iGCSE Mathematics (0580) : E5.2 Carry out calculation involving perimeter and area.iGCSE Style Questions Paper 4

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Question

(a) ABCDEFGH is a regular octagon with sides of length 6 cm.
The diagram shows part of the octagon.
O is the centre of the octagon and M is the midpoint of AB.

(i) (a) Show that angle OAM is 67.5°.

(b) Calculate the area of the octagon.

(ii) Find the area of the circle that passes through the vertices of the octagon.

(b)

The diagram shows a horizontal container for water with a uniform cross-section.
The cross-section is a semicircle.
The radius of the semicircle is 0.45 m and the length of the container is 4 m.

(i) Calculate the volume of the container.

(ii)

The greatest depth of the water in the container is 0.3 m.
The diagram shows the cross-section.

Calculate the number of litres of water in the container.
Give your answer correct to the nearest integer.

▶️ Answer/Explanation

Solution

(a)(i)(a) In a regular octagon, the central angle is 360°/8 = 45°. Angle OAM is half of the interior angle (135°), so 135°/2 = 67.5°.

(a)(i)(b) 174 cm²

First find OM using tan(67.5°) = OM/3 → OM = 3×tan(67.5°) ≈ 7.2426 cm. Area of one triangle is ½×6×7.2426 ≈ 21.728 cm². Total area is 8×21.728 ≈ 174 cm².

(a)(ii) 193 cm²

The circle passes through all vertices, so radius OA = 3/cos(67.5°) ≈ 7.8426 cm. Area = π×7.8426² ≈ 193 cm².

(b)(i) 1.27 m³

Volume = ½×π×0.45²×4 ≈ 1.27 m³ (using π ≈ 3.142).

(b)(ii) 742 or 743 litres

When depth is 0.3m, the water forms a segment. First find the angle θ = 2×cos⁻¹(0.15/0.45) ≈ 2.462 radians. Area of segment = ½×0.45²×(2.462 – sin(2.462)) ≈ 0.1856 m². Volume = 4×0.1856 ≈ 0.742 m³ = 742 litres.

Question

(a) A shape is made from a square \( ABCD \) (side 11 cm) and two equal sectors of a circle.

(i) Calculate the area of the shape.

(ii) Calculate the perimeter of the shape.

(b)

A cube \( ABCDEFGH \) has edge 7 cm. Calculate the angle between \( AG \) and the base of the cube.

▶️ Answer/Explanation

Solution

(a)(i) 311 or 311.0 to 311.1

Square area = 11×11 = 121 cm². Two sectors form a full circle (radius 11 cm) with area π×11² ≈ 380.13 cm². Total area = 121 + 190.06 ≈ 311.1 cm².

(a)(ii) 78.6 or 78.55 to 78.56

Square perimeter = 3 sides (33 cm) + two quarter-circumferences (each 17.28 cm) ≈ 33 + 34.56 + 11 ≈ 78.56 cm.

(b) 35.2 or 35.3

AG is space diagonal (√(7²+7²+7²) ≈ 12.12 cm. Angle with base = arctan(vertical/planar) = arctan(7/√(7²+7²)) ≈ 35.3°.

iGCSE Mathematics (0580) : E5.2 Carry out calculation involving perimeter and area.iGCSE Style Questions Paper 4

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