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A rectangle ABCD with diagonal $14$ cm is inscribed in a circle with centre O as shown in the given figure. If the area of the shaded portion is expressed as $a + b\sqrt{3}$, find the values of $a$ and $b$. Also, find the perimeter of the sector OABO.
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Radius $r = \frac{14}{2} = 7$ cm (1/2 Mark)
nArea of shaded region = $2 \times (\frac{60}{360} \times \frac{22}{7} \times 7 \times 7 - \frac{\sqrt{3}}{4} \times 7 \times 7)$ (1 Mark)
n$= 2 \times (\frac{1}{6} \times 154 - \frac{49\sqrt{3}}{4}) = \frac{154}{3} - \frac{49\sqrt{3}}{2}$ cm$^2$ (1 Mark)
n$a + b\sqrt{3} = \frac{154}{3} - \frac{49\sqrt{3}}{2}$
n$a = \frac{154}{3}$ and $b = -\frac{49}{2}$ (1 Mark)
nPerimeter of the sector OABO = $7 + 7 + \frac{120}{360} \times 2 \times \frac{22}{7} \times 7$ (1 Mark)
n$= 14 + \frac{1}{3} \times 44 = 14 + \frac{44}{3} = \frac{42+44}{3} = \frac{86}{3}$ cm or $28.67$ cm (1/2 Mark)
nArea of shaded region = $2 \times (\frac{60}{360} \times \frac{22}{7} \times 7 \times 7 - \frac{\sqrt{3}}{4} \times 7 \times 7)$ (1 Mark)
n$= 2 \times (\frac{1}{6} \times 154 - \frac{49\sqrt{3}}{4}) = \frac{154}{3} - \frac{49\sqrt{3}}{2}$ cm$^2$ (1 Mark)
n$a + b\sqrt{3} = \frac{154}{3} - \frac{49\sqrt{3}}{2}$
n$a = \frac{154}{3}$ and $b = -\frac{49}{2}$ (1 Mark)
nPerimeter of the sector OABO = $7 + 7 + \frac{120}{360} \times 2 \times \frac{22}{7} \times 7$ (1 Mark)
n$= 14 + \frac{1}{3} \times 44 = 14 + \frac{44}{3} = \frac{42+44}{3} = \frac{86}{3}$ cm or $28.67$ cm (1/2 Mark)