JAMB Mathematics Past Questions & Answers - Page 119

591.

In the figure, PQST is a parallelogram and TSR is a straight line. If the area of \(\bigtriangleup\)QRS is 20cm2, find the area of the trapezium PQRT.

A.

35cm2

B.

65cm2

C.

70cm2

D.

140cm2

Correct answer is C

A\(\bigtriangleup\) = \(\frac{1}{2}\) x 8 x h = 20

= \(\frac{1}{2}\) x 8 x h = 4h

h = \(\frac{20}{4}\)

= 5cm

A\(\bigtriangleup\)(PQTS) = L x H

A\(\bigtriangleup\)PQRT = A\(\bigtriangleup\)QSR + A\(\bigtriangleup\)PQTS

20 + 50 = 70cm\(^2\)

ALTERNATIVE METHOD

A\(\bigtriangleup\)PQRT = \(\frac{1}{2}\) x 5 x 28

= 70cm\(^2\)

592.

In diagram, PQ || ST and < PQR = 120o, < RST = 130o, find the angle marked x

A.

50

B.

65

C.

70

D.

80

Correct answer is C

x + 60o + 50o = 180o

x = 110o = 180o

x = 180o - 110o

= 70o

593.

In the diagram, PR is a diameter of the circle PQRS. PST and QRT are straight lines. Find QRS

A.

20o

B.

25o

C.

30o

D.

35o

Correct answer is B

< PSR = \(\frac{1}{2}\)(180o) = 90o (angle substended by a semi-circle)

∴ < TSR = 90o

< SRT = 90o - 30o = 60o

< PRS = 90o - 35o = 55o

< PRQ = 180 - (60o + 55o) = 180 - 115o = 65o

< SQR = 35o(angles in the same segment)

< QSR + < SRQ + < SQR = 180o

< QSR + 120o + 35o = 180o

< QSR + 155 = 180o

< QSR = 180o - 155o

= 25o

594.

The shaded area represents

A.

x \(\leq\) 0, y \(\leq\) 0, 2y + 3x \(\leq\) 6

B.

x \(\geq\) 0, y \(\geq\) 3, 3x + 2y \(\geq\) 6

C.

x \(\geq\) 2, y \(\geq\) 0, 3x + 2y \(\leq\) 6

D.

x \(\geq\) 0, y \(\geq\) 0, 3x + 2y \(\geq\) 6

Correct answer is A

m = \(\frac{y_2 - y_1}{x_2 - x_2} = \frac{3 - 0}{0 - 2} = \frac{-3}{2}\)

= \(\frac{y - y_1}{x- x_1}\)

m = \(\frac{y - 3}{x}\) \(\geq\) \(\frac{-3}{2}\)

2(y - 3) \(\geq\) - 3x = 2y - 6 \(\geq\) - 3x

= 2y + 3x \(\leq\) 6 ; x \(\leq\) 0, y \(\leq\) 0

595.

In the venn diagram, the shaded region is?

A.

(P \(\cap\) Q) \(\cup\) R

B.

(P \(\cap\) Q) \(\cap\) R

C.

(P \(\cap\) Q1) \(\cap\) R

D.

(P \(\cap\) Q1) \(\cup\) R

Correct answer is C

No explanation has been provided for this answer.