(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...

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 \(9\text{ cm}^2\).