▶️ Answer/Explanation
Solution
The diagram illustrates a metal plate ABCDEF with a circular sector BAFO (radius 20 cm, center O), two semicircles BCD and DEF, and two triangles OBD and ODF. Angles BOD and DOF are both \( \theta \) radians, and point D lies on the circle centered at O.
Part (a): Area of the metal plate with \( \theta = 1.2 \)
Sector BOF angle: \( 2\pi – 2 \theta = 2\pi – 2.4 \)
Area of sector BOF: \( \frac{1}{2} \times 20^2 \times (2\pi – 2.4) = 200 \times (2\pi – 2.4) \approx 776.637 \) cm²
Radius of semicircles BCD and DEF (from triangle OBD, where BD is the diameter): \( r = 20 \sin 0.6 \approx 11.291 \) cm
Area of two semicircles: \( 2 \times \frac{1}{2} \pi \times (20 \sin 0.6)^2 = \pi \times 11.291^2 \approx 400.641 \) cm²
Area of triangles OBD and ODF: \( 2 \times \frac{1}{2} \times 20 \times 20 \times \sin 1.2 = 400 \sin 1.2 \approx 372.811 \) cm²
Total area: \( 776.637 + 400.641 + 372.811 \approx 1550 \) cm² (to 3 significant figures)
Answer: \( 1550 \) cm²
Part (b): Exact perimeter with semicircle area \( 50\pi \) cm²
Semicircle area: \( \frac{1}{2} \pi r^2 = 50\pi \)
\( r^2 = 100 \Rightarrow r = 10 \) cm
In triangle OBD: \( \sin \theta = \frac{10}{20} = 0.5 \Rightarrow \theta = \frac{\pi}{6} \)
Sector BOF angle: \( 2\pi – 2 \cdot \frac{\pi}{6} = 2\pi – \frac{\pi}{3} = \frac{5\pi}{3} \)
Arc length of sector BOF: \( 20 \times \frac{5\pi}{3} = \frac{100\pi}{3} \) cm
Two semicircle arcs: \( 2 \times \pi \times 10 = 20\pi \) cm
Total perimeter (arcs only, as straight segments BD and DF may not contribute to external perimeter in some interpretations): \( \frac{100\pi}{3} + 20\pi = \frac{100\pi + 60\pi}{3} = \frac{160\pi}{3} \) cm
Answer: \( \frac{160\pi}{3} \) cm
▶️ Answer/Explanation
Solution
The diagram illustrates a metal plate OABCDEF formed by three circular sectors with centre O: sectors AOB and EOF with radius \( r \) cm and angle \( \theta \) radians, and sector COD with radius \( 2r \) cm and angle \( 2\theta \) radians. The perimeter is 14 cm, and the area is 10 cm².
Find the values of \( r \) and \( \theta \)
Perimeter (sum of straight and curved segments):
Straight segments: OA, BC, DE, FO = \( r + r + r + r = 4r \) cm
Arcs: AOB (\( r \theta \)), COD (\( 2r \cdot 2\theta = 4r\theta \)), EOF (\( r \theta \)) = \( r \theta + 4r \theta + r \theta = 6r \theta \) cm
Total perimeter: \( 4r + 6r \theta = 14 \quad (1) \)
Area (sum of sector areas):
Sector AOB: \( \frac{1}{2} r^2 \theta \)
Sector COD: \( \frac{1}{2} (2r)^2 \cdot 2\theta = \frac{1}{2} \cdot 4r^2 \cdot 2\theta = 4r^2 \theta \)
Sector EOF: \( \frac{1}{2} r^2 \theta \)
Total area: \( \frac{1}{2} r^2 \theta + 4r^2 \theta + \frac{1}{2} r^2 \theta = 5r^2 \theta = 10 \quad (2) \)
Solve equation (2):
\( 5r^2 \theta = 10 \Rightarrow r^2 \theta = 2 \Rightarrow \theta = \frac{2}{r^2} \)
Substitute into equation (1):
\( 4r + 6r \cdot \frac{2}{r^2} = 4r + \frac{12}{r} = 14 \)
Multiply by \( r \): \( 4r^2 + 12 = 14r \Rightarrow 4r^2 – 14r + 12 = 0 \Rightarrow 2r^2 – 7r + 6 = 0 \)
Solve quadratic: \( (r – 2)(2r – 3) = 0 \Rightarrow r = 2 \text{ or } r = \frac{3}{2} \)
Since \( r > \frac{3}{2} \), discard \( r = \frac{3}{2} \).
For \( r = 2 \): \( \theta = \frac{2}{2^2} = \frac{2}{4} = 0.5 \)
Check constraint: \( \theta = 0.5 < \frac{\pi}{4} \approx 0.785 \), which holds.
Verify:
Perimeter: \( 4 \cdot 2 + 6 \cdot 2 \cdot 0.5 = 8 + 6 = 14 \) cm
Area: \( 5 \cdot 2^2 \cdot 0.5 = 5 \cdot 4 \cdot 0.5 = 10 \) cm²
Answer: \( r = 2 \), \( \theta = 0.5 \)
▶️ Answer/Explanation
Solution
The diagram shows a sector OBC with centre O, radius 15 cm, and angle \( \frac{2}{5}\pi \) radians. An inner arc AD, with radius \( r \) cm, connects points A and D on lines OB and OC. The shaded region is bounded by outer arc BC, inner arc AD, and straight lines AB and DC, with a given area of \( \frac{209}{5}\pi \) cm².
Find the perimeter of the shaded region
Area of sector OBC: \( \frac{1}{2} \cdot 15^2 \cdot \frac{2}{5}\pi = \frac{1}{2} \cdot 225 \cdot \frac{2}{5}\pi = 45\pi \) cm²
Area of sector OAD: \( \frac{1}{2} \cdot r^2 \cdot \frac{2}{5}\pi = \frac{r^2}{5}\pi \) cm²
Shaded area (OBC − OAD): \( 45\pi – \frac{r^2}{5}\pi = \frac{209}{5}\pi \)
\( \frac{225 – r^2}{5}\pi = \frac{209}{5}\pi \Rightarrow 225 – r^2 = 209 \Rightarrow r^2 = 16 \Rightarrow r = 4 \) cm
Perimeter of shaded region:
Straight segments AB and DC: \( AB = DC = 15 – 4 = 11 \) cm
Outer arc BC: \( 15 \cdot \frac{2}{5}\pi = 6\pi \) cm
Inner arc AD: \( 4 \cdot \frac{2}{5}\pi = \frac{8}{5}\pi \) cm
Total perimeter: \( 11 + 6\pi + 11 + \frac{8}{5}\pi = 22 + \frac{30\pi + 8\pi}{5} = 22 + \frac{38}{5}\pi \) cm
Answer: \( 22 + \frac{38}{5}\pi \) cm