In the diagram above, QR is in the diameter of the semicircle QR. Find the area of the figure to the nearest whole number. [\(\pi = \frac{22}{7}\)]

  • A 89 cm\(^2\)
  • B 70 cm\(^2\)
  • C 90 cm\(^2\)
  • D 80 cm\(^2\)
jamb 2006mathematics

The correct answer is A. 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\(^{2}\)

Previous question Next question