Let's break down and solve the problems shown in the image:

Question 9:

A rectangular room was measured to be 5 m long and 3 m wide.

(a) Calculate the area of the room.

The area $A$ of a rectangle is given by: $A = \text{length} \times \text{width}$ $A = 5 \, \text{m} \times 3 \, \text{m} = 15 \, \text{m}^2$

(b) Find the lower and upper bounds of the true length and width.

Assume the measurements were taken to the nearest meter. This means the actual length and width could be ± 0.5 m from the given measurements. The bounds are:

• Length: $5 \, \text{m}$
  • Lower bound: $5 - 0.5 = 4.5 \, \text{m}$
  • Upper bound: $5 + 0.5 = 5.5 \, \text{m}$
• Width: $3 \, \text{m}$
  • Lower bound: $3 - 0.5 = 2.5 \, \text{m}$
  • Upper bound: $3 + 0.5 = 3.5 \, \text{m}$

(c) What are the lower and upper bounds of the true area?

The lower bound of the area is calculated by multiplying the lower bounds of the length and width: $\text{Lower bound} = 4.5 \times 2.5 = 11.25 \, \text{m}^2$ The upper bound of the area is calculated by multiplying the upper bounds of the length and width: $\text{Upper bound} = 5.5 \times 3.5 = 19.25 \, \text{m}^2$

(d) Find the maximum error in the answer to Part (a).

The maximum error is the difference between the area using the measured values and the maximum possible area (the upper bound): $\text{Maximum error} = 19.25 - 15 = 4.25 \, \text{m}^2$

Question 10:

The diameter of a circular pizza tray is measured to be 28.6 cm.

(a) Calculate the area of the tray.

The area $A$ of a circle is given by: $A = \frac{\pi d^2}{4}$ Substitute $d = 28.6 \, \text{cm}$: $A = \frac{\pi \times (28.6)^2}{4} = \frac{\pi \times 817.96}{4} = 643.59 \, \text{cm}^2$ Rounded to one decimal place, the area is: $A \approx 643.6 \, \text{cm}^2$

(b) What are the lower and upper bounds of the true length of the diameter?

Assuming the diameter was measured to the nearest 0.1 cm, the lower and upper bounds are:

• Diameter: $28.6 \, \text{cm}$
  • Lower bound: $28.6 - 0.05 = 28.55 \, \text{cm}$
  • Upper bound: $28.6 + 0.05 = 28.65 \, \text{cm}$

(c) Find the lower and upper bounds of the true area of the tray, correct to one decimal place.

The lower bound of the area uses the lower bound of the diameter: $A_{\text{lower}} = \frac{\pi \times (28.55)^2}{4} = \frac{\pi \times 814.90}{4} = 640.24 \, \text{cm}^2 \approx 640.2 \, \text{cm}^2$ The upper bound of the area uses the upper bound of the diameter: $A_{\text{upper}} = \frac{\pi \times (28.65)^2}{4} = \frac{\pi \times 821.05}{4} = 645.72 \, \text{cm}^2 \approx 645.7 \, \text{cm}^2$

(d) What is the maximum error in the answer to Part (a)?

The maximum error is the difference between the calculated area and the upper bound: $\text{Maximum error} = 645.7 - 643.6 = 2.1 \, \text{cm}^2$

Would you like further details or clarification on any part? Here are five related questions:

1. How do you determine bounds when the measurement precision is different (e.g., nearest cm vs mm)?
2. How does the formula for the area of a circle change if you’re given the radius instead of the diameter?
3. Why does error analysis consider both lower and upper bounds?
4. What would happen to the maximum error if the diameter of the pizza tray were measured more accurately?
5. How would the area calculation change if you were asked for three decimal places instead of one?

Tip: When measuring or calculating physical quantities, always consider the precision of your measuring instruments to estimate the possible error in your results!