Mensuration notes and_ solved problems

Lesson on Mensuration for students studying for WASSCE, NECO & JAMB exams.

1. Mensuration
Lengths and perimeters
Pythagoras Theorem
The Pythagoras theorem states that in any right triangle (also called right angled triangle), the
square of the length of the hypotenuse, which is the side opposite the 90°, is equal to the sum of the
squares of the other two sides.
2 2 2
a b c

Thus, if the lengths of the two smaller sides of a right triangle are 3 units and 4 units, we can find the
length of the hypotenuse by using 2 2 2 2
3 4 9 16 25 25 5 unitsc c c        

Similarly, if the length of one side and the length of the hypotenuse are known in a right triangle, the
length of the other side can be calculated using the same equation.
Sine Rule
The sine rule applies to any triangle. It states that if A, B and C are the three angles in a triangle and
a, b and c are the lengths of the sides opposite to angles A, B and C respectively, then sin sin sin
a b c
A B C



Thus if we know the measure of one angle and the length of the opposite side, given the measure of
another angle we can calculate the length of the side opposite to it or conversely, given the length of
a side we can find the measure of the angle opposite to it.
Cosine Rule
For the triangle as shown above, the cosine rule states that

2. 2 2 2
2 2 2
2 2 2
2 cos
2 cos
2 cos
a b c bc A
b a c ac B
c a b ab C
  
  
  
Parts of a circle
Figures below show the different parts of a circle.

The length of the diameter is twice the length of the radius. A circle is defined by the length of its
radius (or diameter). Note that the diameter is a special chord passing through the centre of the
circle.
Length of an arc and circumference
In any circle, the length of an arc is given by the formula sr where r is the length of the radius, s
is the length of the arc in the same unit as r and θ is the angle in radians (also called central angle)
the arc subtends at the centre of the circle as shown in the figure below.

As there are 2 radians in one full circle, the perimeter or length around a circle, also called its
circumference is given as 2Cr

Also, the angle in a full circle = 360°.
Thus, if an arc subtends an angle θ at the centre where θ is measured in degrees, the length of the
arc is given as 2
360
sr


s and r have the same unit.
Perimeter of a sector
Consider the sector of a circle as shown below:

3. The perimeter is given as the sum of the lengths of the two straight sides, each equal to the radius of
the circle and the length of the arc.
Thus, perimeter of the sector = 22
360
rr

 where θ is the central angle of the sector in
degrees.
Perimeter of a rectangle
A rectangle is a 4 sided closed shape (known as quadrilateral) where the opposite sides are equal in
length. All four angles are each equal to 90° in measure.

The perimeter of a rectangle is equal to the sum of the lengths of all four sides. As shown in the
picture above, if the length is a units and width is b units, the perimeter is 2a + 2b = 2(a + b)
Note that a square is a special type of rectangle in which all four sides measure the same length.
Perimeter of a square = 4a where a is the length of one side.

Perimeter of an irregular polygon on a grid
The perimeter of an irregular polygon can be estimated or calculated. Consider the polygon in the
grid as shown in the figure below.

4. If each grid is a square of side length 1 unit, then the shape can be broken down into several regular
shapes as shown. The lengths of the oblique sides can be calculated using the Pythagoras theorem.
Adding all of them up will give the perimeter of the polygon.

Thus, the perimeter of the shape (in black) is 5 5 3 1 5 2 1 2 7 2 3 5 units         

Unit of perimeter is the same as the unit of length.
Areas

Area is a measure of how big a flat shape is. You can think of it as a measure of the amount of paint
needed to colour the shape.
Area of a triangle
The area of a triangle can be calculated by multiplying half the length of any of its sides with the
perpendicular distance of that side with the opposite vertex. The side is called the base and the
perpendicular distance is referred to as the height or altitude of the triangle.
Please note that you can choose any side of the triangle as its base. According to your choice of the
base, the height is defined.

