Mensuration | Jamb Mathematics

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Hello! As you prepare for your examination on Mensuration, focus on understanding formulas for calculating areas, perimeters, surface areas, and volumes of various geometric shapes. Practice solving problems involving 2D and 3D figures such as triangles, circles, cylinders, cones, and spheres to strengthen your application skills. Reviewing unit conversions and real-life applications of mensuration will also help you tackle a variety of questions effectively.

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Mensuration? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Mensuration together and move one step closer to achieving your exam success! Blissful learning.

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Calculation problems involving perimeters and areas of triangles, quadrilaterals, circles and composite figures

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Triangles

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1. Perimeter of a Triangle

Question: A triangle has sides of length

5

cm,

7

cm, and

9

cm. Find its perimeter.

Solution:
The perimeter of a triangle is the sum of its sides:

P = 5 + 7 + 9 = 21

cm.
Thus, the perimeter is $21$ cm.

2. Area of a Right-Angled Triangle

Question: Find the area of a right-angled triangle with base

6

cm and height

8

cm.

Solution:
Area of a triangle formula:

A = \frac{1}{2} \times \text{base} \times \text{height}

A = \frac{1}{2} \times 6 \times 8 = 24

cm².
Thus, the area is $24$ cm².

3. Area Using Heron's Formula

Question: A triangle has sides of

7

cm,

8

cm, and

9

cm. Find its area.

Solution:
First, compute the semi-perimeter:

s = \frac{7 + 8 + 9}{2} = \frac{24}{2} = 12

cm.

Using Heron’s formula:

A = \sqrt{s(s-a)(s-b)(s-c)}

A = \sqrt{12(12-7)(12-8)(12-9)}

A = \sqrt{12 \times 5 \times 4 \times 3}

A = \sqrt{720} = 26.83

cm².
Thus, the area is approximately $26.83$ cm².

4. Equilateral Triangle Area

Question: Find the area of an equilateral triangle with side

10

cm.

Solution:
Formula for area of an equilateral triangle:

A = \frac{\sqrt{3}}{4} s^2

A = \frac{\sqrt{3}}{4} \times 10^2 = \frac{\sqrt{3} \times 100}{4}

A \approx \frac{173.2}{4} = 43.3

cm².
Thus, the area is approximately $43.3$ cm².

5. Isosceles Triangle Perimeter

Question: An isosceles triangle has equal sides of

6

cm and a base of

8

cm. Find its perimeter.

Solution:

P = 6 + 6 + 8 = 20

cm.
Thus, the perimeter is $20$ cm.

Quadrilaterals

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6. Perimeter of a Square

Question: A square has a side length of

9

cm. Find its perimeter.

Solution:

P = 4 \times 9 = 36

cm.
Thus, the perimeter is $36$ cm.

7. Area of a Square

Question: Find the area of a square with a side length of

7

cm.

Solution:

A = 7^2 = 49

cm².
Thus, the area is $49$ cm².

8. Perimeter of a Rectangle

Question: A rectangle has a length of

12

cm and a width of

5

cm. Find its perimeter.

Solution:

P = 2(l + w) = 2(12 + 5) = 2 \times 17 = 34

cm.
Thus, the perimeter is $34$ cm.

9. Area of a Rectangle

Question: Find the area of a rectangle with a length of

14

cm and a width of

6

cm.

Solution:

A = l \times w = 14 \times 6 = 84

cm².
Thus, the area is $84$ cm².

10. Perimeter of a Parallelogram

Question: A parallelogram has opposite sides of length

10

cm and

6

cm. Find its perimeter.

Solution:

P = 2(a + b) = 2(10 + 6) = 2 \times 16 = 32

cm.
Thus, the perimeter is $32$ cm.

Circles

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11. Circumference of a Circle

Question: Find the circumference of a circle with a radius of

7

cm. (Use

\pi = 3.14

)

Solution:

C = 2\pi r = 2 \times 3.14 \times 7 = 43.96

cm.
Thus, the circumference is $43.96$ cm.

12. Area of a Circle

Question: Find the area of a circle with a diameter of

14

cm. (Use

\pi = 3.14

)

Solution:
Radius:

r = \frac{14}{2} = 7

cm.

A = \pi r^2 = 3.14 \times 7^2 = 3.14 \times 49 = 153.86

cm².
Thus, the area is $153.86$ cm².

