(a) The grid shows part of the net of a cuboid.
Complete the net.
(b) The volume of another cuboid is 60cm3.
Each side is a whole number of centimetres long.
Write down a possible set of dimensions for the cuboid.
(c) Each side of a cube has length 2cm.
Work out the total surface area of the cube.
Give the units of your answer.
(d) Change 9cm2 into mm2.
(e) The diagram shows a triangle.
(i) Calculate the length AB.
(ii) Use trigonometry to calculate angle ACB.
(f) The diameter of the large circle is 13cm.
The radius of the small circle is 2cm.
Calculate the shaded area.
▶️ Answer/Explanation
(a) Ans: correct net drawn
Complete the net by drawing the remaining faces of the cuboid, ensuring all 6 faces are connected correctly.
(b) Ans: Possible dimensions (e.g., 5, 4, 3)
Find all integer triplets \((l, w, h)\) such that \(l \times w \times h = 60\). Examples include \(5 \times 4 \times 3\) or \(10 \times 3 \times 2\).
(c) Ans: 24 cm2
Surface area of a cube is \(6 \times \text{side}^2 = 6 \times 2^2 = 24 \text{cm}^2\).
(d) Ans: 900 mm2
Convert cm2 to mm2 by multiplying by \(100\) (since \(1 \text{cm} = 10 \text{mm}\)). Thus, \(9 \text{cm}^2 = 900 \text{mm}^2\).
(e)(i) Ans: 7.55 m
Use Pythagoras’ theorem: \(AB = \sqrt{5.2^2 + 5.5^2} = \sqrt{27.04 + 30.25} = \sqrt{57.29} \approx 7.55 \text{m}\).
(e)(ii) Ans: 43.3°
Using \(\tan^{-1}\left(\frac{5.5}{5.2}\right) = \tan^{-1}(1.0577) \approx 43.3°\).
(f) Ans: 120.2 cm2
Shaded area = Area of large circle (\( \pi \times 6.5^2 \)) – Area of small circle (\( \pi \times 2^2 \)) = \(132.73 – 12.57 \approx 120.2 \text{cm}^2\).
The scale drawing represents the positions of 3 towns, A, B and C.
The scale is 1 centimetre represents 4 kilometres.
(a) Measure the bearing of B from A.
(b) A transmitter is placed near to the 3 towns.
(i) The transmitter is equidistant from A and B.
Using a straight edge and compasses only, construct the locus of points equidistant from A and B.
(ii) The transmitter is also on the bisector of angle ABC.
Using a straight edge and compasses only, construct the bisector of angle ABC.
(iii) Mark the position, T, of the transmitter on the scale drawing.
(c) Work out the actual distance, in kilometres, of town A from T.
(d) The signal from the transmitter has a range of 30 kilometres in all directions.
On the scale drawing, construct the locus of points 30 kilometres from T.
(e) Would the signal from the transmitter reach town C ?
Give a reason for your answer.
▶️ Answer/Explanation
(a) Ans: 106 to 110
Using a protractor, measure the angle from north at A to the line AB. The bearing should be between 106° and 110°.
(b)(i) Ans: Correct bisector of AB constructed with 2 pairs of arcs.
Draw arcs from A and B intersecting above and below AB. Join these intersections to form the perpendicular bisector.
(b)(ii) Ans: Correct bisector of angle ABC with arcs
Draw an arc centered at B intersecting BA and BC. From these points, draw intersecting arcs to construct the angle bisector.
(b)(iii) Ans: T marked at intersection of their bisectors
The transmitter T is located where the perpendicular bisector of AB and the angle bisector of ∠ABC intersect.
(c) Ans: 24.4[km] to 26.0[km]
Measure AT in cm (e.g., 6.2cm) and multiply by scale factor (4km/cm) to get actual distance (24.8km).
(d) Ans: Circle, radius 7.5(±0.2)cm centre T.
30km ÷ 4km/cm = 7.5cm radius. Draw a circle centered at T with this radius.
(e) Ans: No It is outside the circle. oe
Town C lies outside the 30km range circle, so it won’t receive the signal.