5. Area of a triangle = 1
2
bh where b is the base (length of any one side) and h is the height
(perpendicular distance of that side with the opposite vertex).
If the three angles in a triangle are A, B and C and the corresponding opposite sides are a, b and c
respectively, then the area can also be found as follows:

Area of a triangle = 1 1 1
sin sin sin
2 2 2
bc A ca B ab C
Area of a circle and sector
The area of a circle with radius r is given as 2
Ar

Therefore, the area of a sector that subtends an angle θ in degrees at the centre is 2
360
Ar



Area of a segment
The area of a segment can be found by subtracting the area of the triangle formed by the two points
on the circle and the centre from the area of the sector, as shown in the figure below.

6. Area of the segment shaded = Area of sector – Area of the triangle = 221
sin
360 2
rr


Area of a rectangle
The area of a rectangle is obtained by multiplying its length and width.

Area = ab
Area of a square
The area of a square is obtained by multiplying its length with length (as all four sides are the same
length).

Area =2
a
Area of a parallelogram
A parallelogram is a quadrilateral (four sided polygon) in which the opposite sides are parallel. The
area of a parallelogram is obtained by multiplying any one of its sides (called base) with the distance
between that side and the side parallel to it (called the height).

7. Area = bh
Note that a rectangle is a special type of a parallelogram with all four right angles.
Area of a trapezium
A trapezium is a quadrilateral in which only two sides are parallel.
The area of a trapezium is equal to one half the sum of the lengths of the parallel sides and the
distance between them.

Area of the trapezium = 
1
2
a b h
Area of a rhombus
A rhombus is a quadrilateral with all equal sides.


The area of a rhombus is equal to one half the product of the length of its diagonals.
In the figure above, area of the rhombus = 1
2
ab
Note that a square is a special type of a rhombus with all four right angles.
The unit of area is the square of the unit of length.
Area of an irregular shape
An irregular shape can be broken down into regular shapes. The area of the irregular shape is sum of
the areas of the regular shapes it is made up of.

8. Consider the irregular shape as shown in the figure below. Let each grid be 1 unit in length.

The shape is broken down into a set of regular shapes (triangles and rectangles in this case).


The area of the shape in black
= Area of triangle A + Area of square B + Area of triangle C + Area of rectangle D + Area of triangle E +
Area of triangle F 1 1 1 1
1 2 2 2 2 1 3 1 1 1 1 2
2 2 2 2
11
1 4 1 3 1 10 square units
22
               
      

Even if a grid is not used, to find the area of an irregular shape you need to break it down into
regular shapes, find each area and then add them up.
Consider the figure below. Assume all angles to be right angles.

9. To find its area we can divide it into rectangles A, B and C, as shown in the figure below.

Area of the whole shape = Area of rectangle A + Area of rectangle B + Area of rectangle C
You may have noticed that there are other ways to divide the shape into rectangles.
Note that the unit of area is the square of the unit in which the lengths are measured.
Scaling – Perimeter and Area of similar shapes

When a shape is scaled up (made larger) or scaled down (made smaller), we get a similar shape.
Similar shapes have the same ratio of corresponding sides. The measures of corresponding angles in
similar shapes are identical.

The triangles shown below are similar.

2
1
AB BC CA
PQ QR RP
  


Perimeter of ∆ABC : Perimeter of ∆PQR = (12 + 14 + 10) : (6 + 7 + 5) = 36 : 18 = 2 : 1
Thus, the perimeters of similar shapes have the same ratio as the ratio of corresponding sides.

10. For the same two triangles, the heights will also be in the same ratio, that is 2 : 1

Area of ∆ABC : Area of ∆PQR = 11
12 8: 6 4 48:12 4:1
22
     
Thus, the ratio of areas of similar shapes is the square of the ratio of their corresponding sides.
Surface Area and Volume of 3 dimension shapes

A 3 dimensional shape is one that has space inside it. It is not a planar shape.
The volume of a 3-d shape is the amount of space contained inside.
The surface area of a 3-d shape is the sum of the areas of the faces that make up the shape.
Prism
A prism is a solid with uniform cross-section. It is named based on the shape of the cross-section.
The thickness of the prism is often referred to as its height. A prism has two more faces than the
number of sides on the polygon that make up its cross section. All the faces other than the two ends
are always rectangles. Figure below shows some prisms.

