Question

10 the cross-section of a piece of conduit is an annulus. it has an outer diameter of 60 mm and an inner diameter of 50 mm.
a find the area of the cross-section to the nearest square millimetre.
b what is the increase in cross-sectional area if the outer diameter is increased to 65 mm? answer correct to the nearest square millimetre.

11 an annulus with an inner diameter of 4.2 metres and an outer diameter of 10 metres is divided into two equal parts. what is the area of each part? answer correct to three significant figures.

Explanation:

Calculate initial cross-sectional area

Using the Annulus Area knowledge point.
The outer diameter is \(D_1 = 60\text{ mm}\), so the outer radius is \(R_1 = 30\text{ mm}\).
The inner diameter is \(d = 50\text{ mm}\), so the inner radius is \(r = 25\text{ mm}\).
The area of the annulus is:

$$ A_1 = \pi (R_1^2 - r^2) = \pi (30^2 - 25^2) = \pi (900 - 625) = 275\pi \approx 863.938\text{ mm}^2 $$

Rounded to the nearest square millimetre, the area is \(864\text{ mm}^2\).

Calculate increased cross-sectional area

Using the Annulus Area knowledge point.
The new outer diameter is \(D_2 = 65\text{ mm}\), so the new outer radius is \(R_2 = 32.5\text{ mm}\).
The inner radius remains \(r = 25\text{ mm}\).
The new area of the annulus is:

$$ A_2 = \pi (R_2^2 - r^2) = \pi (32.5^2 - 25^2) = \pi (1056.25 - 625) = 431.25\pi \approx 1354.757\text{ mm}^2 $$

Find the increase in area

Using the Annulus Area knowledge point.
The increase in cross-sectional area is:

$$ \Delta A = A_2 - A_1 = 431.25\pi - 275\pi = 156.25\pi \approx 490.874\text{ mm}^2 $$

Rounded to the nearest square millimetre, the increase is \(491\text{ mm}^2\).

Calculate area of half-annulus

Using the Annulus Area knowledge point.
The outer diameter is \(D = 10\text{ m}\), so the outer radius is \(R = 5\text{ m}\).
The inner diameter is \(d = 4.2\text{ m}\), so the inner radius is \(r = 2.1\text{ m}\).
The total area of the annulus is:

$$ A_{\text{total}} = \pi (R^2 - r^2) = \pi (5^2 - 2.1^2) = \pi (25 - 4.41) = 20.59\pi \approx 64.685\text{ m}^2 $$

Since it is divided into two equal parts, the area of each part is:

$$ A_{\text{part}} = \frac{A_{\text{total}}}{2} = \frac{20.59\pi}{2} \approx 32.342\text{ m}^2 $$

Rounded to three significant figures, the area is \(32.3\text{ m}^2\).

Answer:

Question 10

a \(864\text{ mm}^2\)
b \(491\text{ mm}^2\)

Question 11

\(32.3\text{ m}^2\)