NCERT Class X Chapter 11: Area Related To Circles Exercise 11.1 Question 4
NCERT Class X Chapter 11: Areas Related To Circles
Question:
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment
(ii) major sector.
(Use π = 3.14)
Given:
Radius of circle, \( r = 10 \) cm
Angle subtended by chord at centre, \( \theta = 90^\circ \)
\( \pi = 3.14 \)
To Find:
(i) Area of the minor segment
(ii) Area of the major sector
Formula:
Area of sector of angle \( \theta \) = \( \displaystyle \frac{\theta}{360^\circ} \pi r^2 \)
Area of triangle (when two sides and included angle known) = \( \displaystyle \frac{1}{2} ab \sin \theta \)
Area of segment = Area of sector − Area of triangle
Solution:
Step 1: Find the area of the sector with angle \( 90^\circ \).
$$ \text{Area of sector} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.14 \times (10)^2 = \frac{1}{4} \times 3.14 \times 100 = 78.5 \ \text{cm}^2 $$Step 2: Find the area of the triangle formed by the two radii and the chord.
$$ \text{Area of triangle} = \frac{1}{2} \times 10 \times 10 \times \sin 90^\circ = \frac{1}{2} \times 10 \times 10 \times 1 = 50 \ \text{cm}^2 $$Step 3: Calculate the area of the minor segment.
$$ \text{Area of minor segment} = \text{Area of sector} - \text{Area of triangle} = 78.5 - 50 = 28.5 \ \text{cm}^2 $$Step 4: Find the area of the major sector (angle \( 360^\circ - 90^\circ = 270^\circ \)).
$$ \text{Area of major sector} = \frac{270^\circ}{360^\circ} \times \pi r^2 = \frac{3}{4} \times 3.14 \times 100 = 235.5 \ \text{cm}^2 $$Result:
(i) Area of minor segment = \( 28.5 \) cm2
(ii) Area of major sector = \( 235.5 \) cm2
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