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Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Solution
The sum of the interior angles of a polygon is 20 right angles. How many sides does the polygon have?
A
12
B
20
C
40
D
10
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The formula for the sum of the interior angles of a regular polygon = (2n - 4) x 90^\circ
Given: Sum = 20 right angles
(2n - 4) x 90^\circ = 20 \times 90^\circ
⇒ 2n - 4 = 20
2n = 24; n = 12.
Source: Myschool Past Questions Archive (Record #3378)
Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Solution
A hunter 1.6 m tall, views a bird on top of a tree at an angle of 45^\circ. if the distance between the hunter and the tree is 10.4 m, find the height of the tree.
A
9.0 m
B
12. 0 m
C
8.8 m
D
10.4 m
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
\(\tan 45 = \frac{x}{10.4}\) = 10.4m
\(height of the tree is equal to 1.6m + 10.4m=12m\)
Source: Myschool Past Questions Archive (Record #3380)
Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Solution
A solid hemisphere has a radius of 7 cm. Find the total surface area.
A
400 cm2
B
462 cm2
C
66 cm2
D
308 cm2
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Total surface area of hemisphere = half of TSA of sphere + circumference of circle
= \(2\pi r^{2} + \pi r^{2} = 3\pi r^{2}\)
= \( 3 \times \frac{22}{7} \times 7^{2} = 462cm^{2}\)
Source: Myschool Past Questions Archive (Record #3381)
Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Solution
The locus of a point P which is equidistant from two given points S and T is
A
the perpendicular bisector of ST
B
the angle bisector of PS and ST
C
a perpendicular to ST
D
a line parallel to ST
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
The locus of a point P that is equidistant from two points S and T is the perpendicular bisector of the segment ST. This means that every point on this line is the same distance from both S and T, effectively dividing the segment into two equal parts and forming a right angle with it.
Source: Myschool Past Questions Archive (Record #3383)
Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Solution
The probability of a student passing any exam is 2/3. If the student takes three exams, what is the probability that he will not pass any of them?
A
2/3
B
4/9
C
8/27
D
1/27
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
P(not passing exam) = \(1 - \frac{2}{3} = \frac{1}{3}\)
P(not passing any of the three exams) = \(\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}\)
= \(\frac{1}{27}\)
Source: Myschool Past Questions Archive (Record #3388)
Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Solution
The acres for rice, pineapple, cassava, cocoa, and palm oil in a certain district are given respectively as 2, 5, 3, 11, and 9. What is the angle sector for cassava in a pie chart?
A
108°
B
180°
C
36°
D
60°
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
Total: 2 + 5 + 3 + 11 + 9 = 30
Cassava: \(\frac{3}{30} \times 360^\circ = 36^\circ\)
Source: Myschool Past Questions Archive (Record #3390)
Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram above , XZ is the diameter of the circle XZW, with center O and radius 15/2 cm. If XY = 12 cm, find the area of the triangle XYZ
Attached Figure / Diagram
A
54 cm2
B
45 cm2
C
27 cm2
D
75 cm2
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Since radius = \(\frac{15}{2}\) diameter = 2 * (\(\frac{15}{2}\)) = 15cm
XYZ is a right-angled triangle (since
Source: Myschool Past Questions Archive (Record #7728)
Mathematics
NECO SSCE 2002
Geometry and Trigonometry
Solution
A wire was shaped to form a square of area 81cm\(^2\). An equilateral triangle was formed by the same wire,What is the length of a side of the triangle formed?
