The perimeter of a sector OAB of a circle of centre O is 291 cm. If the radius is 63 cm, calculate, correct to one decimal place, the: (a) angle subtended by the sector; (b) length of the chord AB; (c) area of the sector. [Take π = 22/7]

Here's the solution for Question 10:

Step 1: Calculate the arc length of the sector.
The perimeter of a sector is given by the formula $P = 2r + L$, where $r$ is the radius and $L$ is the arc length.
Given $P = 291$ cm and $r = 63$ cm.

$$291 = 2(63) + L$$ $$291 = 126 + L$$ $$L = 291 - 126 = 165 \text{ cm}$$

Step 2: Calculate the angle subtended by the sector in radians.
The arc length is also given by $L = r\theta$, where $\theta$ is the angle in radians.

$$165 = 63\theta$$ $$\theta = \frac{165}{63} \text{ radians}$$

Step 3: Convert the angle to degrees for part (a).
To convert radians to degrees, multiply by $\frac{180^\circ}{\pi}$. Use $\pi = \frac{22}{7}$.

$$\theta = \frac{165}{63} \times \frac{180^\circ}{\frac{22}{7}}$$ $$\theta = \frac{165}{63} \times \frac{180^\circ \times 7}{22}$$ $$\theta = \frac{165 \times 180^\circ \times 7}{63 \times 22}$$ $$\theta = \frac{165 \times 180^\circ}{9 \times 22}$$ $$\theta = \frac{15 \times 180^\circ}{9 \times 2}$$ $$\theta = \frac{15 \times 90^\circ}{9}$$ $$\theta = 15 \times 10^\circ = 150^\circ$$