(a) 
The diagram shows a shape made from a major sector AOB and triangles OBC and AOD.
OB = 6 cm, BC = 2 cm, obtuse angle AOC = 135° and angle BCO = 90°.
(i) Show that angle BOC = 19.5°, correct to 1 decimal place.
(ii) Calculate the area of the major sector AOB.
(iii) C is the midpoint of OD. Calculate AD.
(iv) Calculate the total area of the shape.
(b) A sector of a circle has radius 8 cm and area 160 cm².
A mathematically similar sector has radius 20 cm.
Calculate the area of the larger sector.
▶️ Answer/Explanation
(a)(i) Using right triangle trigonometry in triangle BOC: sin(BOC) = opposite/hypotenuse = BC/OB = 2/6. Therefore, angle BOC = sin⁻¹(1/3) ≈ 19.5°.
(a)(ii) 64.6 cm². First find sector angle: 360° – 135° – 19.5° = 205.5°. Then area = (205.5/360) × π × 6² ≈ 64.6 cm².
(a)(iii) 16.1 cm. First find OC using Pythagoras: √(6² – 2²) ≈ 5.66 cm, so OD = 11.32 cm. Then use cosine rule in triangle AOD: AD² = 6² + 11.32² – 2×6×11.32×cos(135°) ≈ 259, so AD ≈ √259 ≈ 16.1 cm.
(a)(iv) 94.2 cm². Sum of: sector area (64.6 cm²), triangle AOD area (0.5×6×11.32×sin135° ≈ 24 cm²), and triangle BOC area (0.5×2×5.66 ≈ 5.66 cm²).
(b) 1000 cm². The scale factor is 20/8 = 2.5. Area scales by 2.5² = 6.25. Therefore, larger sector area = 160 × 6.25 = 1000 cm².
(a) 
The diagram shows a sector of a circle that is made into a cone by joining OA to OB.
The sector angle is \( x^\circ \) and the radius of the sector is 7.5 cm.
The base radius of the cone is 1.5 cm.
Calculate the value of \( x \).
(b) 
The diagram shows a cylinder with radius 8 cm inside a sphere with radius 17 cm.
Both ends of the cylinder touch the curved surface of the sphere.
(i) Show that the height of the cylinder is 30 cm.
(ii) Calculate the volume of the cylinder as a percentage of the volume of the sphere.
[The volume, \( V \), of a sphere with radius \( r \) is \( V = \frac{4}{3} \pi r^3 \).]
(c) 
The diagram shows a solid sphere with radius 6 cm inside a cube with side length 20 cm.
The cube contains water to a depth of 15 cm.
The sphere is removed.
Calculate the new depth of water in the cube.
[The volume, \( V \), of a sphere with radius \( r \) is \( V = \frac{4}{3} \pi r^3 \).]
▶️ Answer/Explanation
(a) 72 or 72.0
The arc length of the sector equals the circumference of the cone’s base. Using \( \frac{x}{360} \times 2\pi \times 7.5 = 2\pi \times 1.5 \), we solve for \( x \).
Simplifying gives \( x = \frac{1.5 \times 360}{7.5} = 72^\circ \).
(b)(i)
Using Pythagoras’ theorem: \( 2 \times \sqrt{17^2 – 8^2} \) or \( \sqrt{34^2 – 16^2} \).
This gives \( 2 \times \sqrt{289 – 64} = 2 \times 15 = 30 \) cm for the cylinder’s height.
(b)(ii) 29.3%
Cylinder volume: \( \pi \times 8^2 \times 30 = 1920\pi \). Sphere volume: \( \frac{4}{3}\pi \times 17^3 \approx 6549.3\pi \).
Percentage: \( \frac{1920}{6549.3} \times 100 \approx 29.3\% \).
(c) 12.7 cm
Original water volume: \( 20 \times 20 \times 15 = 6000 \) cm³. Sphere volume: \( \frac{4}{3}\pi \times 6^3 \approx 904.78 \) cm³.
New water volume after removal: \( 6000 – 904.78 = 5095.22 \) cm³. New depth: \( \frac{5095.22}{400} \approx 12.7 \) cm.