In the figure, PQR is a semicircle. Calculate the area of the shaded region
125\(\frac{2}{7}\)2
149\(\frac{2}{7}\)cm2
234\(\frac{1}{7}\)cm2
267\(\frac{1}{2}\)cm2
Correct answer is A
No explanation has been provided for this answer.
In the figure, YXZ = 30o, XYZ = 105o and XY = 8cm. Calculate YZ
16\(\sqrt{2}\)cm
8\(\sqrt{2}\)cm
4\(\sqrt{2}\)cm
22cm
Correct answer is C
yzx + 105o + 30o = 180o
yzx = 180 - 155 = 45o
\(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o}\)
yz = \(\frac{8 \sin 30}{sin 45}\)
= 8(\(\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1}\)
= 4 \(\div\) \(\frac{1}{\sqrt{2}}\)
= 4\(\sqrt{2}\)cm
In triangles XYZ and XQP, XP = 4cm, XQ = 5cm and PQ = QY = 3cm. Find ZY
8cm
6cm
4cm
3cm
Correct answer is B
No explanation has been provided for this answer.
In the figure, PS = QS = RS and QSR - 100o, find QPR
40o
50o
80o
100o
Correct answer is B
Since PS = QS = RS
S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius
QPR \(\pm\) \(\frac{100^o}{2}\) = 50o
14%
40%
46\(\frac{3}{4}\)%
53\(\frac{1}{3}\)%
Correct answer is C
This histogram is transferred into this frequency table
\(\begin{array}{c|c} Marks & 20 & 40 & 60 & 80 & 100 \\ \hline students & 9 & 7 & 6 & 6 & 2\end{array}\)
Students who scored more than 40 = 6 + 6 + 2 = 14
i.e. \(\frac{14}{30}\) x 100% = 46\(\frac{3}{4}\)%
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