Year : 
1985
Title : 
Mathematics (Core)
Exam : 
JAMB Exam

Paper 1 | Objectives

41 - 47 of 47 Questions

# Question Ans
41.

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

A. 190 sq.cm

B. 20 sq.cm

C. 210 sq.cm

D. 160sq.cm

E. 320sq.cm

Detailed Solution

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
42.

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 = 86o and NQP = 122o find (xo, yo)

A. 28o, 36o

B. 36o, 28o

C. 43o, 61o

D. 61o, 43o

E. 36o, 43o

Detailed Solution

yo = 180o - (86o + 58o)

180 - 144 = 36o

xo = 180 - (94 + 58)

180 -152 = 28

(xo, yo) = (28o, 36o)
43.

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

A. 25o

B. 35o

C. 45o

D. 125o

E. 55o

Detailed Solution

< 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
44.

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

A. \(\sqrt{3a}\)

B. 3a

C. 3a2

D. a\(\sqrt{2}\)

E. a\(\sqrt{3}\)

Detailed Solution

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
45.

In the figure, PQ is the tangent from P to the circle QRS with SR as its diameter. If QRS = \(\theta\)oand RQP = \(\phi\)o, which of the following relationships between \(\theta\)o and \(\phi\)o is correct

A. \(\theta\)o + \(\phi\)o = 902

B. \(\phi\)o = 902 - 2\(\theta\)o

C. \(\theta\)o = \(\phi\)o

D. \(\phi\)o = 2\(\theta\)o

E. \(\theta\)o + 2\(\phi\)o

Detailed Solution

180 - \(\phi\)o = \(\theta\)o + \(\phi\)o (Sum of opposite interior angle equal to its exterior angle)

180 = 2\(\phi\) + \(\theta\)o
46.

In the figure, the area of the shaded segment is

A. 3\(\pi\)

B. 9\(\frac{\sqrt{3}}{4}\)

C. 3 \(\pi - 3 \frac{\sqrt{3}}{4}\)

D. \(\frac{(\sqrt{3 - \pi)}}{4}\)

E. \(\pi + \frac{9 \sqrt{3}}{4}\)

Detailed Solution

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 \(\pi - 3 \frac{\sqrt{3}}{4}\)
47.

In the figure, find angle x

A. 100o

B. 120o

C. 60o

D. 110o

E. 140o

Detailed Solution

In the figure, angle x = 20o + 80o + 40o

= 140o
41.

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

A. 190 sq.cm

B. 20 sq.cm

C. 210 sq.cm

D. 160sq.cm

E. 320sq.cm

Detailed Solution

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
42.

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 = 86o and NQP = 122o find (xo, yo)

A. 28o, 36o

B. 36o, 28o

C. 43o, 61o

D. 61o, 43o

E. 36o, 43o

Detailed Solution

yo = 180o - (86o + 58o)

180 - 144 = 36o

xo = 180 - (94 + 58)

180 -152 = 28

(xo, yo) = (28o, 36o)
43.

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

A. 25o

B. 35o

C. 45o

D. 125o

E. 55o

Detailed Solution

< 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
44.

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

A. \(\sqrt{3a}\)

B. 3a

C. 3a2

D. a\(\sqrt{2}\)

E. a\(\sqrt{3}\)

Detailed Solution

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
45.

In the figure, PQ is the tangent from P to the circle QRS with SR as its diameter. If QRS = \(\theta\)oand RQP = \(\phi\)o, which of the following relationships between \(\theta\)o and \(\phi\)o is correct

A. \(\theta\)o + \(\phi\)o = 902

B. \(\phi\)o = 902 - 2\(\theta\)o

C. \(\theta\)o = \(\phi\)o

D. \(\phi\)o = 2\(\theta\)o

E. \(\theta\)o + 2\(\phi\)o

Detailed Solution

180 - \(\phi\)o = \(\theta\)o + \(\phi\)o (Sum of opposite interior angle equal to its exterior angle)

180 = 2\(\phi\) + \(\theta\)o
46.

In the figure, the area of the shaded segment is

A. 3\(\pi\)

B. 9\(\frac{\sqrt{3}}{4}\)

C. 3 \(\pi - 3 \frac{\sqrt{3}}{4}\)

D. \(\frac{(\sqrt{3 - \pi)}}{4}\)

E. \(\pi + \frac{9 \sqrt{3}}{4}\)

Detailed Solution

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 \(\pi - 3 \frac{\sqrt{3}}{4}\)
47.

In the figure, find angle x

A. 100o

B. 120o

C. 60o

D. 110o

E. 140o

Detailed Solution

In the figure, angle x = 20o + 80o + 40o

= 140o