Find the area of the shaded portion of the semicircular figure.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option B
Explanation:
Asector = \(\frac{60}{360} \times \pi r^2\)
= \(\frac{1}{6} \pi r^2\)
A\(\bigtriangleup\) = \(\frac{1}{2}r^2 \sin 60^o\)
\(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\)
A\(\text{shaded portion}\) = Asector -
A\(\bigtriangleup\)
= (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\)
= \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\)
= \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)
Asector = \(\frac{60}{360} \times \pi r^2\)
= \(\frac{1}{6} \pi r^2\)
A\(\bigtriangleup\) = \(\frac{1}{2}r^2 \sin 60^o\)
\(\frac{1}{2} r^2 \times \frac{\sqrt{3}}{2} = \frac{r^2\sqrt{3}}{4}\)
A\(\text{shaded portion}\) = Asector -
A\(\bigtriangleup\)
= (\(\frac{1}{6} \pi r^2 - \frac{r^2\sqrt{3}}{4})^3\)
= \(\frac{\pi r^2}{2} - \frac{3r^2\sqrt{3}}{4}\)
= \(\frac{r^2}{4}(2 \pi - 3 \sqrt{3})\)