47cm
61cm
88cm
231cm
Correct answer is B
Perimeter of the sector = \(2r + \frac{\theta}{360} \times 2\pi r\)
= \(2(14) + \frac{135}{360} \times 2 \times \frac{22}{7} \times 14\)
= \(28 + 33\)
= 61 cm
16cm
17cm
18cm
20cm
Correct answer is B
From the diagram
\(r^2 = B^2 + 15^2\\
= 64 + 225\\
r = \sqrt{289} = 17cm\)
\(9\pi cm^2\)
\(15\pi cm^2\)
\(16\pi cm^2\)
\(20\pi cm^2\)
Correct answer is B
Curved surface area or a cone \(=\pi rl\)
from the information \(l^2 = 4^2 + 3^2 = 16+9\\
l = \sqrt{25} = 5; ∴ C.S.A\hspace{1mm} = \frac{22}{7}\times 3 \times 5\\
Since \frac{22}{7}=\pi ∴ C.S.A\hspace{1mm} =\hspace{1mm}15\pi\)
\(5\frac{1}{3}cm^3\)
\(10\frac{2}{3}cm^3\)
\(16cm^3\)
\(32cm^3\)
Correct answer is B
No explanation has been provided for this answer.
8m
9m
10m
13m
Correct answer is A
Given, Length of minor arc = 5.5m
Angle subtended by minor arc = 360° - 210° = 150°
\(\therefore 5.5 = \frac{150}{360} \times 2 \times \frac{22}{7} \times r \)
\(\frac{55r}{21} = 5.5\)
\(r = \frac{5.5 \times 21}{55}\)
r = 2.1m
Length of major arc = \(\frac{210}{360} \times 2 \times \frac{22}{7} \times 2.1\)
= \(7.7m \approxeq 8m\) (to the nearest metre)
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