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Mathematics
UTME / JAMB 1990
Geometry and Trigonometry
Solution
Find the length of a side of a rhombus whose diagonals are 6cm and 8cm
A
8cm
B
5cm
C
4cm
D
3cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The diagonal of a rhombus is a line segment that joins any two non-adjacent vertices.
A rhombus has two diagonals that bisect each other at right angles.
i.e this splits 6cm into 3cm each AND 8cm to 4cm
Using Hyp\(^2\) = adj\(^2\) + opp\(^2\)
Hyp\(^2\) = 3\(^2\) + 4\(^2\)
Hyp\(^2\) = 25
Hyp = 5
∴ Length (L) is 5cm because a rhombus is a quadrilateral with 4 equal lengths
Source: WAEC / JAMB / NECO Classroom Archive (Record #31476)
Mathematics
UTME / JAMB 1990
Geometry and Trigonometry
Solution
The angle of a sector of s circle, radius 10.5cm, is 48°, Calculate the perimeter of the sector
A
8.8cm
B
25.4cm
C
25.6cm
D
29.8cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Length of Arc AB = \(\frac{\theta}{360}\) 2\(\pi\)r
= \(\frac{48}{360}\) x 2\(\frac{22}{7}\) x \(\frac{21}{2}\)
= \(\frac{4 \times 22 \times \times 3}{30}\) \(\frac{88}{10}\) = 8.8cm
Perimeter = 8.8 + 2r
= 8.8 + 2(10.5)
= 8.8 + 21
= 29.8cm
Source: WAEC / JAMB / NECO Classroom Archive (Record #31475)
Mathematics
UTME / JAMB 1990
Geometry and Trigonometry
Solution
Each of the interior angles of a regular polygon is 140°. How many sides has the polygon?
A
9
B
8
C
7
D
5
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
For a regular polygon of n sides
n = \(\frac{360}{\text{Exterior angle}}\)
Exterior < = 180^\circ - 140^\circ
= 40^\circ
n = \(\frac{360}{40}\)
= 9 sides
Source: WAEC / JAMB / NECO Classroom Archive (Record #31477)
Mathematics
UTME / JAMB 1990
Geometry and Trigonometry
Solution
The locus of a point which moves so that it is equidistant from two intersecting straight lines is the?
A
perpendicular bisector of the two lines
B
angle bisector of the two lines
C
bisector of the two lines
D
line parallel to the two lines
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The required locus is angle bisector of the two lines
Source: WAEC / JAMB / NECO Classroom Archive (Record #31481)
Mathematics
UTME / JAMB 1990
Geometry and Trigonometry
Solution
4, 16, 30, 20, 10, 14 and 26 are represented on a pie chart. Find the sum of the angles of the bisectors representing all numbers equals to or greater than 16
A
48o
B
84o
C
92o
D
276o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Given that 4, 16, 30, 20, 10, 14 and 26
Adding up = 120
nos \(\geq\) 16 are 16 + 30 + 20 + 26 = 92
The requires sum of angles = \(\frac{92}{120}\) x \(\frac{360^o}{1}\)
= 276o
Source: WAEC / JAMB / NECO Classroom Archive (Record #31483)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Solution
Express 0.00562 in standard form
A
5.62 x 10-3
B
5.62 x 10-2
C
562 x 10-2
D
5.62 x 102
E
5.62 x 103
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
0.00562 = 5.62 x 10\(^{-3}\)
Source: Myschool Past Questions Archive (Record #22601)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Solution
An arc of length 22cm subtends an angle of \theta at the center of the circle. What is the value of \theta if the radius of the circle is 15cm?[Take \pi = 22/7]
A
70o
B
84o
C
96o
D
156o
E
168o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(2\pi r = 2 \times 3.142 \times 15 = 94.28cm\)
\(94.28cm = 360^\circ\)
\(1cm = \frac{360^\circ}{94.28}\)
\(\therefore 22cm = \frac{360^\circ}{94.28} \times 22\)
= 84^\circ
Source: Myschool Past Questions Archive (Record #22643)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Solution
