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PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 1
1 MENSURATION(part1)
GAYAZA HIGH SCHOOL
S4 MATHEMATICS ( MAY)2020
Mensuration is the branch of mathematics which studies the measurement of the geometric
shapes and the calculation of their parameters like area, perimeter, surface area, volume etc.
Types of Mensuration
Plane mensuration deals with the sides, perimeters and areas of plane figures of
different shapes.
Solid mensuration deals with the surface areas and volumes of solid objects.
The shapes exist in either 2 dimensions (2D) or 3 dimensions (3D).
Differences between 2D and 3D shapes
2D shape 3D shape
This is a shape surrounded by three
or more straight lines in a plane.
These shapes are plane figures such
as the triangle, square, rectangle,
trapezium, parallelograms , rhombus
,kite , circle etc.
These shapes have lengths in two
directions .
We can measure and calculate their
area and perimeter
This is a shape surrounded by a number of
surfaces or planes.
These shapes are called solids such as the
prisms( cube,cuboid, cylinder , triangular , etc
),cone,sphere , pyramids etc
These shapes have lengths in three different
directions.
We can measure and calculate their volume and
total surface area.
AREA OF TRIANGLE:
Area of triangle when given the base and the perpendicular height.
Find the area of the following triangles
8cm 10cm 20cm
12cm
9cm
16cm
3.2 m
4.5 m
PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 2
2 MENSURATION(part1)
In addition to the formula π¨πππ ππ ππππππππ = π
πππ , there are two other useful formulae
Area of a triangle when two sides and an included angle are given
Example
Find the area of the triangle
π¨πππ =π
πΓ π Γ π Γ πππππΒ° = ππ. πππππ
Example
Find the area of a triangle ABC such that π΄πΆ = 6ππ, π΅πΆ = 9ππ and < π΅πΆπ΄ = 32Β° Solution
π΄ =1
2πππ πππΆ
=1
2Γ 6 Γ 9 Γ π ππ32Β°
= 14.31ππ2
Example
The area of a triangle is 18.1ππ2, if the two of its sides are 7cm and 9cm, find the
included angle . Solution
π΄ =1
2πππ πππΆ
18.1 =1
2Γ 7 Γ 9 Γ π πππ
π πππ = 0.5746 β΄ π = π ππβ10.5746 = 35.07Β°
Area of a triangle when all the three sides are given
We use Heronβs formula
Example
Find the area of the triangle
Given that π, π and π are the sides of the triangle ABC, then
ππππ ππ ππππππππ = βπ(π β π)(π β π)(π β π) where π =π
π(π + π + π)
π¨πππ ππ ππππππππ =π
πππππππ¨ =
π
πππππππ© =
π
πππππππͺ
PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 3
3 MENSURATION(part1)
Solution π =π
π(π + π + π) =
1
2(π + π + π) = ππππ
π¨πππ ππ ππππππππ = βπ(π β π)(π β π)(π β π)
= βππ(ππ β π)(ππ β π)(ππ β π)
= βπππ
= ππ.πππππ
EXERCISE
2. In a fitness exercise, students run round the three sides of a triangular field
PQR. Given that ππ = 95π, ππ = 120π and ππ = 145π. Find the area
of the field.
3. A fishing boat travelled 3.2km from a lighthouse L to a point M. It then
travelled from M to a point N and then back to the lighthouse. If
ππ = 1.7ππ, πΏπ = 2.8ππ, find the area covered by the boat.
4. A farmer marks off a triangular piece of land for growing vegetables. Given
that π΄π΅ = 39.5π, π΅πΆ = 68.6π and < π΄π΅πΆ = 43Β°, calculate the area of the
piece of land.
5. In an isosceles triangle ABC in which π΄π΅ = 12ππ, π΄πΆ = π΅πΆ = π₯ππ and angle
π΄πΆπ΅ = 120Β°. Find the (a) value of x (b)area of the triangle
PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 4
4 MENSURATION(part1)
Further worked examples
Example 1 In the given triangle ABC, the shaded area is 20cπ2. Given that π΄πΆ = 10ππ, π΅π = π₯ππ and ππΆ = 2π₯ cm , find the area of the unshaded region.
Example 2 In the figure ABCD is a rectangle in which AD= 5π₯ ππ and AB= 3π₯ ππ. M and N are the mid points of BC and CD respectively.
