Question

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the
area of the corresponding : ( 1 ) minor segment ( 2 ) major sector. (Use a = 3.14)​

Answer 1

Answer:

Step-by-step explanation:

Ar of minor segment = ar of sector - area of triangle

Ar of major sector = ar of circle (360-theta÷360)

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Answer 2

Answer:

Step by step explanation:-

$area \: of \: segment = r^{2} \: (\pi \times \alpha \div 360 \: - \sin( \alpha \div 2 ) \times \cos( \alpha \div 2) ) \\ = \: 100(\pi \times 90 \div 360 \: - \sin(45) \times \cos(45) \\ = 100(22 \div 7 \div 4 \: - 1 \div \sqrt{2} \times 1 \div \sqrt{2} ) \\ = 100(88 \div 7 - 1 \div 4) \\ = 100((352 - 7) \div 28) \\ = 100(345 \div 28) \\ (25 \times 345) \div 7 \\ =( 8625 \div 7)\: cm ^{2}$

$ar \: of \: sector = \alpha \div 360 \times \pi \: r^{2} \\ = 270 \div 360 \times 22 \div 7 \times 10 \times 10 \\ = 3 \div 4 \times 22 \div 7 \times 10 \times 10 \\ (11 \times 3 \times 5 \times 10) \div 7 \\ 150 \times \: 11 \div 7 \\ = (1650 \div 7)cm^{2}$

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