NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 11: Area Related To Circles - Exercise 11.4Question 16
Class 10 - Mathematics - CHAPTER 11: Area Related To Circles - Exercise 11.4
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 \(49\pi\,\text{cm}^2\).