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)

Answer 1

Let AB be the chord of the circle subtending 90° angle at centre O of the circle.

(ii) Area of major sector OADB = \((\frac{360^{\degree} - 90^{\degree}}{360^{\degree}}) \times \pi r^2\) = \((\frac{270 ^{\degree}}{360^{\degree}})\)\(\pi r^2\)

= \(\frac{3}{4} \times  3.14 \times 10 \times 10\)
= \(235.5 \, cm^2\)

Area of minor sector OACB = \(\frac{90^{\degree}}{360 ^{\degree}} \times \pi r^2\)

 = \(\frac {1}4 \times 3.14 \times 10 \times 10\)
= \(78.5\, cm^2\)

Area of ΔOAB = \(\frac{1} 2\times  OA \times OB =  \frac{1} 2\times  10 \times 10 = 50 \,cm^2\)

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(i) Area of minor segment ACB = Area of minor sector OACB - Area of ΔOAB
                                              = 78.5 - 50 = 28.5 \(cm^2\)