The volume of a prism is the product of its cross sectional area and height.
If the prism is a square prism (with all equal sides, also called a cube) then the volume = 3
a where a
is the length of each side.
If the dimensions of a rectangular prism are , and ,l b h its volume = lbh

If the area of each rectangular face in a triangular prism is A and the height of the prism is h , then
its volume is Ah
Note that a cylinder is a prism with circular cross section, as shown in the figure below.

11. Volume of a cylinder with radius r and height h = 2
rh
The surface area of a square prism of side length a = 2
6a (as there are six faces each with area = 2
a )

The surface area of a rectangular prism with dimensions , and l b h  2 2 2 2lb bh lh lb bh lh     


The surface area of a cylinder with radius r and height h = 2 x area of each circular face + area of the
curved portion which is a rectangle with length = h and width = 2πr
=
2
2 2 2r rh r r h    
Pyramid
A pyramid is a shape that has got a flat base in the shape of a polygon and slanted sides that meet at
a point called the apex of the pyramid.
A pyramid has one more faces than the number of sides on the polygon that make up its base. All
faces except the base on a pyramid are triangles. The height of a pyramid is the distance from the
centre of its base to the apex. Figure below shows some pyramids.

The volume of a pyramid is equal to the product of one-third the area of its base and its height.
The surface area of a prism is equal to the sum of the areas of its base and other faces.
Note that a cone is a pyramid with a circular base. Figure below shows a cone.

12. The slant height of a cone is the distance of the tip (apex) to any point on its circular base.
If the height of a cone is h , its slant height is l and the radius of its base is r , then applying the
Pythagorean theorem we have 2 2 2
l h r
The volume of a cone, like any other pyramid, is equal to the product of one-third the area of its
base and its height. As the base is a circle with radius r , volume = 21
3
rh
The surface area of a cone is equal to the sum of the areas of the circular base and the curved face.
To find the curved surface area of a cone, consider opening up the cone along any line connecting
the tip and a point on the circular base. Once opened up, the cone will like the following figure.

The opened up cone without the circular base is a portion of a circle whose radius is the same as the
slant height l
The arc length is equal to the circumference of the circular base of radius r.
Thus, 2
2 Area of the sector with radius
2
rl
ll



Area of the sector with radius l =rl = curved surface area of the cone
Total surface area of the cone = curved surface area + area of circular base = 
2
rl r r l r    
Sphere
A sphere is like a ball – a circle in 3 dimension. Like a circle, a sphere can be defined by its radius,
which is the distance of its centre to any point on the sphere.

13. The surface area of a sphere of radius r is 2
4r and the volume of a sphere of radius r is 34
3
r
The Earth as a sphere
For most practical purposes, we can assume the earth to be a sphere (although in reality, it is not an
exact sphere). Let R represent the radius of the earth.

Any point on the earth’s surface is characterized by its longitude and latitude, which are similar to
the coordinates of a point on the coordinate grid.
Latitudes are horizontal circles that run parallel and have varying radii from R (at the equator) to 0
(at the poles). Thus, the latitude of a place on the equator is 0°. The latitude of the North Pole is 90°
north and the latitude of the South Pole is 90° south. The latitude of any place is a number in
between 0 and 90 in degrees, north or south.

The longitudes are circles of same size running north to south that intersect at the poles. They are
also referred to as ‘great circles’. Greenwich in the UK has a longitude of 0°. As there are 360° in a
full circle, a place to the east of Greenwich has a longitude less than 180° east and a place to the
west of Greenwich has a longitude of less than 180° west, thus covering an overall angle of 360°
round the earth.

