In the diagram above, QR is in the diameter of the semicircle QR. Find the area of the figure to the nearest whole number. [\(\pi = \frac{22}{7}\)]
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Correct Answer: Option A
Explanation:
Area of rectangle PQRS = \(10 \times 7 = 70cm^2\)
Area of semi-circle: \(\frac{\pi r^{2}}{2}\)
r = \(\frac{7}{2} cm\)
Area of semi-circle = \(\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
= \(\frac{77}{4} = 18.75 cm^{2}\)
Area of figure = \((70 + 18.75) cm^{2}\)
= 88.75 cm\(^{2} \approxeq\) 89 cm\(^{2}\)
Area of rectangle PQRS = \(10 \times 7 = 70cm^2\)
Area of semi-circle: \(\frac{\pi r^{2}}{2}\)
r = \(\frac{7}{2} cm\)
Area of semi-circle = \(\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)
= \(\frac{77}{4} = 18.75 cm^{2}\)
Area of figure = \((70 + 18.75) cm^{2}\)
= 88.75 cm\(^{2} \approxeq\) 89 cm\(^{2}\)