The diagram is a circle with centre O Find the ... - JAMB Mathematics 1992 Question
A
9(\(\pi - 2) cm^2\)
B
18\(\pi cm^2\)
C
9\(\pi cm^2\)
D
36\(\pi cm^2\)
correct option: a
Area of the quadrant = \(\frac{1}{4} \pi r^2 = \frac{1}{4} \pi (6)^2\)
= \(\frac{36 \pi}{4} = 9 \pi \)
Area of the triangle = \(\frac{1}{2} \times 6 = 18 \times \sin 90^o = 18\)
Area of shaded portion =(9 \(\pi \) - 18)cm^2\) =
9(\(\pi - 2)cm^2\).
= \(\frac{36 \pi}{4} = 9 \pi \)
Area of the triangle = \(\frac{1}{2} \times 6 = 18 \times \sin 90^o = 18\)
Area of shaded portion =(9 \(\pi \) - 18)cm^2\) =
9(\(\pi - 2)cm^2\).
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