Question

15 find the angle at the centre of these sectors to the nearest degree.
a area of \\(104.72\text{ cm}^2\\) and a radius of \\(10\text{ cm}\\)
b area of \\(139.63\text{ m}^2\\) and a radius of \\(40\text{ m}\\)

16 an area to be landscaped is a sector with a radius of \\(19\text{ metres}\\) and an angle at the centre of \\(145^\circ\\). the area is to be covered with turf at \\(\\$12.50\\) per square metre and then top-dressed with soil at \\(\\$2.30\\) per square metre. find the total cost of the landscaping to the nearest dollar.

Explanation:

Set up the sector area formula

Using the Sector Area knowledge point, the area \(A\) of a sector with radius \(r\) and central angle \(\theta\) in degrees is given by:

$$A = \frac{\theta}{360^\circ} \times \pi r^2$$

Rearranging this formula to solve for the central angle \(\theta\):

$$\theta = \frac{360^\circ \times A}{\pi r^2}$$

Solve Question 15a

Using the Sector Area knowledge point with \(A = 104.72\text{ cm}^2\) and \(r = 10\text{ cm}\):

$$\theta = \frac{360^\circ \times 104.72}{\pi \times 10^2}$$

$$\theta = \frac{37700^\circ}{\pi \times 100} \approx 120.001^\circ$$

Rounding to the nearest degree gives \(120^\circ\).

Solve Question 15b

Using the Sector Area knowledge point with \(A = 139.63\text{ m}^2\) and \(r = 40\text{ m}\):

$$\theta = \frac{360^\circ \times 139.63}{\pi \times 40^2}$$

$$\theta = \frac{50266.8^\circ}{\pi \times 1600} \approx 10.0001^\circ$$

Rounding to the nearest degree gives \(10^\circ\).

Calculate the area for Question 16

Using the Sector Area knowledge point with \(r = 19\text{ m}\) and \(\theta = 145^\circ\):

$$A = \frac{145^\circ}{360^\circ} \times \pi \times 19^2$$

$$A = \frac{145}{360} \times \pi \times 361 \approx 456.787\text{ m}^2$$

Calculate the total cost for Question 16

The total cost per square metre is the sum of the turf cost and the top-dressing cost:

$$\text{Cost per m}^2 = \$12.50 + \$2.30 = \$14.80$$

The total cost is:

$$\text{Total Cost} = 456.787\text{ m}^2 \times \$14.80/\text{ m}^2 \approx \$6760.45$$

Rounding to the nearest dollar gives \(\$6760\).

Answer:

Question 15a

\(120^\circ\)

Question 15b

\(10^\circ\)

Question 16

\(\$6760\)