25o
65o
115o
130o
Correct answer is B
x = 65o (An interior angle of a cyclic quadrilateral = opposite exterior angle).
1:2
2:1
1:2
3:1
Correct answer is B
From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A1, the area of \(\bigtriangleup\)PXY
= Area of \(\bigtriangleup\)YZR = A2
Area of \(\bigtriangleup\)PQR = A = A1 + 2A2
But from similarity of triangles
\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)
\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)
A = 4A2 But, A = A1 + 2A
A1 = 4A2 - 2A2
A1 = 2A2
\(\frac{A_1}{A_2}\) = 2
A1:A2 = 2:1
Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1
In the diagram, PQ//RS, QU//PT and < PSR = 42o. Find angle x.
84o
48o
42o
32o
Correct answer is C
From the diagram, < QPS = xo (Corresponding angles)
Also, < QPS = < PSR(Alternate angles)
x = 42o
In the figure shown, PQs is a straight line. What is the value of < PRQ?
128o
108o
98o
78o
Correct answer is D
< QPR + < PRQ = < RQS
(Sum of two interior angles of a triangle = Opposite exterior angles)
70o + < PRQ = 148
< PRQ = 148o - 70o
= 78o
34cm
52cm
56cm
96cm
Correct answer is B
For a rhombus, all the sides are equal and diagonals bisect each other at 90o. Hence, the triangles formed are congruent: (under RHS).
Thus: \(\bar{PS}^2\) = 52 + 122
\(\bar{PS} = \sqrt{25 + 144}\)
= \(\sqrt{169}\)
\(\bar{PS}\) = 13cm
perimeter = 4 x length of a side
= 4 x 13cm
= 52cm
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