13. Find Radius from Circumference

Question: The circumference of a circle is

62.8

cm. Find its radius. (Use

\pi = 3.14

)

Solution:

C = 2\pi r

62.8 = 2 \times 3.14 \times r

r = \frac{62.8}{6.28} = 10

cm.
Thus, the radius is $10$ cm.

14. Find Diameter from Area

Question: The area of a circle is

78.5

cm². Find its diameter. (Use

\pi = 3.14

)

Solution:

A = \pi r^2

78.5 = 3.14 r^2

r^2 = \frac{78.5}{3.14} = 25

r = 5

cm.
Thus, diameter = $2r = 10$ cm.

Composite Figures

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15. Area of a Composite Shape

Question: A rectangle (

8

\times

6

cm) has a semicircle (

\text{diameter } 6

cm) on top. Find the total area. (Use

\pi = 3.14

)

Solution:
Rectangle area:

A = 8 \times 6 = 48

cm².
Semicircle area:

A = \frac{1}{2} \pi r^2 = \frac{1}{2} \times 3.14 \times 3^2 = 14.13

cm².
Total area =

48 + 14.13 = 62.13

cm².
Thus, the total area is $62.13$ cm².

Triangles

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16. Find Base Given Area and Height

Question: The area of a triangle is

48

cm², and its height is

8

cm. Find the base.

Solution:
Using the formula:

A = \frac{1}{2} \times \text{base} \times \text{height}

48 = \frac{1}{2} \times b \times 8

b = \frac{48 \times 2}{8} = 12

cm.
Thus, the base is $12$ cm.

17. Find Height Given Area and Base

Question: The area of a triangle is

36

cm², and its base is

9

cm. Find the height.

Solution:

A = \frac{1}{2} \times b \times h

36 = \frac{1}{2} \times 9 \times h

h = \frac{36 \times 2}{9} = 8

cm.
Thus, the height is $8$ cm.

18. Right Triangle Hypotenuse

Question: A right triangle has legs

9

cm and

12

cm. Find its hypotenuse.

Solution:
Using Pythagoras' Theorem:

c^2 = a^2 + b^2 = 9^2 + 12^2

c^2 = 81 + 144 = 225

c = \sqrt{225} = 15

cm.
Thus, the hypotenuse is $15$ cm.

19. Isosceles Triangle Area Given Perimeter

Question: An isosceles triangle has a perimeter of

36

cm, with equal sides of

12

cm each. Find its area.

Solution:
Base =

36 - (12 + 12) = 12

cm.
Using Pythagoras' theorem to find height:

h^2 = 12^2 - 6^2 = 144 - 36 = 108

h = \sqrt{108} \approx 10.39

cm.

A = \frac{1}{2} \times 12 \times 10.39 = 62.34

cm².
Thus, the area is approximately $62.34$ cm².

Quadrilaterals

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20. Area of a Rhombus Given Diagonals

Question: A rhombus has diagonals of

12

cm and

16

cm. Find its area.

Solution:

A = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 12 \times 16 = 96

cm².
Thus, the area is $96$ cm².

21. Find Side Length of a Square Given Perimeter

Question: A square has a perimeter of

40

cm. Find its side length.

Solution:

P = 4s \Rightarrow s = \frac{40}{4} = 10

cm.
Thus, the side length is $10$ cm.

22. Find Diagonal of a Rectangle Given Sides

Question: A rectangle has sides of

8

cm and

6

cm. Find its diagonal.

Solution:
Using Pythagoras' Theorem:

d^2 = 8^2 + 6^2 = 64 + 36 = 100

d = \sqrt{100} = 10

cm.
Thus, the diagonal is $10$ cm.

23. Trapezium Area Given Heights and Bases

Question: A trapezium has bases of

8

cm and

12

cm, and height

5

cm. Find its area.

Solution:

A = \frac{1}{2} \times (a + b) \times h

A = \frac{1}{2} \times (8 + 12) \times 5 = \frac{1}{2} \times 20 \times 5 = 50

cm².
Thus, the area is $50$ cm².

Circles

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24. Find Radius Given Area

Question: A circle has an area of

314

cm². Find its radius. (Use

\pi = 3.14

)

Solution:

A = \pi r^2

314 = 3.14 \times r^2

r^2 = \frac{314}{3.14} = 100

r = \sqrt{100} = 10

cm.
Thus, the radius is $10$ cm.