14. To find the distance between two points P and Q on the same longitude, we can use the formula sR
where R is the radius of the earth and  is the difference of latitudes expressed in radians
of the two points. As latitudes are given in degrees, we need to convert them to radians by using the
relation 180° = π radians.

To find the distance between two points P and Q on the same latitude, say θ, we first need to find
the radius R’ of the circle at that latitude by applying simple trigonometry as shown in the figure
below.
' sin and 'R R s R
where again  is the difference of longitudes of the two points
expressed in radians. As longitudes are given in degrees, we need to convert them to radians by
using the relation 180° = π radians.
Volume of a compound shape
Just like area, the volume of a compound shape can be obtained by breaking it down into smaller
regular shapes and adding the volumes up.

Consider the shape as shown in the figure below.

15. To find the volume of this shape, you can break it down into rectangular prisms as shown in the
figure below.


Volume of the shape = Volume of rectangular prism A + Volume of rectangular prism B

Note that the unit of volume is the cube of the unit in which the lengths are measured.
Scaling – Volume of similar shapes

When a 3 dimensional shape is scaled up (made larger) or scaled down (made smaller), we get a
similar shape. Similar shapes have the same ratio of corresponding sides. The measures of
corresponding (solid) angles in similar shapes are identical.
Consider the two similar rectangular prisms in the figure below.


Ratio of lengths of corresponding sides = 864
2:1
432
  
Hence the two prisms are similar.

Volume of the smaller prism = 4 x 3 x 2 = 24 cubic units.
Volume of the larger prism = 8 x 6 x 4 = 192 cubic units.

16. Volume of the larger prism : Volume of smaller prism = 192 : 24 = 8 : 1

Thus, the ratio of volumes of similar shapes is the cube of the ratio of their corresponding sides.

Nets of 3 dimension shapes
When a 3 dimensional shape is opened up on a flat surface, we get a net of the shape. Following
figure shows nets of some common shapes.


Problems
1. Find the area of a square inscribed in a circle of radius r.
Answer: 2
2r
Solution:


If the radius of the circle is r then its diameter = 2r
As seen in the figure above, the length of a diagonal of the square = 2r
If each side of the square = x, then by applying the Pythagoras theorem we have 
2
2 2 2 2 2 2
2 2 4 2x x r x r x r     

Area of the inscribed square = 2
2r

17. 2. Find the area of an equilateral triangle with side length = 10 cm.
Answer: 2
25 3 cm
Solution:
All three angles in an equilateral triangle measures 60° each. Figure below shows an equilateral
triangle with side length = 10 cm


Area of a triangle = 21 1 50 3
sin 10 10 sin 60 = 25 3 cm
2 2 2
bc A     
3. Moscow and Copenhagen are both situated almost at the same latitude of 55.7° N. The longitude
of Moscow is 37.62° E and the longitude of Copenhagen is 12.57° E. Find the distance between the
two cities rounded to the nearest hundred km. Radius of the earth = 6371 km.
Answer: 2300 km
Solution:
Let R’ be the radius of the circle that connects all points on the earth with a latitude of 55.7° N
R’ = Rsin(55.7°) as shown in the figure below.

Difference in longitudes of the two cities = 37.62° - 12.57° = 25.05° = 25.05
180
 radians
∴ Distance between the two cities = 6,371,000 x sin(55.7°) x 25.05
3.14
180
 metres = 2299874.85 m
= 2299.87 km ≈ 2300 km (rounded to the nearest hundred km)

18. 4. The length of each side of a rhombus is 13 cm and one of the diagonals is 24 cm long. Find the
area of the rhombus.
Answer: 2
120 cm
Solution:
Area of a rhombus = 12
1
2
dd where 1
d and 2
d are the lengths of the diagonals.
Let the length of the unknown diagonal be 2x cm.

Figure below shows the rhombus in question.

