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JAMB Mathematics Past Questions & Answers - Page 129

641.

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

642.

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

643.

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 $\pi - 3 \frac{\sqrt{3}}{4}$

644.

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

View answer & explanation

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

645.

In the figure, find the value of x

110^o

100^o

90^o

80^o

View answer & explanation

Correct answer is C

Z = X = Y = 180^o .........(i)

2Z + Y + Y = 190 = 2Z + 2Y = 180

Z + Y = 90^o........(ii)

hence x + (z + y) = 180

x + 90^o = 180^o

x = 180^o - 90^o

= 90^o