NCERT Class X Chapter 11: Area Related To Circles Exercise 11.1 Question 6
NCERT Class X Chapter 11: Area Related To Circles
Question:
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use \( \pi = 3.14 \) and \( \sqrt{3} = 1.73 \))
Given:
Radius of the circle, \( r = 15\,\text{cm} \)
Angle subtended by the chord at the centre, \( \theta = 60^\circ \)
\( \pi = 3.14 \)
\( \sqrt{3} = 1.73 \)
To Find:
Areas of the minor and major segments of the circle.
Formula:
Area of sector: \( \displaystyle \text{Area}_{\text{sector}} = \frac{\theta}{360^\circ}\pi r^2 \)
Area of triangle (using sine rule): \( \displaystyle \text{Area}_{\triangle} = \frac{1}{2} r^2 \sin\theta \)
Area of segment: \( \displaystyle \text{Area}_{\text{segment}} = \text{Area}_{\text{sector}} - \text{Area}_{\triangle} \)
Solution:
Step 1: Find the area of the sector with angle \( 60^\circ \).
$$ \text{Area}_{\text{sector}} = \frac{60^\circ}{360^\circ} \times \pi \times (15)^2 = \frac{1}{6} \times 3.14 \times 225 $$ $$ = 0.5233 \times 225 = 117.75\,\text{cm}^2 $$Step 2: Find the area of the triangle formed by the two radii and the chord.
$$ \text{Area}_{\triangle} = \frac{1}{2} r^2 \sin\theta = \frac{1}{2} \times (15)^2 \times \sin 60^\circ $$ $$ = \frac{1}{2} \times 225 \times \frac{\sqrt{3}}{2} $$ $$ = \frac{225 \sqrt{3}}{4} $$ $$ = \frac{225 \times 1.73}{4} = \frac{389.25}{4} = 97.3125\,\text{cm}^2 $$Step 3: Find the area of the minor segment.
$$ \text{Area}_{\text{minor segment}} = \text{Area}_{\text{sector}} - \text{Area}_{\triangle} $$ $$ = 117.75 - 97.3125 = 20.4375\,\text{cm}^2 $$ $$ \approx 20.44\,\text{cm}^2 $$Step 4: Find the area of the circle.
$$ \text{Area}_{\text{circle}} = \pi r^2 = 3.14 \times (15)^2 = 3.14 \times 225 = 706.5\,\text{cm}^2 $$Step 5: Find the area of the major segment.
$$ \text{Area}_{\text{major segment}} = \text{Area}_{\text{circle}} - \text{Area}_{\text{minor segment}} $$ $$ = 706.5 - 20.44 = 686.06\,\text{cm}^2 $$Result:
Area of minor segment \( \approx 20.44\,\text{cm}^2 \)
Area of major segment \( \approx 686.06\,\text{cm}^2 \)
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