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

Consider a circle with center O and radius 10 cm. Let AB be a chord subtending an angle of 90° at the center.

(ii) The area of the major sector OADB is calculated as: \((\frac{360^{\degree} - 90^{\degree}}{360^{\degree}}) \times \pi r^2\) = \((\frac{270 ^{\degree}}{360^{\degree}})\)\(\pi r^2\)

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

The area of the minor sector OACB is calculated as: \(\frac{90^{\degree}}{360 ^{\degree}} \times \pi r^2\)

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

The area of triangle OAB is calculated as: \(\frac{1} 2\times OA \times OB = \frac{1} 2\times 10 \times 10 = 50 \,cm^2\)

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(i) The area of the minor segment ACB is the difference between the area of the minor sector OACB and the area of triangle OAB.
                                            = 78.5 - 50 = 28.5 \(cm^2\)