25. Find Arc Length Given Radius and Angle

Question: A circle has a radius of

14

cm, and a sector subtends an angle of

60^\circ

. Find the arc length. (Use

\pi = 3.14

)

Solution:

L = \frac{\theta}{360^\circ} \times 2\pi r

L = \frac{60}{360} \times 2 \times 3.14 \times 14

L = \frac{1}{6} \times 87.92 = 14.65

cm.
Thus, the arc length is $14.65$ cm.

26. Find Sector Area Given Radius and Angle

Question: A sector of a circle has a radius of

10

cm and an angle of

45^\circ

. Find its area. (Use

\pi = 3.14

)

Solution:

A = \frac{\theta}{360^\circ} \times \pi r^2

A = \frac{45}{360} \times 3.14 \times 10^2

A = \frac{1}{8} \times 314 = 39.25

cm².
Thus, the sector area is $39.25$ cm².

Composite Figures

27. Find Perimeter of a Semi-Circle

Question: A semicircle has a diameter of

20

cm. Find its perimeter. (Use

\pi = 3.14

)

Solution:

P = \pi r + 2r

P = 3.14 \times 10 + 20 = 51.4

cm.
Thus, the perimeter is $51.4$ cm.

28. Find Area of a Quarter Circle

Question: A quarter-circle has a radius of

7

cm. Find its area. (Use

\pi = 3.14

)

Solution:

A = \frac{1}{4} \times \pi r^2

A = \frac{1}{4} \times 3.14 \times 7^2

A = \frac{1}{4} \times 3.14 \times 49 = 38.47

cm².
Thus, the area is $38.47$ cm².

29. Composite Shape with Rectangle and Triangle

Question: A rectangle

8

\times

6

cm has a right-angled triangle on top with a base of

8

cm and height of

4

cm. Find the total area.

Solution:
Rectangle area:

A = 8 \times 6 = 48

cm².
Triangle area:

A = \frac{1}{2} \times 8 \times 4 = 16

cm².
Total area:

48 + 16 = 64

cm².
Thus, the total area is $64$ cm².

Arc Length

1. Find Arc Length Given Radius and Angle

Question: A circle has a radius of

10

cm, and a sector subtends an angle of

60^\circ

. Find the arc length. (Use

\pi = 3.14

)

Solution:
Using the formula for arc length:

L = \frac{\theta}{360^\circ} \times 2\pi r

L = \frac{60}{360} \times 2 \times 3.14 \times 10

L = \frac{1}{6} \times 62.8

L = 10.47

cm
Thus, the arc length is $10.47$ cm.

2. Find Arc Length Given Diameter and Angle

Question: A sector of a circle has a diameter of

14

cm and an angle of

45^\circ

. Find the arc length. (Use

\pi = 3.14

)

Solution:
Radius:

r = \frac{14}{2} = 7

cm.

L = \frac{45}{360} \times 2\pi r

L = \frac{1}{8} \times 2 \times 3.14 \times 7

L = \frac{1}{8} \times 43.96

L = 5.5

cm
Thus, the arc length is $5.5$ cm.

3. ### Find Angle Given Arc Length and Radius

Question: The arc length of a sector is

15.7

cm, and the radius is

10

cm. Find the angle subtended at the center. (Use

\pi = 3.14

)

Solution:

L = \frac{\theta}{360^\circ} \times 2\pi r

15.7 = \frac{\theta}{360} \times 2 \times 3.14 \times 10

15.7 = \frac{\theta}{360} \times 62.8

\theta = \frac{15.7 \times 360}{62.8}

\theta = 90^\circ

Thus, the angle is $90^\circ$ .

Chord Length

4. Find Chord Length Given Radius and Angle

Question: A chord subtends an angle of

60^\circ

at the center of a circle with a radius of

12

cm. Find the length of the chord.

Solution:
Using the chord length formula:

c = 2r \sin\left(\frac{\theta}{2}\right)

c = 2 \times 12 \times \sin\left(\frac{60}{2}\right)

c = 24 \times \sin(30^\circ)

c = 24 \times 0.5

c = 12

cm
Thus, the chord length is $12$ cm.

5. Find Chord Length Given Radius and Perpendicular Distance

Question: A chord in a circle of radius

10

cm is

8

cm away from the center. Find its length.

