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JAMB Mathematics Past Questions & Answers - Page 114

566.

TQ is tangent to circle XYTR, < YXT = 32o, RTQ = 40o. find < YTR

108^o

121^o

140^o

148^o

View answer & explanation

Correct answer is A

< TWR = < QTR = 40o (alternate segment)

< TWR = < TXR = 40o(Angles in the same segments)
< YXR = 40o + 32o = 72o

< YXR + < YTR = 180o(Supplementary)

72o + < YTR = 180o

< YTR = 180o - 72o

= 108o

567.

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.

35cm²

65cm²

70cm²

140cm²

View answer & explanation

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 $^2$

568.

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

View answer & explanation

Correct answer is C

x + 60o + 50o = 180o

x = 110o = 180o

x = 180o - 110o

= 70o

569.

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

20^o

25^o

30^o

35^o

View answer & explanation

Correct answer is B

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

∴ < TSR = 90^o

< SRT = 90^o - 30^o = 60^o

< PRS = 90^o - 35^o = 55^o

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

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

< QSR + < SRQ + < SQR = 180^o

< QSR + 120^o + 35^o = 180^o

< QSR + 155 = 180^o

< QSR = 180^o - 155^o

= 25^o

570.

The shaded area represents

x $\leq$ 0, y $\leq$ 0, 2y + 3x $\leq$ 6

x $\geq$ 0, y $\geq$ 3, 3x + 2y $\geq$ 6

x $\geq$ 2, y $\geq$ 0, 3x + 2y $\leq$ 6

x $\geq$ 0, y $\geq$ 0, 3x + 2y $\geq$ 6

View answer & explanation

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