Year 9 Measurement: free practice test with answers

Sample measurement questions with answers

Try each one before you open the answer.

1. Errors and accuracy in measurement

   A student measures the length of a book as 28 cm. The true length is 27.5 cm. Find the absolute error, the relative error and the percentage error.

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   Key idea: Absolute error is how far the measurement is from the true value. Relative error compares that gap to the true value. Percentage error converts the relative error to a percentage.

   1. Find the absolute error: $\text{absolute error} = \lvert \text{measured} - \text{true} \rvert = \lvert 28 - 27.5 \rvert = 0.5 \text{ cm}$
   2. Find the relative error: $\text{relative error} = \frac{\text{absolute error}}{\text{true value}} = \frac{0.5}{27.5} = \frac{1}{55}$
   3. Convert to a percentage error: $\text{percentage error} = \frac{1}{55} \times 100 \approx 1.82\%$
   4. Answer: The absolute error is $0.5$ cm, the relative error is $\dfrac{1}{55}$, and the percentage error is approximately $1.82\%$.

2. Length and perimeter

   Convert:

   (a) $3.5$ km to metres (b) $248$ mm to centimetres (c) $65$ cm to metres

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   Key idea: $1$ km $= 1000$ m, $1$ cm $= 10$ mm and $1$ m $= 100$ cm. Multiply when changing to a smaller unit and divide when changing to a larger unit.

   1. (a) $3.5 \times 1000 = 3500$
   2. (b) $248 \div 10 = 24.8$
   3. (c) $65 \div 100 = 0.65$

   Answer: (a) $3500$ m (b) $24.8$ cm (c) $0.65$ m

3. Circle circumference and perimeter of a sector

   Find the circumference of a circle with a diameter of $14$ cm. Give an exact answer and an answer correct to $2$ decimal places.

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   Key idea: Circumference $C = \pi d$, or $C = 2\pi r$.

   1. $C = \pi \times 14 = 14\pi$
   2. $14\pi = 43.982\ldots \approx 43.98$

   Answer: $14\pi$ cm $\approx 43.98$ cm

4. Area

   Find the area of:

   (a) a triangle with base $14$ cm and perpendicular height $9$ cm

   (b) a parallelogram with base $12$ m and perpendicular height $7.5$ m

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   Key idea: Triangle: $A = \frac{1}{2}bh$. Parallelogram: $A = bh$, using the perpendicular height.

   1. (a) $\frac{1}{2} \times 14 \times 9 = 63$
   2. (b) $12 \times 7.5 = 90$

   Answer: (a) $63 \text{ cm}^2$ (b) $90 \text{ m}^2$

5. Perimeter and area of composite shapes

   The front wall of a garden shed is a rectangle $8$ m wide and $3$ m high, with a triangle on top that has a base of $8$ m and a perpendicular height of $2.5$ m. Find the total area of the front wall.

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   Key idea: Split the composite shape into simple shapes, find each area, then add.

   1. Rectangle: $8 \times 3 = 24$
   2. Triangle: $\frac{1}{2} \times 8 \times 2.5 = 10$
   3. Total: $24 + 10 = 34$

   Answer: $34 \text{ m}^2$

6. Surface area of prisms and pyramids

   Mia wants to paint the four walls and the ceiling of a rectangular room. The room is $6$ m long, $4$ m wide and $2.5$ m high. The room has one door measuring $2$ m by $0.9$ m and one window measuring $1.2$ m by $1$ m, which are not painted. One litre of paint covers $12$ m² and costs $8.50. How much will Mia spend on paint? Round up to the nearest whole litre when buying paint.

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   Key idea: Find the total area to be painted by calculating the area of all four walls and the ceiling, then subtract the unpainted door and window. Use the coverage rate to find litres needed, then calculate the cost.

   1. Find the area of the four walls:\

   $$A_{\text{walls}} = 2(lh + wh) = 2(6 \times 2.5 + 4 \times 2.5) = 2(15 + 10) = 50 \text{ m}^2$$

   2. Find the area of the ceiling:\

   $$A_{\text{ceiling}} = 6 \times 4 = 24 \text{ m}^2$$

   3. Subtract the door and window:\

   $$A_{\text{painted}} = 50 + 24 - (2 \times 0.9) - (1.2 \times 1) = 74 - 1.8 - 1.2 = 71 \text{ m}^2$$

   4. Find litres needed, rounding up:\

   $$\text{Litres} = \lceil 71 \div 12 \rceil = \lceil 5.916\ldots \rceil = 6 \text{ L}$$

   5. Calculate the cost:\

   $$\text{Cost} = 6 \times \text{\textdollar}8.50 = \text{\textdollar}51.00$$

   Answer: $51.00

7. Surface area of cylinders

   A cylinder has a radius of $4$ cm and a height of $10$ cm. Find its total surface area. Give your answer correct to one decimal place.

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   Key idea: The total surface area of a cylinder is the curved surface (a rectangle when unrolled) plus two circular ends: $\text{TSA} = 2\pi r h + 2\pi r^2$.

   1. Write the formula: $\text{TSA} = 2\pi r h + 2\pi r^2$
   2. Substitute $r = 4$ and $h = 10$:\

   $$\text{TSA} = 2\pi(4)(10) + 2\pi(4)^2$$

   3. Simplify each term:\

   $$\text{TSA} = 80\pi + 32\pi = 112\pi$$

   4. Evaluate:\

   $$\text{TSA} = 112\pi \approx 351.9 \text{ cm}^2$$

   Answer: $351.9$ cm²

8. Volume of prisms

   A composite solid is formed by placing a smaller rectangular prism on top of a larger rectangular prism. The larger prism has a length of $10$ cm, a width of $6$ cm and a height of $4$ cm. The smaller prism has a length of $5$ cm, a width of $6$ cm and a height of $3$ cm, and sits flush against one end of the top of the larger prism. Find the total volume of the composite solid.

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   Key idea: The total volume of a composite solid made from two prisms is the sum of their individual volumes.

   1. Find the volume of the larger prism:\

   $$V_{\text{large}} = 10 \times 6 \times 4 = 240 \text{ cm}^3$$

   2. Find the volume of the smaller prism:\

   $$V_{\text{small}} = 5 \times 6 \times 3 = 90 \text{ cm}^3$$

   3. Add the two volumes:\

   $$V_{\text{total}} = 240 + 90 = 330 \text{ cm}^3$$

   Answer: $330$ cm³

Free worksheets for this chapter

Worked examples and a free PDF with answers for each topic.

• Year 9 measurement error worksheet
• Year 9 surface area and volume worksheet