The area of a semi-circle is 5024 cm2. a) Calculate the length of the straight lined side. b) What is the circumference of the semi-circle? Give your answer to the nearest centimetre.

$$\text{Area} = \frac{1}{2} \pi r^2$$ $$5024 = \frac{1}{2} \pi r^2$$ $$10048 = \pi r^2$$ $$r^2 = \frac{10048}{\pi}$$ $$r = \sqrt{\frac{10048}{\pi}}$$ $$r \approx \sqrt{\frac{10048}{3.1415926535}}$$ $$r \approx \sqrt{3200.0000000000004}$$ $$r \approx 56.5685 \text{ cm}$$

$$\text{Diameter} = 2r$$ $$\text{Diameter} \approx 2 \times 56.5685 \text{ cm}$$ $$\text{Diameter} \approx 113.137 \text{ cm}$$

Rounding to two decimal places:

$$\text{Diameter} \approx 113.14 \text{ cm}$$

The length of the straight lined side is $\boxed{\text{113.14 cm}}$.

b) What is the circumference of the semi-circle? Give your answer to the nearest centimetre.
The circumference of a semi-circle includes the curved arc and the straight diameter.

$$\text{Circumference} = \pi r + 2r$$ $$\text{Circumference} = r(\pi + 2)$$

Using the calculated value of $r \approx 56.5685 \text{ cm}$:

$$\text{Circumference} \approx 56.5685 \times (3.1415926535 + 2)$$ $$\text{Circumference} \approx 56.5685 \times 5.1415926535$$ $$\text{Circumference} \approx 290.999 \text{ cm}$$

$$\text{Circumference} \approx 291 \text{ cm}$$

The circumference of the semi-circle is $\boxed{\text{291 cm}}$.

Question 6:

a) The area of a netball court is $465.125 \text{ m}^2$. If the breadth of the court is $15.25 \text{ m}$, what is the length?

$$465.125 \text{ m}^2 = \text{Length} \times 15.25 \text{ m}$$ $$\text{Length} = \frac{465.125 \text{ m}^2}{15.25 \text{ m}}$$ $$\text{Length} = 30.5 \text{ m}$$

The length of the court is $\boxed{\text{30.5 m}}$.

b) A netball court is divided into thirds. What will the area of a third be? Give your answer to the nearest square metre.

$$\text{Area of one third} = \frac{1}{3} \times \text{Total Area}$$ $$\text{Area of one third} = \frac{1}{3} \times 465.125 \text{ m}^2$$ $$\text{Area of one third} = 155.04166... \text{ m}^2$$

$$\text{Area of one third} \approx 155 \text{ m}^2$$

The area of a third of the court is $\boxed{\text{155 m}^2}$.