14. Find area of minor segment made by a chord which subtends right-angle at the centre of a circle of radius 10cm
Answers
Answer:
62 is anwer
Step-by-step explanation:
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tion:
We use the formula for the area of a sector of the circle to solve the problem.
In a circle with radius r and the angle at the centre with degree measure θ,
(i) Area of the sector = θ/360° × πr2
(ii) Area of the segment = Area of the sector - Area of the corresponding triangle
Area of the right triangle = 1/2 × base × height
Let's draw a figure to visualize the area to be calculated.
A chord of a circle of radius 10 cm
Here, Radius, r = 10 cm, θ = 90°
Visually it’s clear from the figure that,
AB is the chord that subtends a right angle at the centre.
(i) Area of minor segment APB = Area of sector OAPB - Area of right triangle AOB
(ii) Area of major segment AQB = πr² - Area of minor segment APB
Area of the right triangle ΔAOB = 1/2 × OA × OB
(i) Area of minor segment APB = Area of sector OAPB - Area of right ΔAOB
= θ/360° × πr2 - 1/2 × OA × OB
= 90°/360° × πr2 - 1/2 × r × r
= 1/4 πr2 -1/2r2
= r2 (1/4 π - 1/2)
= r2 (3.14 - 2)/4
= (r2 × 1.14)/4
= (10 × 10 × 1.14)/4 cm² (Since radius r is given as 10 cm)
= 28.5 cm²
(ii) Area of major sector AOBQ = πr2 - Area of minor sector OAPB
= πr2 - θ/360° × πr2
= πr2 (1 - 90°/360°)
= 3.14 × (10 cm)2 × 3/4
= 235.5 cm2