A
3cm
B
6cm
C
9cm
D
12cm
E
27cm
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
Area of a square = L X L = 81
: L = 9
Perimeter of square = 4 X L
4 X 9 = 36cm
An equilateral triangle with equal sides:
36/3 = 12cm
Source: WAEC / JAMB / NECO Classroom Archive (Record #42139)
Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram above are two concentric circles of radii r and R respectively with center O. If r = 2/3R, express the area of the shaded portion in terms of π and R
Attached Figure / Diagram
A
21/25πR2
B
9/25πR2
C
21/23πR2
D
5/9πR2
medium Difficulty
Verified Correct Answer: Option D
Step-by-Step Mathematical Solution
r = \(\frac{2}{3}\)R
∴R = \(\frac{3}{3}\)R
Area of small circle = πr2
= π(\(\frac{2R}{3}\))2
Area of the big circle πr2 = π\(\frac{(3R)^2}{3}\)
Area of shaded portion = π(\(\frac{3R}{3}\))2 - π(\(\frac{2R}{3}\))2
= π[(\(\frac{3R}{3}\))2 - (\(\frac{2R}{3}\))2]
= π[(\(\frac{3R}{3}) + (\frac{2R}{3}) - (\frac{3R}{3}\)) - (\(\frac{2R}{3}\))]
= π[(\(\frac{5R}{3}\)) (\(\frac{R}{3}\))]
= π x \(\frac{5R}{3}\) x \(\frac{R}{3}\)
= 5/9πR2
Source: Myschool Past Questions Archive (Record #10928)
Mathematics
UTME / JAMB 2002
Geometry and Trigonometry
Diagram
Solution
The triangle PQR above is
Attached Figure / Diagram
A
an obtuse-angled triangle
B
a scalene triangle
C
an isosceles triangle
D
an equilateral triangle
medium Difficulty
Verified Correct Answer: Option C
Step-by-Step Mathematical Solution
∠PQR = 180 - 128 (∠s on a straight line)
= 52o
∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52o
∴ΔPQR is an isosceles triangle
Source: Myschool Past Questions Archive (Record #10930)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram O is the center of the circle. Reflex angle XOY = 210^\circ and the length of the minor arc is 5.5m. Find, correct to the nearest meter, the length of the major arc.
Attached Figure / Diagram
A
8m
B
9m
C
10m
D
13m
medium Difficulty
Verified Correct Answer: Option A
Step-by-Step Mathematical Solution
Given, Length of minor arc = 5.5m
Angle subtended by minor arc = 360^\circ - 210^\circ = 150^\circ
\(\therefore 5.5 = \frac{150}{360} \times 2 \times \frac{22}{7} \times r \)
\(\frac{55r}{21} = 5.5\)
\(r = \frac{5.5 \times 21}{55}\)
r = 2.1m
Length of major arc = \(\frac{210}{360} \times 2 \times \frac{22}{7} \times 2.1\)
= \(7.7m \approxeq 8m\) (to the nearest metre)
Source: Myschool Past Questions Archive (Record #26511)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
In the diagram above, ∠PQU=36^\circ, ∠QRT = 29^\circ, PQ||RT. Find ∠PQR
Attached Figure / Diagram
A
94o
B
65o
C
61o
D
54o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
< UQR = 29^\circ (alternate angles)
< PQR = < PQU + < UQR
= 36^\circ + 29^\circ
= 65^\circ
Source: Myschool Past Questions Archive (Record #26521)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Diagram
Solution
Find the value of x in the diagram
Attached Figure / Diagram
A
31o
B
35o
C
37o
D
41o
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
Sum of exterior angle of any polygon is 360o
(2x+5)o + 2xo + (x-20)o + xo + (3x+10)o + (x + 15)o = 360o; 10x = 350
x = 35
Source: Myschool Past Questions Archive (Record #26526)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Solution
The length, in cm, of the sides of a right angled triangle are x, (x+2) and (x+1) where x > 0. Find , in cm, the length of its hypotenuse
A
4
B
5
C
13
D
17
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
The longest side of a right angle triangle is the hypotenuse.
\((x + 2)^{2} = x^{2} + (x + 1)^{2}\)
\(x^2 + 4x + 4 = x^2 + x^2 + 2x + 1\)
\(2x^{2} - x^{2} + 2x - 4x + 1 - 4 = 0\)
\(x^{2} - 2x - 3 = 0\)
\(x^{2} - 3x + x - 3 = 0 \implies x(x - 3) + 1(x - 3) = 0\)
\((x - 3)(x + 1) = 0 \implies \text{x = 3 or -1}\)
\(x > 0 \implies x = 3\)
The longest side = 3 + 2 = 5.
Source: Myschool Past Questions Archive (Record #26531)
Mathematics
WAEC / WASSCE 2002
Geometry and Trigonometry
Solution
Points X, Y and Z are located in the same horizontal plane such that Y is 12 km north of X and Z is on a bearing of 270^\circ from X. If |XZ| = 6km, calculate, correct to one decimal place, lYZl
A
18km
B
13.4km
C
13km
D
10.4km
medium Difficulty
Verified Correct Answer: Option B
Step-by-Step Mathematical Solution
ZY\(^2\) = 12\(^2\) + 6\(^2\)
ZY\(^2\) = 144 + 36 = 180
ZY = \(\sqrt{180}\)
= \(13.416\)
= 13.4 km
Source: Myschool Past Questions Archive (Record #26533)