JAMB Mathematics Past Questions & Answers - Page 144

716.

In the figure, FGHJ is a circle of radius 3cm centre O. FOH, GOJ are perpendicular diameters. With G as centre of an arc of a circle is drawn to pass through F and H. Find the length of the perimeter of the lunar portion shaded.

A.

3\(\pi\)\(\sqrt{2 - 1}\)cm

B.

\(\frac{9}{2}\)\(\pi\)cm

C.

3 \(\pi\)(1 + \(\frac{2}{2}\))

D.

3(1 + \(\frac{2}{2}\))

Correct answer is C

Perimeter of lunar portion = 3\(\pi\) + \(\frac{3\pi \sqrt{2}}{2}\)

= 3 \(\pi\)(1 + \(\frac{2}{2}\))

717.

FG is a given straight line and H is a fixed point. The construction marks shown in the diagram indicate the

A.

perpendicular bisector of the line

B.

perpendicular bisector of th eline JK

C.

perpendicular bisector of the line HI

D.

perpendicular from H to the line FG

Correct answer is D

No explanation has been provided for this answer.

718.

In the figure, PS bisects angle QPR. Find the ratio SR:QR.

A.

1:2

B.

1:3

C.

1:4

D.

1:5

E.

1:6

Correct answer is C

From Internal bisection theorem, \(\frac{QP}{PR}\) = \(\frac{QS}{SR}\)

= \(\frac{4}{1}\)

= 1:4

719.

In the figure, If PT is parallel to RS, PQ = PT, and angle SQT = 90o, Find x

A.

35o

B.

50o

C.

55o

D.

70o

E.

80o

Correct answer is D

No explanation has been provided for this answer.

720.

In the figure, PQ\\SR, ST\\, ST\\RQ, PS = 7cm, PT = 7cm, SR = 4cm. Find the ratio of the area QRST to the area of PQRS.

A.

56.77

B.

56.105

C.

28:105

D.

28:49

E.

56:49

Correct answer is B

Area of QRST = 4 x 7 = 28

Area of PQRS = \(\frac{1}{2}\)(4 + 11) x 7

= \(\frac{7}{2}\) x \(\frac{15}{1}\)

= 28:52.5

= 56:105