In the diagram above, O is the center of the circle PQRS and ∠QPS = 36°. Find ∠QOS
36o
72o
108o
144o
288o
Correct answer is B
< QOS = 2 < QPS (angle subtended at the centre)
\(\therefore\) < QOS = 2 x 36° = 72°
In the diagram above, TR = TS and ∠TRS = 60o. Find the value of x
30
45
60
120
150
Correct answer is C
∠TRS = ∠RST = 60o
∠PSR + ∠PQR = 180o, 120o + xo = 180o
x = 60o
In the diagram above, O is the center of the circle and ∠POR = 126°. Find ∠PQR
234o
117o
72o
63o
54o
Correct answer is B
Reflex < POR = 360° - 126° = 234°
\(\therefore\) < PQR = \(\frac{1}{2} \times 234°\)
= 117°
48o
45o
37o
32o
30o
Correct answer is C
∠RQT = 180° - (95° + 48°) = 73°
∠OQR = 90° - 37° = 53°
∠QOR = 180° - (53° + 53°) = 74°
QSR = 74°/2 = 37°
5/12cm2
7/12cm2
11/6cm2
2 1/3cm2
6 1/6cm2
Correct answer is B
Area of \(\Delta OXY\) = \(12\frac{1}{4} cm^2\)
Area of sector OXY = \(\frac{30}{360} \times \frac{22}{7} \times 7 \times 7\)
= \(\frac{77}{6} = 12\frac{5}{6} cm^2\)
Area of the shaded portion = \(12\frac{5}{6} - 12\frac{1}{4}\)
= \(\frac{7}{12} cm^2\)
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