100o
120o
60o
110o
140o
Correct answer is E
In the figure, angle x = 20o + 80o + 40o
= 140o
In the figure, the area of the shaded segment is
3\(\pi\)
9\(\frac{\sqrt{3}}{4}\)
3 \(\pi - 3 \frac{\sqrt{3}}{4}\)
\(\frac{(\sqrt{3 - \pi)}}{4}\)
\(\pi + \frac{9 \sqrt{3}}{4}\)
Correct answer is C
Area of sector = \(\frac{120}{360} \times \pi \times (3)^2 = 3 \pi\)
Area of triangle = \(\frac{1}{2} \times 3 \times 3 \times \sin 120^o\)
= \(\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9\sqrt {3}}{4}\)
Area of shaded portion = 3\(\pi - \frac{9\sqrt {3}}{4}\)
= 3 \(\pi - 3 \frac{\sqrt{3}}{4}\)
The figure is an example of the construction of a
perpendicular bisector of a given straight line QR
perpendicular from a given point to a given line QR
perpendicular to a line from a given point P on that line
given angle
Correct answer is B
QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR
In the figure, find the value of x
110o
100o
90o
80o
Correct answer is C
Z = X = Y = 180o .........(i)
2Z + Y + Y = 190 = 2Z + 2Y = 180
Z + Y = 90o........(ii)
hence x + (z + y) = 180
x + 90o = 180o
x = 180o - 90o
= 90o
210o
150o
105o
50o
Correct answer is D
Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the nonagon = 9
where 3 are equal and 6 other angles = 1110o
( 2 x 9 - 4)90o = (18 - 4)90o
= 14 x 90o = 1260o
9 angles = 12600, 6 angles = 1110o
Remaining 3 angles = 1260o - 1110o = 150o
size of one of the3 angles \(\frac{150}{3}\) = 50o
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