2010 - WAEC Mathematics Past Questions and Answers - page 5
41
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 option: 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
Users' Answers & Comments< POS = \(\frac{1}{2}\) < ROS = \(\frac{1}{2}\) x 66o
3x = 33o
x = \(\frac{33^o}{3}\)
= 11o
42
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 option: 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
Users' Answers & Comments< 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
43
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 option: 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
Users' Answers & Comments3x = 2 x 60
x = \(\frac{2 \times 60}{3}\) = 40o
2 x 60 = 2y
y = 60o
x + y = 40 + 60
= 100o
44
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 option: 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
Users' Answers & Commentswhere 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
45
In the diagram, GI is a tangent to the circle at H. If EF//GI, calculate the size of < EHF
A
126o
B
72o
C
64cmo
D
32cmo
correct option: b
F = 54o (Alternate triangle angles)
< GHE = F = 54o(Angle between a chord and a tangent = angles in the alternate segment)
Now, < GHE + < EHF + < IHF = 180o(Angles on a straight line)
54o + < EHF + 54o = 180o
< EHF = 180o - 108o
= 72o
Users' Answers & Comments< GHE = F = 54o(Angle between a chord and a tangent = angles in the alternate segment)
Now, < GHE + < EHF + < IHF = 180o(Angles on a straight line)
54o + < EHF + 54o = 180o
< EHF = 180o - 108o
= 72o
46
The diagram is a net right rectangular pyramid. Calculate the total surface area
A
208cm2
B
112cm2
C
92cm2
D
76cm2
correct option: c
Total surface area = sum of the area of the \(\bigtriangleup\) S + Area of the rectangle
= 2 x Area of \(\bigtriangleup\) PTQ + 2 x Area of \(\bigtriangleup\) QUR + Area of rectangle PQRS
2 x \(\frac{1}{2}\)(8 x 4) + 2 x \(\frac{1}{2}\)(5 x 4) + 8 x 5
= 32 + 20 + 40
= 92cm2
Users' Answers & Comments= 2 x Area of \(\bigtriangleup\) PTQ + 2 x Area of \(\bigtriangleup\) QUR + Area of rectangle PQRS
2 x \(\frac{1}{2}\)(8 x 4) + 2 x \(\frac{1}{2}\)(5 x 4) + 8 x 5
= 32 + 20 + 40
= 92cm2
47
The diagram shows a rectangular cardboard from which a semi-circle is cut off. Calculate the area of the remaining part
A
44cm2
B
99cm2
C
154cm2
D
165cm2
correct option: b
Area of remaining = Area of rectangle = Area of semi-circle
22 x 8 - \(\frac{1}{2}\)xar2
ehere r - \(\frac{14}{2}\)cm = 7cm
Area of remaining = 176 - \(\frac{1}{2}\) x \(\frac{22}{4}\) x 7 x 7
= 176 - 77
= 99cm2
Users' Answers & Comments22 x 8 - \(\frac{1}{2}\)xar2
ehere r - \(\frac{14}{2}\)cm = 7cm
Area of remaining = 176 - \(\frac{1}{2}\) x \(\frac{22}{4}\) x 7 x 7
= 176 - 77
= 99cm2
48
In the diagram, 0 is the centre of the circle. Find the value x
A
34
B
29
C
17
D
14
correct option: d
POQ in a straight line
Hence, < POQ + < QOR = 180o
56o + < QOR = 180o
< QOR = 180o - 56o
= 124o
Now, in \(\bigtriangleup\) QOR OR = OQ = Radius
< ORQ = < OQR = 2x (Base angles of an Isosceles \(\bigtriangleup\))
2x + 124 + 2x = 180o
4x + 124 = 180
4x = 180 - 124
4x = 56
x = \(\frac{56}{4}\)
x = 14o
Users' Answers & CommentsHence, < POQ + < QOR = 180o
56o + < QOR = 180o
< QOR = 180o - 56o
= 124o
Now, in \(\bigtriangleup\) QOR OR = OQ = Radius
< ORQ = < OQR = 2x (Base angles of an Isosceles \(\bigtriangleup\))
2x + 124 + 2x = 180o
4x + 124 = 180
4x = 180 - 124
4x = 56
x = \(\frac{56}{4}\)
x = 14o
49
In the diagram, < WOX = 60o, < YOE = 50o and < OXY = 30o. What is the bearing of x from y?
A
300o
B
240o
C
190o
D
150o
correct option: a
The bearing of x from y = 270o + \(\theta\)
where \(\theta\) + 50o = y
in \(\bigtriangleup\) OXY
O + X + Y = 180o
Where O = 40o + 30o = 70o
70o + 30o + y = 180o
y + 100o = 180o
y = 180o - 100o = 30o
\(\theta\) + 50o = 80o
80o - 50o = 30o
The bearing of x from y = 270o + 30o = 300o
Users' Answers & Commentswhere \(\theta\) + 50o = y
in \(\bigtriangleup\) OXY
O + X + Y = 180o
Where O = 40o + 30o = 70o
70o + 30o + y = 180o
y + 100o = 180o
y = 180o - 100o = 30o
\(\theta\) + 50o = 80o
80o - 50o = 30o
The bearing of x from y = 270o + 30o = 300o