Solution:
Using Pythagoras' theorem:

r^2 = d^2 + \left(\frac{c}{2}\right)^2

10^2 = 8^2 + \left(\frac{c}{2}\right)^2

100 = 64 + \frac{c^2}{4}

\frac{c^2}{4} = 36

c^2 = 144

c = 12

cm
Thus, the chord length is $12$ cm.

Sector Perimeter

6. Find Perimeter of a Sector

Question: A sector has a radius of

14

cm and an angle of

90^\circ

. Find its perimeter. (Use

\pi = 3.14

)

Solution:

L = \frac{90}{360} \times 2\pi r

L = \frac{1}{4} \times 2 \times 3.14 \times 14

L = 21.98

cm
Total perimeter =

L + 2r = 21.98 + 28 = 49.98

cm
Thus, the perimeter is $49.98$ cm.

Sector Area

7. Find Area of a Sector Given Radius and Angle

Question: Find the area of a sector with a radius of

10

cm and an angle of

45^\circ

. (Use

\pi = 3.14

)

Solution:

A = \frac{\theta}{360} \times \pi r^2

A = \frac{45}{360} \times 3.14 \times 10^2

A = \frac{1}{8} \times 314

A = 39.25

cm²
Thus, the sector area is $39.25$ cm².

8. Find Angle Given Sector Area and Radius

Question: A sector has an area of

78.5

cm² and a radius of

10

cm. Find the angle. (Use

\pi = 3.14

)

Solution:

A = \frac{\theta}{360} \times \pi r^2

78.5 = \frac{\theta}{360} \times 3.14 \times 100

\theta = \frac{78.5 \times 360}{314}

\theta = 90^\circ

Thus, the angle is $90^\circ$ .

Segment Area

9. Find Area of a Minor Segment

Question: A circle has a radius of

7

cm, and a chord subtends an angle of

60^\circ

at the center. Find the area of the minor segment. (Use

\pi = 3.14

)

Solution:
Sector area:

A_s = \frac{60}{360} \times \pi r^2

A_s = \frac{1}{6} \times 3.14 \times 49

A_s = 25.66

cm²

Triangle area:

A_t = \frac{1}{2} \times r^2 \times \sin(60^\circ)

A_t = \frac{1}{2} \times 49 \times 0.866

A_t = 21.22

cm²

Segment area:

A = A_s - A_t = 25.66 - 21.22 = 4.44

cm²
Thus, the segment area is $4.44$ cm².

10. Find Area of a Major Segment

Question: Using the above problem, find the major segment area.

Solution:
Total circle area:

A = \pi r^2 = 3.14 \times 49 = 153.86

cm²
Major segment area = Total circle area - Minor segment area

A = 153.86 - 4.44 = 149.42

cm²
Thus, the major segment area is $149.42$ cm².

Arc Length Problems

11. Find Arc Length Given Central Angle and Circumference

Question: A circle has a circumference of

125.6

cm. Find the arc length subtended by an angle of

72^\circ

. (Use

\pi = 3.14

)

Solution:

L = \frac{\theta}{360^\circ} \times C

L = \frac{72}{360} \times 125.6

L = \frac{1}{5} \times 125.6

L = 25.12

cm
Thus, the arc length is $25.12$ cm.

12. Find Arc Length Given Chord Length and Radius

Question: A chord of a circle has a length of

10

cm, and the radius of the circle is

8

cm. Find the arc length subtended by the chord. (Use

\pi = 3.14

)

Solution:
Using the formula for the central angle:

\theta = 2 \arcsin\left(\frac{c}{2r}\right)

\theta = 2 \arcsin\left(\frac{10}{16}\right)

\theta \approx 2 \times 38.68^\circ = 77.36^\circ

Now, arc length:

L = \frac{77.36}{360} \times 2\pi r

L = \frac{77.36}{360} \times 2 \times 3.14 \times 8

L \approx 10.8

cm
Thus, the arc length is approximately $10.8$ cm.

Chord Length Problems

13. Find Chord Length Given Central Angle and Radius

Question: A circle has a radius of

9

cm, and a chord subtends a central angle of

120^\circ

. Find the chord length.

