(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 < PM...
Question 1 Report
(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 \(\pi = \frac{22}{7}\)].
Answer Details
(a) Solve \((x-2)(x-3)=12\).
Expand the left side:
\[x^2-5x+6=12\]Bring every term to one side:
\[x^2-5x-6=0\]Factorise. Two numbers whose product is \(-6\) and whose sum is \(-5\) are \(-6\) and \(+1\):
\[(x-6)(x+1)=0\]Hence \(x-6=0\) or \(x+1=0\), giving:
\[x=6 \quad\text{or}\quad x=-1\](b) Area of the shaded (overlap) portion.
From the diagram the two circles have centre \(M\) and \(N\), each of radius \(r=7\text{ cm}\), and \(\angle PMQ=\angle PNQ=60^{\circ}\). The shaded lens is the region common to both circles. It is made up of two equal circular segments cut off by the common chord \(PQ\): one from circle \(M\) and one from circle \(N\).
Area of one segment = area of sector \(-\) area of triangle.
Area of sector \(PMQ\):
\[\frac{60}{360}\times\frac{22}{7}\times 7^2=\frac{1}{6}\times\frac{22}{7}\times 49=\frac{154}{6}=25.667\text{ cm}^2\]Area of triangle \(PMQ\) (two radii with included angle \(60^{\circ}\)):
\[\frac{1}{2}\times 7\times 7\times\sin 60^{\circ}=24.5\times 0.8660=21.217\text{ cm}^2\]Area of one segment:
\[25.667-21.217=4.450\text{ cm}^2\]The shaded lens is two such segments (one per circle):
\[2\times 4.450=8.90\text{ cm}^2\]Correct to the nearest whole number, the area of the shaded portion is \(9\text{ cm}^2\).
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