WAEC Mathematics Past Questions & Answers - Page 124

616.

The shaded portion in the diagram is the solution of

A.

x + y \(\leq\) 3

B.

x + y < 3

C.

x + y > 3

D.

x + y \(\geq\) 3

Correct answer is B

Using \(\frac{x}{a} + \frac{y}{b}\) < 1for the equation of the time

where a = intercept on x-axis and b = intercept on y - axis

\(\frac{x}{3} + \frac{y}{3} = 1\)

= \(\frac{x + y}{3} = 1\)

= x + y < 3

617.

In the diagram, O is the centre of the circle, < SQR = 60o, < SPR = y and < SOR = 3x. Find the value of (x + y)

A.

110o

B.

100o

C.

80o

D.

70o

Correct answer is B

3x = 2 x 60 = 2y (Angle at centre = 2 x angle at circumference)

3x = 2 x 60

x = \(\frac{2 \times 60}{3}\) = 40o

2 x 60 = 2y

y = 60o

x + y = 40 + 60

= 100o

618.

In the diagram, the tangent MN makes an angle of 55o with the chord PS. IF O is the centre of the circle, find < RPS

A.

55o

B.

45o

C.

35o

D.

25o

Correct answer is C

Join SR

< PRS = 90o(Angle in a semicircle)

< PRS = 55o (Angle between a chord and a tangent = Angle in the alternate segment)

< PSR + < PRS + < RSP = 180o

90v + 55o + < RSp = 180o

< RSP = 180o - 145o

= 35o

619.

In the diagram, < ROS = 66o and < POQ = 3x. some construction lines are shown. Calculate the value of x.

A.

10o

B.

11o

C.

22o

D.

35o

Correct answer is B

From the diagram, OP bisects < ROS

< POS = \(\frac{1}{2}\) < ROS = \(\frac{1}{2}\) x 66o

3x = 33o

x = \(\frac{33^o}{3}\)

= 11o

620.

In the diagram, triangles HKL and HIJ are similar. Which of the following ratios is equal to \(\frac{LH}{JH}\)

A.

\(\frac{KL}{JI}\)

B.

\(\frac{HK}{JK}\)

C.

\(\frac{JI}{KL}\)

D.

\(\frac{HK}{LK}\)

Correct answer is A

\(\bigtriangleup\) is similar to \(\bigtriangleup\) HIJ

< HKL = HJI = xo

Hence, \(\frac{LH}{JH} = \frac{KH}{JH} \frac{KL}{IJ}\)

\(\frac{LH}{JH} = \frac{KL}{JI}\)