Let J be the set of positive integers, If H = {x: x∈J, x\(^2\) < 3 and x ≠ 0}, then
A
H = {1}
B
H is an infinite set
C
H = {0, 1, 2}
D
H = {}
E
J ≤ H
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
H = {x: x is a positive integer, x\(^2\) < 3 and x \(\neq\) 0}
H = {1}
Source: Myschool Past Questions Archive (Record #22645)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Diagram
Solution
In the diagram above, O is the center of the circle with radius 10cm, and ∠ABC = 30^\circ. Calculate, correct to 1 decimal place, the length of arc AC [Take \pi = 22/7]
Attached Figure / Diagram
A
5.2cm
B
10.5cm
C
13.2cm
D
20.6cm
E
31.4cm
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
In the figure, < AOC = 2 x < ABC = 60^\circ (angle subtended at the centre)
\(\therefore\) Arc AC = \(\frac{60}{360} \times 2 \times 10 \times \frac{22}{7}\)
= \(\frac{31.4}{3}\)
= 10.466 cm \(\approxeq\) 10.5 cm
Source: Myschool Past Questions Archive (Record #22650)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Solution
A sector of a circle of radius 7cm has an area of 44cm2. Calculate the angle of the sector correct to the nearest degree [Take π = 22/7]
A
6o
B
26o
C
52o
D
103o
E
206o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
πr2 = 360o
44cm2 = θ; ∴ θ = 44/πr2 x 360o = 102.9o = 103o
Source: Myschool Past Questions Archive (Record #22664)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Diagram
Solution
In the diagram above PS||RQ, |RQ| = 6.4cm and perpendicular PH = 3.2cm. Find the area of SQR
Attached Figure / Diagram
A
5.12cm2
B
9.60cm2
C
10.24cm2
D
20.48cm2
E
40.96cm2
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Area of a triangle = \(\frac{1}{2} \times b \times h\)
Area of \(\Delta\) SQR = \(\frac{1}{2} \times 6.4 \times 3.2\)
= 10.24 cm\(^2\)
Source: Myschool Past Questions Archive (Record #22675)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Diagram
Solution
Which angle is equal to ∠VWZ?
Attached Figure / Diagram
A
∠VXZ
B
∠VYX
C
∠XZW
D
∠YXZ
E
∠VYW
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
< VWZ = < VXZ (angles in the same segment)
Source: Myschool Past Questions Archive (Record #22676)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Diagram
Solution
In the diagram below, O is the center of the circle if ∠QOR = 290o, find the size ∠QPR
Attached Figure / Diagram
A
110o
B
70o
C
55o
D
35o
E
20o
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
360º - 290º = 70º
\(\frac{70}{2}\) = 35º
Source: Myschool Past Questions Archive (Record #22685)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Diagram
Solution
In the diagram above, PQ is a tangent at T to the circle ABT. ABC is a straight line and TC bisects ∠BTO. Find x.
Attached Figure / Diagram
A
20o
B
30o
C
35o
D
40o
E
50o
medium Difficulty
Verified Correct Answer: Option E
Step-by-Step Mathematical Solution
From the figure < TAB = < BTQ = 40^\circ (alternate segment)
\(\therefore\)< ATB = 180^\circ - (70^\circ + 40^\circ) = 70^\circ (angle on a straight line)
< BTC = \(\frac{40^\circ}{2} = \frac{< BTQ}{2}\)
\(\therefore < BTQ = 40^\circ\)
x^\circ = 180^\circ - (40^\circ + 70^\circ + 20^\circ)
= 50^\circ
Source: Myschool Past Questions Archive (Record #22687)
Mathematics
WAEC / WASSCE 1990
Geometry and Trigonometry
Diagram
Which of the sketches above gives a correct method for constructed an angle of 120o at the point P?
Attached Figure / Diagram
A
I only
B
II only
C
III only
D
I and II only
E
I, II and III
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #22690)