As the diagonals of a rhombus bisect each other and intersect at right angles, we can use the
Pythagoras theorem to find x. 2 2 2 2 2 2
13 12 13 12 169 144 25
5 cm 2 10 cm
xx
xx
       
   

Area of the trapezium = 21
24 10 120 cm
2
  
5. A and B are closed semi-circles in the shape of the letter ‘D’. The area of A thrice the area of B.
Find the ratio of the perimeter of A to that of B.
Answer:3 :1
Solution:
Let the radius of semi-circle A be a
r and that of semi-circle B be b
r . 2
22
2
11
3 3 3
22
aa
ab
bb
rr
rr
rr
     

Perimeter of semi-circle A = 22
a a a
r r r  
Similarly, perimeter of semi-circle B = 22
b b b
r r r  
Perimeter of A : Perimeter of B = 

2 3
21
a a
bb
r r
rr




 =3 :1
6. A flexible wire is first bent in the shape of a square with area 484 sq.cm. It is then opened up and
bent in the shape of a circle. What is the area of the circle?

19. Answer: 2
615.44 cm
Solution:
Side length of a square with area = 484 sq.cm = 484 = 22 cm.
Perimeter of the square = length of the wire = circumference of the circle = 22 x 4 = 88 cm
Let the radius of the circle be r 88
2 88 cm 14 cm
2 3.14
rr   


Area of a circle with radius 14 cm = 
2
2
14 3.14 196 615.44 cm   
7. From a piece of wood in the form of a trapezium PQRS, a quarter of a circle with radius 4 cm has
been cut out as shown in the figure below. If PQ = QR and if ST = 2 cm, find the area of the remaining
piece.

Answer: 2
7.44 cm
Solution:
The central angle on a quarter circle = 90°. Hence 90mR


PQ = QR = TR = radius of the circle = 4 cm

Area of the trapezium = 
21
4 6 4 20 cm
2
   
Area of the quarter circle = 
2
2
4
12.56 cm
4


Area of the remaining portion = Area of the trapezium – Area of the cut out portion (quarter circle) = 2 2 2
20 cm 12.56 cm 7.44 cm

8. Find the length of the longest ladder that can fit inside a cubical room that measures 4 m on all
sides.
Answer: 4 3 m
Solution:
The length of the longest ladder that can fit in the room is the length of the longest diagonal. Figure
below represents the cubical room of side length 4 m.

20. Length of the longest diagonal = AH (shown in red) = BG = ED = FC
As all angles are right angles, we can use the Pythagoras theorem.
In right angled triangle ACH, 2 2 2
AH AC CH


Again, in right angled triangle CGH, 2 2 2
CH CG GH


Combining the two we have 2 2 2 2 2 2 2 2 2
4 4 4 3 4 mAH AC CG GH       
4 3 mAH
9. Find the volume of a hollow pipe, 30 cm long, with internal diameter 8 cm and thickness 1 cm.
Answer: 3
847.8 cm
Solution:
Volume of the hollow pipe = Volume of the outer cylinder – Volume of the inner cylinder as shown in
the figure below.

As the internal diameter is 8 cm, internal radius = 4 cm.
As thickness of the pipe = 1 cm, external radius = 5 cm
Volume of a cylinder is given by 2
rh where r is the radius and h is the height.
∴ Volume of the hollow pipe =   
2 2 2 2 3
5 30 4 30 30 5 4 30 3.14 9 847.8 cm        

21. 10. A right circular metallic cone is 5.2 cm high and the radius of its base is 2.5 cm. It is melted and
recast into a sphere. What is the surface area of the sphere?
Answer: 2
50.24 cm
Solution:
When the cone is melted and recast into the sphere, volume of the cone = volume of the sphere
Volume of a cone = 21
3
rh where r is the radius of its base and h is the height.
∴ Volume of the cone = 
2
31
2.5 5.2 cm
3

Volume of a sphere of radius r = 34
3
r
Thus, 34
3
r = 
21
2.5 5.2
3
 3 3 3
3
4 2.5 2.5 5.2 8.125 cm
8.125 2 cm
rr
r
     
  

Surface area of a sphere with radius 2 cm = 
2
2
4 2 50.24 cm