JAMB Mathematics Past Questions & Answers - Page 123

611.

In this figure, PQ = PR = PS and SRT = 68<sup>o</sup>. Find QPS

136^o

124^o

12^o

68^o

View answer & explanation

Correct answer is A

Since PQRS is quadrilateral 2y + 2x = QPS = 360^o

i.e. 2(y + x) + QPS = 360^o

QPS = 360^o - 2(y + x)

But x + y + 68^o = 180^o

x + y = 180^o - 68^o = 180^o

x + y = 180^o - 68^o

= 112^o

QPS = 360^o - 2(112^o)

360^o - 224^o = 136^o

612.

In the figure above, find the value of y

28^o

122^o

150^o

152^o

View answer & explanation

Correct answer is B

ADB = 180^o - 152^o = 28^o

28^o + y + 30^o = 180^o

y = 180^o - 58^o

= 122^o

613.

PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PNR = 110^o and PMQ = 55^o, find MPQ

40^o

30^o

25^o

15^o

View answer & explanation

Correct answer is D

PQM = 110^o(Ext. < of a cyclic quad)

MPQ = 180^o - (110 + 55)

= 180^o - 165^o

= 15^o

614.

PMN and PQR are two secants of the circle MQTRN and PT is a tangent. If PM = 5cm, PN = 12cm and PQ = 4.8cm, calculate the respective lengths of PR and PT in centimeters

7.3, 5.9

7.7, 12.5

12.5, 7.7

5.9, 7.3

View answer & explanation

Correct answer is C

\(\frac{PQ}{PN} = \frac{PM}{PR} = \frac{QM}{NR}\)

\(\frac{4.8}{12} = \frac{5}{PR}\)

PR = \(\frac{5 \times 12}{4.8} = \frac{50}{4}\)

= 12.5

\(\frac{PQ}{PN} = \frac{PM}{PT} = \frac{TM}{NT}\)

\(\frac{PT}{12} = \frac{5}{PR}\)

PT² = 60

PT = \(\sqrt{60}\)

= 7.746

= 7.7