Year 10 and 10A Measurement and surds: free practice test with answers

Sample measurement and surds questions with answers

Try each one before you open the answer.

1. Irrational numbers including surds

   Write $0.\overline{45}$ as a fraction in simplest form.

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   Key idea: Multiply by a power of 10 that moves one full repeating block, then subtract.

   1. Let $x = 0.4545\ldots$, so $100x = 45.4545\ldots$
   2. $99x = 45$, so $x = \frac{45}{99}$
   3. $x = \frac{5}{11}$

   Answer: $\frac{5}{11}$

2. Adding and subtracting surds

   Simplify $\sqrt{50} + \sqrt{8}$.

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   Key idea: Break each surd into a perfect-square factor, simplify, then collect like surds.

   1. Simplify $\sqrt{50}$: $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$
   2. Simplify $\sqrt{8}$: $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$
   3. Add like surds: $5\sqrt{2} + 2\sqrt{2} = 7\sqrt{2}$

   Answer: $7\sqrt{2}$

3. Rationalising the denominator

   Rationalise the denominator and simplify.

   (a) $\dfrac{6}{\sqrt{3}}$

   (b) $\dfrac{5}{2\sqrt{10}}$

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   Key idea: Multiply the numerator and denominator by the surd in the denominator, since $\sqrt{a} \times \sqrt{a} = a$.

   (a)

   1. $\frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$

   (b)

   1. $\frac{5}{2\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{5\sqrt{10}}{2 \times 10} = \frac{5\sqrt{10}}{20} = \frac{\sqrt{10}}{4}$

   Answer: (a) $2\sqrt{3}$ (b) $\dfrac{\sqrt{10}}{4}$

4. Review of length

   Convert each length to the unit shown in brackets.

   (a) 3.75 km (m)

   (b) 48 mm (cm)

   (c) 0.62 m (mm)

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

   (a)

   1. $3.75 \times 1000 = 3750$ m

   (b)

   1. $48 \div 10 = 4.8$ cm

   (c)

   1. $1 \text{ m} = 1000 \text{ mm}$, so $0.62 \times 1000 = 620$ mm

   Answer: (a) 3750 m (b) 4.8 cm (c) 620 mm

5. Review of area

   Find the area of a trapezium with parallel sides of 8.4 cm and 13.6 cm and a perpendicular height of 5 cm.

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   Key idea: Area of a trapezium $= \dfrac{1}{2}(a + b)h$, where $a$ and $b$ are the parallel sides.

   1. $a + b = 8.4 + 13.6 = 22$
   2. $A = \frac{1}{2} \times 22 \times 5 = 55$

   Answer: 55 cm²

6. Surface area of prisms and cylinders

   A solid is made by placing a triangular prism on top of a rectangular prism. The rectangular prism is 10 cm long, 6 cm wide and 4 cm tall. The triangular prism sits exactly on the top face of the rectangular prism and runs the full 10 cm length. Its cross-section is an isosceles triangle with base 6 cm and perpendicular height 3 cm; each sloping face has a slant height of 3.35 cm. Find the total surface area of the composite solid.

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   Key idea: Add the surface areas of each part, then subtract any faces that are joined (hidden) and therefore no longer on the outside.

   1. Rectangular prism faces. The two ends ($6 \times 4$), two long sides ($10 \times 4$), and the bottom ($10 \times 6$). The top face is hidden where the triangular prism sits, so omit it.\

   $$\text{SA}_{\text{rect}} = 2(6 \times 4) + 2(10 \times 4) + (10 \times 6) = 48 + 80 + 60 = 188 \text{ cm}^2$$

   2. Triangular prism faces. Two triangular ends and two sloping rectangles. The base of the triangular prism is hidden (joined to the rectangular prism), so omit it.\

   $$\begin{gathered}\text{Area of each triangle} = \tfrac{1}{2} \times 6 \times 3 = 9 \text{ cm}^2 \\ \text{Two sloping rectangles} = 2 \times (3.35 \times 10) = 67 \text{ cm}^2 \\ \text{SA}_{\text{tri}} = 2(9) + 67 = 18 + 67 = 85 \text{ cm}^2\end{gathered}$$

   3. Total surface area:\

   $$\text{SA} = 188 + 85 = 273 \text{ cm}^2$$

   Answer: $273 \text{ cm}^2$

7. Surface area of pyramids and cones

   A right pyramid has a square base with side length 8 m and a perpendicular height of 3 m. Find the total surface area of the pyramid.

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   Key idea: The slant height of each triangular face is found using Pythagoras' theorem with the perpendicular height and the distance from the centre of the base to the midpoint of a side.

   1. Distance from centre of base to midpoint of a side: $\frac{8}{2} = 4$ m
   2. Slant height: $l = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ m
   3. Area of one triangular face: $\frac{1}{2} \times 8 \times 5 = 20$ m²
   4. Total area of 4 triangular faces: $4 \times 20 = 80$ m²
   5. Base area: $8^2 = 64$ m²
   6. Total surface area: $64 + 80 = 144$ m²

   Answer: $144$ m²

8. Volume of pyramids and cones

   A grain silo is made up of a cylinder topped with a right cone. The cylinder has a radius of 3 m and a height of 10 m. The cone sits on top of the cylinder, has the same radius of 3 m, and has a perpendicular height of 4 m. Find the total volume of the silo. Give your answer correct to one decimal place.

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   Key idea: The total volume is the volume of the cylinder plus the volume of the cone, using $V_{\text{cone}} = \frac{1}{3}\pi r^2 h$.

   1. Volume of cylinder: $\pi \times 3^2 \times 10 = 90\pi$ m³
   2. Volume of cone: $\frac{1}{3} \times \pi \times 3^2 \times 4 = \frac{1}{3} \times 36\pi = 12\pi$ m³
   3. Total volume: $90\pi + 12\pi = 102\pi \approx 320.4$ m³

   Answer: $320.4$ m³

Free worksheets for this chapter

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

• Year 10A surds and fractional indices worksheet
• Year 10 trigonometry worksheet
• Year 10 geometric reasoning and proofs worksheet
• Year 10A pythagoras and trigonometry in 3d worksheet
• Year 10 direct and inverse proportion worksheet
• Year 10 surface area and volume worksheet
• Year 10A surface area and volume of pyramids, cones and spheres worksheet