A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segme

Question

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)

Accepted Answer

Given:
Radius ($r$) = 10 cm
Central angle ($	heta$) = 90° 
Step 1: Find the area of the corresponding sector.
Area of sector = $\frac{	heta}{360^{\circ}} 	imes \pi r^2$
Area of sector = $\frac{90^{\circ}}{360^{\circ}} 	imes 3.14 	imes (10)^2$
Area of sector = $\frac{1}{4} 	imes 3.14 	imes 100$
Area of sector = $0.25 	imes 314 = 78.5 	ext{ cm}^2$ 
Step 2: Find the area of the triangle formed by the two radii and the chord.
Since the angle is 90°, the triangle is a right-angled isosceles triangle with base and height equal to the radius ($r$).
Area of triangle = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$
Area of triangle = $\frac{1}{2} 	imes 10 	imes 10 = 50 	ext{ cm}^2$ 
Step 3: Calculate the area of the minor segment.
Area of minor segment = Area of sector - Area of triangle
Area of minor segment = $78.5 - 50 = 28.5 	ext{ cm}^2$ 
Therefore, the area of the corresponding minor segment is $28.5 	ext{ cm}^2$.

Areas Related to Circles

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