Solution:
Using the chord length formula:

c = 2r \sin\left(\frac{\theta}{2}\right)

c = 2 \times 9 \times \sin(60^\circ)

c = 18 \times 0.866

c = 15.59

cm
Thus, the chord length is $15.59$ cm.

14. Find Perpendicular Distance from Center to Chord

Question: A chord of length

12

cm is drawn in a circle with a radius of

10

cm. Find the perpendicular distance from the center to the chord.

Solution:
Using Pythagoras' theorem:

r^2 = d^2 + \left(\frac{c}{2}\right)^2

10^2 = d^2 + 6^2

100 = d^2 + 36

d^2 = 64

d = 8

cm
Thus, the perpendicular distance is $8$ cm.

Sector Perimeter Problems

15. Find Perimeter of a Quarter-Circle Sector

Question: A sector of a circle has a radius of

14

cm and an angle of

90^\circ

. Find its perimeter. (Use

\pi = 3.14

)

Solution:
Arc length:

L = \frac{90}{360} \times 2\pi r

L = \frac{1}{4} \times 2 \times 3.14 \times 14

L = 21.98

cm
Total perimeter:

P = L + 2r = 21.98 + 28 = 49.98

cm
Thus, the perimeter is $49.98$ cm.

16. Find Perimeter of a Sector With a 60-Degree Angle

Question: A sector of a circle has a radius of

12

cm and subtends an angle of

60^\circ

at the center. Find its perimeter. (Use

\pi = 3.14

)

Solution:
Arc length:

L = \frac{60}{360} \times 2\pi r

L = \frac{1}{6} \times 2 \times 3.14 \times 12

L = 12.56

cm
Total perimeter:

P = L + 2r = 12.56 + 24 = 36.56

cm
Thus, the perimeter is $36.56$ cm.

Sector Area Problems

17. Find Area of a Half-Circle Sector

Question: A semicircle has a radius of

10

cm. Find its area. (Use

\pi = 3.14

)

Solution:

A = \frac{1}{2} \pi r^2

A = \frac{1}{2} \times 3.14 \times 10^2

A = 157

cm²
Thus, the sector area is $157$ cm².

18. Find Sector Area Given Arc Length

Question: A sector of a circle has an arc length of

14

cm and a radius of

7

cm. Find its area. (Use

\pi = 3.14

)

Solution:
Using the area formula:

A = \frac{1}{2} L r

A = \frac{1}{2} \times 14 \times 7

A = 49

cm²
Thus, the sector area is $49$ cm².

Segment Area Problems

19. Find Area of a Minor Segment Given Radius and Chord Length

Question: A circle has a radius of

8

cm, and a chord of length

10

cm. Find the area of the minor segment. (Use

\pi = 3.14

)

Solution:
Find central angle:

\theta = 2 \arcsin\left(\frac{c}{2r}\right)

\theta = 2 \arcsin\left(\frac{10}{16}\right)

\theta \approx 76.5^\circ

Sector area:

A_s = \frac{76.5}{360} \times \pi r^2

A_s = \frac{76.5}{360} \times 3.14 \times 64

A_s \approx 42.8

cm²

Triangle area:

A_t = \frac{1}{2} \times 10 \times 7.48

A_t = 37.4

cm²

Segment area:

A = A_s - A_t = 42.8 - 37.4 = 5.4

cm²
Thus, the segment area is $5.4$ cm².

20. Find Area of a Major Segment

Question: Using the previous problem, find the area of the major segment.

Solution:
Total circle area:

A = \pi r^2 = 3.14 \times 64 = 201.06

cm²
Major segment area = Total circle area - Minor segment area

A = 201.06 - 5.4 = 195.66

cm²
Thus, the major segment area is $195.66$ cm².

Calculation problems involving total surface areas and volumes of cuboids, cylinders. cones, pyramids, prisms, spheres and composite figures;

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Cuboid Problems

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1. Find Surface Area of a Cuboid

Question: A cuboid has dimensions

6

\times

4

\times

8

cm. Find its total surface area.

Solution:
Total surface area of a cuboid:

A = 2 (lw + lh + wh)

A = 2 (6 \times 4 + 6 \times 8 + 4 \times 8)

A = 2 (24 + 48 + 32)

A = 2 \times 104 = 208

cm²
Thus, the surface area is $208$ cm².

2. Find Volume of a Cuboid

Question: A cuboid has a length of

5

cm, width of

3

cm, and height of

10

cm. Find its volume.

