A chord of a circle of radius 10 cm subtends a right angle at the centre.

Question:

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use π = 3.14.]

 

Solution:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB

$=\frac{\theta}{360^{\circ}} \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$

$=\frac{90^{\circ}}{360^{\circ}} \times 3.14(10)^{2}-\frac{1}{2} \times 10 \times 10$

$=78.5-50$

$=28.5 \mathrm{~cm}^{2}$

Hence, the area of minor segment is 28.5 cm2

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