In a circle of radius 21 cm, an arc subtends an angle of $60^o$ at the centre. Find
(i) the length of the arc.
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord.

AcademicMathematicsNCERTClass 10

Given:

Radius of the circle $r=21 \mathrm{~cm}$.

Angle subtended by the arc $=60^{\circ}$

To do:

We have to find

(i) the length of the arc.

(ii) area of the sector formed by the arc.

(iii) area of the segment formed by the corresponding chord.

Solution:

(i) Let the length of the arc be $l$.

We know that,

Length of arc $=2 \pi r(\frac{\theta}{360^{\circ}})$

Therefore,

Length of the arc $l=2 \times \frac{22}{7} \times 21 \times \frac{60^{\circ}}{360^{\circ}} \mathrm{cm}$

$=132 \times \frac{1}{6} \mathrm{cm}$

$=22 \mathrm{cm}$

The length of the arc is $22 \mathrm{~cm}$.

(ii) We know that,

Area of the sector $=\pi r^{2} \times \frac{\theta}{360^{\circ}}$

Therefore,

Area of the sector formed by the arc$=\frac{22}{7}(21)^{2} \times \frac{60^{\circ}}{360^{\circ}}$

$=\frac{22}{7} \times 21 \times 21 \times \frac{1}{6}$

$=231 \mathrm{~cm}^{2}$

The area of the sector is $231 \mathrm{~cm}^{2}$.  

(iii) Area of the segment formed by the corresponding chord $=$ Area of the sector $-$ Area of the triangle formed between the chord and the radius of the circle

$=231-(\frac{1}{2} r^{2} \sin \theta)$

$=231-\frac{1}{2} \times(21)^{2} \times \sin 60^{\circ}$

$=231-\frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2}$

$=231-\frac{441 \sqrt{3}}{4}$

$=231-190.95$

$=40.05 \mathrm{~cm}^{2}$

The area of the segment formed by the corresponding chord is $40.05 \mathrm{~cm}^{2}$.

raja
Updated on 10-Oct-2022 13:23:55

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