Mathematics Questions & Answers - Free Online Practice

1,751.

In the figure, PQ and QR are chords of the circle PQR. QRS is a straight line and PR is equal to RS, < PSR is 20o. What is the size of <POQ.

70^o

90^o

80^o

40^o

60^o

View answer & explanation

Correct answer is C

Isosceles Triangle PSR:

RSP ≡ RPS → 20º

That is PRQ = 40º

POQ = 2 * 40 = 80º (Angle subtended by chord PQ at centre is twice angle subtended at circumference).

The size of POQ = 80º

1,752.

If O is the centre of the circle, < POS equls

70^o

75^o

105^o

140^o

150^o

View answer & explanation

Correct answer is E

Since O is the centre of the circle < POS = 150^o

i.e. < substended at the centre is twice that substended at any part of the circumference

1,753.

In this figure, PQRS is a parallelogram, PS = PT and < PST = 55\(^o\). The size of <PQR is

125^o

120^o

115^o

110^o

10^o

View answer & explanation

Correct answer is D

Both pairs of opp. angles are equal

< STP = 55\(^o\) - isosceles angle

< TSR = 55\(^o\) - alternate angle to < STP

Hence, < PSR = 55\(^o\) + 55\(^o\) = 110\(^o\)

\(\therefore\) < PQR = 110\(^o\)

1,754.

In the figure, \(\bigtriangleup\) ABC are in adjacent planes. AB = AC = 5cm, BC = 6cm and o then AE is equal to

3\(\sqrt{2}\)

2\(\sqrt{3}\)

\(\frac{\sqrt{3}}{2}\)

\(\frac{2}{\sqrt{3}}\)

View answer & explanation

Correct answer is B

BC = 6 : DC = \(\frac{6}{2}\) = 3cm

By construction < EDE = 180^o(90^o + 60^o) = 180^o - 150^o

= 30^o(angle on a strt. line)

From rt < triangle ADC, AD² = 5² - 3²

= 25 - 9 = 6

AD = 4

From < AEC, let AS = x

\(\frac{x}{sin 60^o}\) - \(\frac{4}{sin 90^o}\)

sin 90^o = 1

sin 60^o = \(\frac{\sqrt{3}}{2}\)

x = 4sin 60^o

x = 3 x \(\frac{\sqrt{3}}{2}\)

= 2\(\sqrt{3}\)