Work out the size of the angle QON.

Angle QON = 60^o (Triangle ABO is an equilateral triangle)

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Calculate the area of the sector ONQ, using the formula: $A = \frac{θ °}{360 °} \times π r^{2}$ , where $r$ is the radius and $θ$ is the angle of the sector.

$\begin{array}{rcl} A & = & \frac{60}{360} \times π \times 11^{2} \end{array}$

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$\begin{array}{rcl} A & = & \frac{121}{6} π {cm}^{2} \end{array}$

Calculate the area of the triangle ABO, using the formula: $A = \frac{1}{2} a b \sin C$ , where $a$ and $b$ are two sides of the triangle and $C$ is the angle between the two sides.

$A = \frac{1}{2} \times 7 \times 7 \times \sin 60$

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$A = \frac{49 \sqrt{3}}{4} {cm}^{2}$

Work out the area of the shaded section by subtracting the area of triangle ABO from the area of the sector QON.

$\begin{array}{rcl} A_{s h a d e d} & = & \frac{121}{6} π - \frac{49 \sqrt{3}}{4} \\ = & 42.1378 . . . {cm}^{2} \end{array}$

Divide the shaded area by the total area of the sector ONQ and multiply by 100% to find the percentage of the total shape that is shaded.

$\begin{array}{rcl} \frac{42.1378 . . .}{\frac{121}{6} π} \times 100 \end{array}$

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$\begin{array}{rcl} = & 66.510186 . . . \end{array}$

Round to 1 decimal place.

$66.5 %$ [1]

Circles, Arcs & Sectors (CIE IGCSE Maths: Extended)

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