JAMB Mathematics Past Questions & Answers - Page 136

676.

In the figure POQ is the diameter of the circle PQR. If PSR = 145o, find xo

25^o

35^o

45^o

125^o

55^o

View answer & explanation

Correct answer is E

< PRQ = \(\frac{1}{2}\) < POQ = 90o

< PSR + < PQR = 180o

< PQR = 180o - 145o = 35o

\(\bigtriangleup\)PQR is a right angled triangle

x = 90 - < PQR

= 90o - 35o

= 55o

677.

In the figure, MNQP is a cyclic quadrilateral. MN and Pq are produced to meet at X and NQ and MP are produced to meet at Y. If MNQ = 86^o and NQP = 122^o find (x^o, y^o)

28^o, 36^o

36^o, 28^o

43^o, 61^o

61^o, 43^o

36^o, 43^o

View answer & explanation

Correct answer is A

y^o = 180^o - (86^o + 58^o)

180 - 144 = 36^o

x^o = 180 - (94 + 58)

180 -152 = 28

(x^o, y^o) = (28^o, 36^o)

678.

In \(\bigtriangleup\) XYZ, XKZ = 90O, XK = 15cm, XZ = 25cm and YK = 8cm. Find the area of \(\bigtriangleup\)XYZ

190 sq.cm

20 sq.cm

210 sq.cm

160sq.cm

320sq.cm

View answer & explanation

Correct answer is C

By Pythagoras, KZ2 = 252 - 52

KZ2 = (25 + 15)(25 - 15) = 400

KZ = \(\sqrt{400}\) = 20

area of XYZ = \(\frac{1}{2} \times 28 \times 15\)

= 210 sq. cm

679.

XYZ is a triangle and XW is perpendicular to YZ at = W. If XZ = 5cm and WZ = 4cm, Calculate XY.

5\(\sqrt{3}\)cm

3\(\sqrt{5}\)cm

3\(\sqrt{3}\)cm

5cm

6cm

View answer & explanation

Correct answer is B

by Pythagoras theorem, XW = 3cm

Also by Pythagoras theorem, XY² = 6² + 3²

XY² = 36 + 9 = 45

XY = \(\sqrt{45} = 3 \sqrt{3}\)

680.

In the figure, O is the centre of circle PQRS and PS//RT. If PRT = 135, then PSO is

67\(\frac{1}{2}\)^o

22\(\frac{1}{2}\)^o

33\(\frac{1}{2}\)^o

45^o

90^o

View answer & explanation

Correct answer is D

< R = 180o - 45o (sum of angles on a straight line)

< R = < P = 45o (corresponding angles)

< PSO = < P = 45o (\(\bigtriangleup\)PSO is a right angle)