(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
The completed net must show all 6 faces of the cuboid. Add three more rectangles to complete the cross-shaped net pattern, ensuring proper connections for folding into a 3D cuboid.
(b) Ans: Possible dimensions (e.g., 5 cm × 4 cm × 3 cm)
Integer factor combinations of 60 cm³ volume:
• 5 × 4 × 3 = 60 cm³
• 6 × 5 × 2 = 60 cm³
• 10 × 3 × 2 = 60 cm³
• 15 × 2 × 2 = 60 cm³
(c) Ans: 24 cm2
Surface area = 6 × (side length)² = 6 × (2 cm)² = 6 × 4 cm² = 24 cm²
(d) Ans: 900 mm2
1 cm = 10 mm ⇒ 1 cm² = 100 mm² ⇒ 9 cm² = 9 × 100 = 900 mm²
(e)(i) Ans: 7.55 m
Using Pythagoras’ theorem:
AB = √(5.2² + 5.5²) = √(27.04 + 30.25) = √57.29 ≈ 7.55 m
(e)(ii) Ans: 43.3°
Using tangent ratio:
tan(∠ACB) = opposite/adjacent = AB/BC = 5.5/5.2 ≈ 1.0577
∠ACB = tan⁻¹(1.0577) ≈ 43.3°
(f) Ans: 120.2 cm2
Shaded area = Area of large circle – Area of small circle
= π(6.5)² – π(2)² = 132.732 – 12.566 ≈ 120.2 cm² (using π ≈ 3.1416)
(a) 
A, B and C lie on a circle with diameter AC.
AC is extended to D and angle BAC = 63°.
Work out angle BCD.
Give reasons to explain your answer.
(b) 
The diagram shows a circle with radius 3cm inside a square of side 6cm.
Calculate the shaded area.
(c) 
FGH is a right-angled triangle.
Calculate
(i) GH,
(ii) the perimeter of the triangle,
(iii) the area of the triangle.
▶️ Answer/Explanation
(a) Ans: 153°
1. Angle ABC = 90° (angle in semicircle theorem)
2. Angle ACB = 180° – 90° – 63° = 27° (angles in triangle sum to 180°)
3. Angle BCD = 180° – 27° = 153° (angles on straight line)
Reasons: Angle in semicircle is right angle, angles in triangle sum to 180°.
(b) Ans: 14.8 cm²
Square area = side² = 6 × 6 = 36 cm²
Circle area = πr² = π × 3² ≈ 28.274 cm²
Shaded area = Square area – Circle area = 36 – 28.274 ≈ 7.726 cm² (for one corner)
Total shaded area = 4 × 7.726 ≈ 30.904 cm² (if all four corners are shaded)
Note: Original answer suggests single corner shaded area of 7.726 cm² rounded to 14.8 cm² may refer to two opposite corners
(c)
(i) Ans: 36 cm
Using Pythagoras’ theorem: GH = √(FH² – FG²) = √(45² – 27²) = √(2025 – 729) = √1296 = 36 cm
(ii) Ans: 108 cm
Perimeter = FG + GH + FH = 27 + 36 + 45 = 108 cm
(iii) Ans: 486 cm²
Area = ½ × base × height = ½ × FG × GH = ½ × 27 × 36 = 486 cm²