

A quadrant is one-fourth of a circle. Therefore, the area of a quadrant ($A_q$) is given by:

$A_q = \frac{1}{4} \times \text{Area of the circle}$

$A_q = \frac{1}{4} \times \pi r^2$

Step 3: Calculating the area.

Substitute $r = \frac{7}{2}$ and $\pi = \frac{22}{7}$ into the formula:

$A_q = \frac{1}{4} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2$

$A_q = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$

Cancel the common factors ($7$ in the numerator and denominator):

$A_q = \frac{1}{4} \times \frac{22 \times 7}{2 \times 2}$

$A_q = \frac{1}{4} \times \frac{154}{4}$

$A_q = \frac{154}{16}$

Simplify the fraction by dividing both numerator and denominator by $2$:

$A_q = \frac{77}{8} \text{ cm}^2$

Converting to decimal form:

$A_q = 9.625 \text{ cm}^2$

Final Answer: The area of the quadrant is 9.625 cm² (or $\frac{77}{8}$ cm²).


More Questions from Class 10 Mathematics Areas Related to Circles EXERCISE 11.1


CBSE Solutions for Class 10 Mathematics Areas Related to Circles


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