JAMB Mathematics Past Questions & Answers - Page 131

651.

In the figure, GHIJKLMN is a cube of side a. Find the length of HN.

\(\sqrt{3a}\)

3a²

a\(\sqrt{2}\)

a\(\sqrt{3}\)

View answer & explanation

Correct answer is E

HJ² = a² + a² = 2a²

HJ = \(\sqrt{2a^2} = a \sqrt{2}\)

HN² = a² + (a\(\sqrt{2}\))² = a² + 2a² = 3a²

HN = \(\sqrt{3a^2}\)

= a\(\sqrt{3}\)cm

652.

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

653.

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)

654.

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