NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 11: Area Related To Circles - Exercise 11.4
Question 16

Question. 16

The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively \(120^\circ\) and \(40^\circ\). Find the areas and arc lengths of the two sectors. What do you observe?

Answer:

Areas: \(\dfrac{49\pi}{3}\,\text{cm}^2\) and \(49\pi\,\text{cm}^2\). Arc lengths: both \(\dfrac{14\pi}{3}\,\text{cm}\).

Detailed Answer with Explanation:

Step 1: Recall the formulas

  • Area of a sector = \(\dfrac{\theta}{360^\circ} \times \pi r^2\)
  • Arc length of a sector = \(\dfrac{\theta}{360^\circ} \times 2\pi r\)
  • Here, \(r\) = radius of the circle, \(\theta\) = central angle of the sector.

Step 2: For the first circle (radius = 7 cm, angle = 120°)

Area = \(\dfrac{120}{360} \times \pi \times 7^2\)

= \(\dfrac{1}{3} \times \pi \times 49\)

= \(\dfrac{49\pi}{3}\,\text{cm}^2\)

Arc length = \(\dfrac{120}{360} \times 2\pi \times 7\)

= \(\dfrac{1}{3} \times 14\pi\)

= \(\dfrac{14\pi}{3}\,\text{cm}\)

Step 3: For the second circle (radius = 21 cm, angle = 40°)

Area = \(\dfrac{40}{360} \times \pi \times 21^2\)

= \(\dfrac{1}{9} \times \pi \times 441\)

= \(49\pi\,\text{cm}^2\)

Arc length = \(\dfrac{40}{360} \times 2\pi \times 21\)

= \(\dfrac{1}{9} \times 42\pi\)

= \(\dfrac{14\pi}{3}\,\text{cm}\)

Step 4: Observation

The arc lengths are the same (\(\dfrac{14\pi}{3}\,\text{cm}\)), even though the radii and angles are different. However, the areas are different: one is \(\dfrac{49\pi}{3}\,\text{cm}^2\), the other is \(49\pi\,\text{cm}^2\).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 11: Area Related To Circles – Exercise 11.4 | Detailed Answers