1990 - JAMB Mathematics Past Questions and Answers - page 5

41
In triangles XYZ and XQP, XP = 4cm, XQ = 5cm and PQ = QY = 3cm. Find ZY
A
8cm
B
6cm
C
4cm
D
3cm
correct option: b
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42
In the figure, YXZ = 30o, XYZ = 105o and XY = 8cm. Calculate YZ
A
16\(\sqrt{2}\)cm
B
8\(\sqrt{2}\)cm
C
4\(\sqrt{2}\)cm
D
22cm
correct option: c

yzx + 105o + 30o = 180o

yzx = 180 - 155 = 45o

(\frac{yz}{sin 30^o} = \frac{8}{sin 45^o})

yz = (\frac{8 \sin 30}{sin 45})

= 8((\frac{1}{2}) = \frac{8}{1} \times \frac{1}{2} \times \frac{\sqrt{2}}{1})

= 4 (\div) (\frac{1}{\sqrt{2}})

= 4(\sqrt{2})cm

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43
In the figure, PQR is a semicircle. Calculate the area of the shaded region
A
125\(\frac{2}{7}\)2
B
149\(\frac{2}{7}\)cm2
C
234\(\frac{1}{7}\)cm2
D
267\(\frac{1}{2}\)cm2
correct option: a
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44
Find the curved surface area of the frustrum in the figure
A
16\(\pi \sqrt{10}\)cm2
B
20\(\pi \sqrt{10}\)cm2
C
24\(\pi \sqrt{10}\)cm2
D
36\(\pi \sqrt{10}\)cm2
correct option: b

(\frac{x}{4} = \frac{6 + x}{6})

6x = 4(6 + x) = 24 + 4x

x = 12 = c = (\pi RL - \pi L)

= (\pi (6) \sqrt{18^2} + 6^2 - \pi \times 4 \times \sqrt{12^2} + 4^2)

= 6(\pi \sqrt{360} - 4 \pi \sqrt{160})

= 36(\pi \sqrt{10} - 16 \pi \sqrt{10})

= 20(\pi \sqrt{10})cm2

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