In the diagram above, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find the value of ∠QPR
72o
64o
52o
32o
26o
Correct answer is C
From the figure, < PSR = 32° (base angles of an isos. triangle)
\(\therefore\) < PRS = 180° - (32° + 32°) = 116° (sum of angles in a triangle)
< QRP = 180° - 116° = 64° (angle on a straight line)
< PQR = 64° (base angles of an isos. triangle)
< QPR = 180° - (64° + 64°) = 52°
35cm3
154cm3
220cm3
528cm3
770cm3
Correct answer is E
Volume = \(\pi r^2 h\)
= \(\frac{22}{7} \times 7^2 \times 5\)
= \(770 cm^3\)
35cm2
154cm2
220cm2
528cm2
770cm2
Correct answer is D
TSA of a cylinder = \(2\pi r^2 + 2\pi rh\)
= \(2\pi r (r + h)\)
= \(2 \times \frac{22}{7} \times 7 (7 + 5)\)
= \(44 \times 12\)
= \(528 cm^2\)
75o
60o
45o
30o
0o
Correct answer is D
\(\theta_{1} = \theta_{2} = 30°\)
calculate the surface area of a sphere of radius 7cm [Take π = 22/7]
86cm2
154cm2
616cm2
1434cm2
4312cm2
Correct answer is C
Surface area of a sphere = \(4\pi r^2\)
= \(4 \times \frac{22}{7} \times 7 \times 7\)
= \(616 cm^2\)
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