Solution:

V = lwh

V = 5 \times 3 \times 10

V = 150

cm³
Thus, the volume is $150$ cm³.

Find Edge Length Given Volume

Question: A cube has a volume of

216

cm³. Find its edge length.

Solution:
Volume of a cube:

V = s^3

s^3 = 216

s = \sqrt[3]{216} = 6

cm
Thus, the edge length is $6$ cm.

Cylinder Problems

4. Find Surface Area of a Cylinder

Question: A cylinder has a radius of

7

cm and a height of

10

cm. Find its total surface area. (Use

\pi = 3.14

)

Solution:

A = 2\pi r (r + h)

A = 2 \times 3.14 \times 7 \times (7 + 10)

A = 2 \times 3.14 \times 7 \times 17

A = 748.58

cm²
Thus, the surface area is $748.58$ cm².

5. Find Volume of a Cylinder

Question: A cylinder has a height of

12

cm and a base radius of

5

cm. Find its volume. (Use

\pi = 3.14

)

Solution:

V = \pi r^2 h

V = 3.14 \times 5^2 \times 12

V = 3.14 \times 25 \times 12

V = 942

cm³
Thus, the volume is $942$ cm³.

Cone Problems

6. Find Surface Area of a Cone

Question: A cone has a slant height of

13

cm and base radius of

5

cm. Find its total surface area. (Use

\pi = 3.14

)

Solution:

A = \pi r (r + l)

A = 3.14 \times 5 \times (5 + 13)

A = 3.14 \times 5 \times 18

A = 282.6

cm²
Thus, the surface area is $282.6$ cm².

7. Find Volume of a Cone

Question: A cone has a height of

9

cm and a base radius of

6

cm. Find its volume. (Use

\pi = 3.14

)

Solution:

V = \frac{1}{3} \pi r^2 h

V = \frac{1}{3} \times 3.14 \times 6^2 \times 9

V = \frac{1}{3} \times 3.14 \times 36 \times 9

V = 339.12

cm³
Thus, the volume is $339.12$ cm³.

Pyramid Problems

8. Find Surface Area of a Square Pyramid

Question: A square pyramid has a base length of

10

cm and a slant height of

13

cm. Find its total surface area.

Solution:

A = b^2 + 2bl

A = 10^2 + 2 \times 10 \times 13

A = 100 + 260

A = 360

cm²
Thus, the surface area is $360$ cm².

Find Volume of a Square Pyramid

Question: A square pyramid has a base side of

6

cm and height of

10

cm. Find its volume.

Solution:

V = \frac{1}{3} b^2 h

V = \frac{1}{3} \times 6^2 \times 10

V = \frac{1}{3} \times 36 \times 10

V = 120

cm³
Thus, the volume is $120$ cm³.

Sphere Problems

10. Find Surface Area of a Sphere

Question: A sphere has a radius of

7

cm. Find its surface area. (Use

\pi = 3.14

)

Solution:

A = 4\pi r^2

A = 4 \times 3.14 \times 7^2

A = 4 \times 3.14 \times 49

A = 615.44

cm²
Thus, the surface area is $615.44$ cm².

11. Find Volume of a Sphere

Question: A sphere has a diameter of

12

cm. Find its volume. (Use

\pi = 3.14

)

Solution:
Radius:

r = \frac{12}{2} = 6

V = \frac{4}{3} \pi r^3

V = \frac{4}{3} \times 3.14 \times 6^3

V = \frac{4}{3} \times 3.14 \times 216

V = 904.32

cm³
Thus, the volume is $904.32$ cm³.

Composite Figure Problems

12. Find Volume of a Hemisphere

Question: A hemisphere has a radius of

5

cm. Find its volume. (Use

\pi = 3.14

)

Solution:

V = \frac{1}{2} \times \frac{4}{3} \pi r^3

V = \frac{1}{2} \times \frac{4}{3} \times 3.14 \times 5^3

V = \frac{1}{2} \times \frac{4}{3} \times 3.14 \times 125

V = 261.67

cm³
Thus, the volume is $261.67$ cm³.

13. Find Surface Area of a Hemisphere

Question: A hemisphere has a radius of

7

cm. Find its total surface area. (Use

\pi = 3.14

)

Solution:

A = 3\pi r^2

A = 3 \times 3.14 \times 7^2

A = 3 \times 3.14 \times 49

A = 461.58

cm²
Thus, the surface area is $461.58$ cm².

