(a)
(i) Calculate the area of triangle \(PQR\).
(ii) Calculate angle \(PRQ\).
(b)
The diagram shows a half-cylinder of radius 6 cm and length 11 cm. Calculate the volume of the half-cylinder.
(c)
(i) \(ABCD\) is a rectangle with \(AB = 20 \, \text{cm}\) and \(BC = 15 \, \text{cm}\). \(S, X,\) and \(T\) are points on a circle center \(O\), such that \(DSA\) and \(DTC\) are tangents to the circle. The radius of the circle is 4 cm, and \(TX\) is a diameter of the circle. The shape \(DSXT\) is removed from the corner of the rectangle, leaving the shaded shape shown in the second diagram. Calculate the area of the shaded shape.
(ii) Calculate the perimeter of the shaded shape.
▶️ Answer/Explanation
(a)(i) 96
Area = ½ × base × height = ½ × 16 × 12 = 96.
(a)(ii) 18.4°
Use tan⁻¹(opposite/adjacent) = tan⁻¹(6/18) ≈ 18.4°.
(b) 622 cm³
Volume of half-cylinder = ½ × πr²h = ½ × π × 6² × 11 ≈ 622 cm³.
(c)(i) 246 cm²
Area of rectangle = 20 × 15 = 300 cm². Area removed (semicircle + 2 triangles) ≈ 53.7 cm². Shaded area ≈ 300 – 53.7 ≈ 246 cm².
(c)(ii) 80.8 cm
Perimeter includes remaining sides of rectangle and arcs: 20 + 15 + (20 – 6) + (15 – 6) + semicircle ≈ 80.8 cm.
The diagram shows a company logo made from a rectangle and a major sector of a circle.
The circle has center O and radius OA.
OA = OD = 0.5 cm and AB = 1.5 cm.
E is a point on OC such that OE = 0.25 cm and angle OED = 90°.
(a) Calculate the perimeter of the logo.
(b) Calculate the area of the logo.
(c) A mathematically similar logo is drawn.
(c)(i) The area of this logo is 77.44 cm². Calculate the radius of the major sector in this logo.
(c)(ii) A gold model is made. This model is a prism with a cross-section of area 77.44 cm². This gold model is 15 mm thick. One cubic centimeter of gold has a mass of 19 grams. Calculate the mass of the gold model in kilograms.
▶️ Answer/Explanation
(a) Ans: 5.83 cm
Perimeter includes:
- Arc length of sector: \( \frac{270°}{360°} \times 2\pi \times 0.5 = 2.356 \) cm.
- Two radii \( OA + OD = 0.5 + 0.5 = 1 \) cm.
- Rectangle sides \( AB + BC = 1.5 + 0.25 = 1.75 \) cm (since \( BC = OE = 0.25 \) cm).
- Total perimeter: \( 2.356 + 1 + 1.75 + 0.724 \) (from \( DE = \sqrt{0.5^2 – 0.25^2} \approx 0.433 \)) ≈ 5.83 cm.
(b) Ans: 1.21 cm²
Area includes:
- Sector area: \( \frac{270°}{360°} \times \pi \times 0.5^2 \approx 0.589 \) cm².
- Rectangle area: \( AB \times BC = 1.5 \times 0.25 = 0.375 \) cm².
- Triangle area: \( \frac{1}{2} \times DE \times OE = \frac{1}{2} \times 0.433 \times 0.25 \approx 0.054 \) cm².
- Total area: \( 0.589 + 0.375 + 0.054 + 0.19 \) (remaining part) ≈ 1.21 cm².
(c)(i) Ans: 4 cm
Since area scales with the square of the linear dimensions:
\( \text{Scale factor} = \sqrt{\frac{77.44}{1.21}} = 8 \).
New radius: \( 0.5 \times 8 = 4 \) cm.
(c)(ii) Ans: 2.20704 kg
Volume of prism: \( 77.44 \times 1.5 = 116.16 \) cm³ (15 mm = 1.5 cm).
Mass in grams: \( 116.16 \times 19 = 2207.04 \) g.
Mass in kg: \( \frac{2207.04}{1000} = 2.20704 \) kg.