(a) Solve : ((x – 2)(x – 3) = 12).
(b) In the diagram, M and N are the centres of two circles of equal radii 7cm. The circle intercept at P and Q. If < PMQ = < PNQ = 60°, calculate, correct to the nearest whole number, the area of the shaded portion. [Take (pi = frac{22}{7})].
Explanation
(a) ((x – 2)(x – 3) = 12)
(x^{2} – 3x – 2x + 6 = 12 implies x^{2} – 5x + 6 – 12 = 0)
(x^{2} – 5x – 6 = 0)
(x^{2} – 6x + x – 6 = 0 implies x(x – 6) + 1(x – 6) = 0)
( (x – 6)(x + 1) = 0 implies text{x = 6 or -1})
(b) The shaded portion comprises two shaded segments labelled A and B.
Area of segment A = (frac{60°}{360°} times pi r^{2} – frac{1}{2} r^{2} sin 60°)
(frac{1}{6} times frac{22}{7} times 7 times 7 – frac{1}{2} times 7 times 7 times 0.8660)
= (25.667 – 21.217)
= (4.45 cm^{2})
Also, the area of the segment B = (4.45 cm^{2}).
Hence, the area of the shaded portion = (4.45 cm^{2} + 4.45 cm^{2})
= (8.9 cm^{2})
(approxeq 9 cm^{2}) (to the nearest whole number).