Calculation problems involving the distance between two points on the earth’s surface

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Key Formulas Used:

Great Circle Distance (Haversine Formula)
Given two points with latitudes and longitudes:
- Point 1: $(\phi_1, \lambda_1)$
- Point 2: $(\phi_2, \lambda_2)$
  The great-circle distance is given by:
$d = 2R \arcsin\left(\sqrt{\sin^2\left(\frac{\Delta\phi}{2}\right) + \cos(\phi_1) \cos(\phi_2) \sin^2\left(\frac{\Delta\lambda}{2}\right)}\right)$

where:
- $\Delta\phi = \phi_2 - \phi_1$
- $\Delta\lambda = \lambda_2 - \lambda_1$
- $R$ is Earth's radius ( $6371$ km or $3958.8$ miles).
Approximate Formula for Short Distances
If the latitude difference is small, use:

$d \approx R \sqrt{(\Delta\phi)^2 + (\cos\phi \cdot \Delta\lambda)^2}$

1. Find Distance Between Two Cities

Question: Find the great-circle distance between New York City (40.7128° N, 74.0060° W) and Los Angeles (34.0522° N, 118.2437° W). Use Earth's radius as 6371 km.

Solution:
Given:

\phi_1 = 40.7128^\circ, \lambda_1 = -74.0060^\circ

\phi_2 = 34.0522^\circ, \lambda_2 = -118.2437^\circ

\Delta\phi = 34.0522 - 40.7128 = -6.6606^\circ

\Delta\lambda = -118.2437 + 74.0060 = -44.2377^\circ

Convert degrees to radians:

\Delta\phi = -6.6606 \times \frac{\pi}{180} = -0.1163

rad

\Delta\lambda = -44.2377 \times \frac{\pi}{180} = -0.7721

rad

Using the Haversine formula:

a = \sin^2\left(\frac{\Delta\phi}{2}\right) + \cos(40.7128^\circ) \cos(34.0522^\circ) \sin^2\left(\frac{\Delta\lambda}{2}\right)

a = \sin^2(-0.05815) + \cos(40.7128) \cos(34.0522) \sin^2(-0.38605)

a \approx 0.09241

c = 2 \arctan2\left(\sqrt{a}, \sqrt{1-a}\right)

c \approx 0.6178

d = 6371 \times 0.6178

d \approx 3936

Thus, the distance is approximately 3936 km.

2. Find Distance Between Two Points on the Same Longitude

Question: Find the distance between Paris (48.8566° N, 2.3522° E) and London (51.5074° N, 2.3522° E).

Solution:
Since both cities have the same longitude, the distance is calculated using:

d = R \times |\Delta\phi|

\Delta\phi = 51.5074 - 48.8566 = 2.6508^\circ

Convert to radians:

\Delta\phi = 2.6508 \times \frac{\pi}{180} = 0.04626

rad

d = 6371 \times 0.04626

d \approx 295

Thus, the distance is approximately 295 km.

3. Find Distance Across the Equator

Question: Find the distance along the equator between (0° N, 10° W) and (0° N, 50° W).

Solution:
At the equator,

d = R \times |\Delta\lambda|

\Delta\lambda = |50 - 10| = 40^\circ

Convert to radians:

\Delta\lambda = 40 \times \frac{\pi}{180} = 0.6981

rad

d = 6371 \times 0.6981

d \approx 4448

Thus, the distance is approximately 4448 km.

4. Find Distance Between Two Opposite Points (Antipodes)

Question: What is the maximum possible distance between two points on Earth?

Solution:
The maximum distance is the Earth's diameter (half of the full circumference):

d = 2R = 2 \times 6371 = 12742

Thus, the maximum possible distance is 12,742 km.

5. Find Distance Along a Meridian

Question: Find the distance between Cairo (30.0444° N, 31.2357° E) and Nairobi (1.2921° S, 36.8219° E).

Solution:

\Delta\phi = 30.0444 + 1.2921 = 31.3365^\circ

Convert to radians:

\Delta\phi = 31.3365 \times \frac{\pi}{180} = 0.5470

rad

d = 6371 \times 0.5470

d \approx 3486

Thus, the distance is approximately 3486 km.

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