Mathematics Questions & Answers - Free Online Practice

1,686.

PQ and PR are tangents from P to a circle centre O as shown in the figure. If QRP = 34^o, find the angle marked x

34^o

56^o

68^o

112^o

View answer & explanation

Correct answer is C

From the circle centre 0, if PQ & PR are tangents from P and QRP = 34^o

Then the angle marked x i.e. QOP

34^o x 2 = 68^o

1,687.

The figure is a solid with the trapezium PQRS as its uniform cross-section. Find its volume

120m²

576m³

816m³

1056m³

View answer & explanation

Correct answer is C

Volume of solid = cross section x H

Since the cross section is a trapezium

= \(\frac{1}{2} (6 + 11) \times 12 \times 8\)

= 6 x 17 x 8 = 816m³

1,688.

In the diagram, PQ and RS are chords of a circle centre O which meet at T outside the circle. If TP = 24cm. TQ = 8cm and TS = 12cm, find TR.

16cm

14cm

12cm

8cm

View answer & explanation

Correct answer is A

PT x QT = TR x TS

24 x 8 = TR x 12

TR = \(\frac{24 \times 8}{12}\)

= = 16cm

1,689.

In the figure, find angle x

100^o

120^o

60^o

110^o

140^o

View answer & explanation

Correct answer is E

In the figure, angle x = 20^o + 80^o + 40^o

= 140^o

1,690.

In the figure, the area of the shaded segment is

3\(\pi\)

9\(\frac{\sqrt{3}}{4}\)

3 \(\pi - 3 \frac{\sqrt{3}}{4}\)

\(\frac{(\sqrt{3 - \pi)}}{4}\)

\(\pi + \frac{9 \sqrt{3}}{4}\)

View answer & explanation

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 \(\pi - 3 \frac{\sqrt{3}}{4}\)