Solution (a ) πππ‘ππ π’ππ βππππ ππππ = ππππ ππ βπ΄π΅π + ππππ ππ βπππΆ + ππππ ππ βπ΄π·π
= [1
2(3π₯) (
5
2π₯)] + [
1
2(3
2π₯) (
5
2π₯)] + [
1
2(5π₯) (
3
2π₯)]
= 15π₯2
4+
15π₯2
8+
15π₯2
4
= 75π₯2
8 ππ2
β π΄πππ ππ π βππππ ππππππ = ππππ ππ ππππ‘πππππ π΄π΅πΆπ· β π‘ππ‘ππ π’ππ βππππ ππππ
= (5π₯ Γ 3π₯) β [75π₯2
8]
A
C P
B x 2 x
10 cm
Solution Shaded area =area of ABC- area of ACP
20 = 1
2(3π₯) Γ 10 β
1
2(2π₯) Γ 10
20 = 15π₯ β 10π₯ = 5π₯
5π₯ = 20
β΄ π₯ = 4ππ
Area of unshaded region = area of ACP=10π₯ = 10 Γ 4 = 40ππ2
U
PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 5
5 MENSURATION(part1)
= 15π₯2 β75π₯2
8=
120π₯2 β 75π₯2
8
β΄ π΄πππ ππ π βππππ ππππππ =45π₯2
8ππ2
(π) π΄πππ ππ βπππΆ =15π₯2
8= 30 β π₯2 = 16 β΄ π₯ = 4ππ
Dimensions of the rectangle are 5 Γ 4 = 20ππ and 3 Γ 4 = 12ππ
(π)π‘πππ =π΄π΅
π΅π=
12
10= 1.2 βπ = π‘ππβ11.2 = 50.19Β°
EXERCISE
1.
2.
3.
PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 6
6 MENSURATION(part1)
Area of Quadrilaterals A quadrilateral is a plane figure bounded by four line segments.Examples
include square,parallelogram, kite, trapezium,rhombus, rectangle etc .
Activity: PROPERTIES OF QUADRILATERALS
Study the picture below and write down the properties of each of the
quadrilaterals basing on their sides, angles and diagonals,
PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 7
7 MENSURATION(part1)
Example
ABCD is a trapezium in which AD is parallel to BC. Given that π΄π· = 25ππ, π΅πΆ = 15ππ, π΄π΅ =
12.8ππ and angle π·π΄π΅ = 40Β°. Calculate the area of the trapezium.
Solution π ππ40Β° =
π
12.8 βπ = 12.8π ππ40Β° = 12.8 Γ 0.6428 = 8.2278ππ
π΄πππ ππ π‘βπ π‘πππππ§ππ’π = 1
2Γ 8.2278(15 + 25) = 164.556ππ2
Example
Figure(i) shows a triangle ABC in which π΄πΆ = 8ππ, π΅πΆ = πππ and angle π΄πΆπ΅ = 30Β°. Figure (ii)
shows a trapezium PQRS in which ππ = 7ππ, ππ = 3ππ, ππ is parallel to ππ and the distance
between them is π ππ. Given that the triangle and trapezium have the same area , determine
the ratio of π: π.
Solution
But the triangle
and trapezium have the same area β5π = 2π βπ
π=
2
5 β΄ π: π = 2: 5
Exercise 1. Find the area of the figures below
2.ππππ is a trapezium whose area is 25ππ2. Given that ππ = 6ππ, ππ = 4.8ππ and π π = 8.4ππ.
Find (i) PT (ii) angle πππ
3. The longer side of a trapezium is three times as long as the shorter parallel side. The
perpendicular distance between the parallel sides is 15cm. If the area of the trapezium is
180ππ2.Calculate the length of its longer parallel side.
PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 8
8 MENSURATION(part1)
4. The diagram below shows a trapezium PQRS in which PS is parallel to QR, ππ = 10ππ and angle πππ = 90Β°.
T is a point on PQ such that ππ = 6ππ, angle πππ = 30Β°and angle πππ = 60Β°.
(a) Find the size of angle πππ
(b) Calculate the length of (i) TR (ii) QR (iii) PS (iv) PQ
(c) Determine the area of the trapezium PQRS
Qns 1-6 Find the area of the figures above
Additional Exercise
PREPARED BY MRS ASSUMPTA KASAMBA@MATHSDEPTGHS 0772937519 9
9 MENSURATION(part1)
7. In the quadrilateral ABCD, < π΅ =< π· = 90Β°, < π΅π΄πΆ = 10Β° and < πΆπ΄π· = 60Β°. Given
that π΄πΆ = 10ππ, find the (i) length of AD (ii) length of AB
(iii) area of the quadrilateral
8. (a)Given that π΄π΅ = 15ππ, π΅πΆ = 13ππ, πΆπΈ = 10ππ Calculate the area of the
parallaelogram ABCD
(b) find the length of the perpendicular from A to BC.
9. In the parallelogram PQRS, < πππ = 90Β°, ππ = 9ππ and ππ = 12ππ. Find the length
of the diagonal PR. Hence find the area of the
parallelogram.
10. PQRS is a trapezium in which ππ = 7ππ,ππ = 17ππ, π π = 15ππ and < πππ = 90Β°
and PQ is parallel to SR. Calculate the (i) length of PS (ii)
area of the trapezium.