MENSURATION
Mensuration
is the art of determining the value of the length ,width .perimeter area volumes
and surface area of plane object or solid object as the case may be. To o this
effectively it will be better to know the
formula to be apply whenever problem are given in any of the said
dimension. Also the various properties of the figures in question are to be
known. And this are hereby tabulated below.First we study the plane shape and
thereafter the Soli shape
Properties of the plane shape object and
their formular for area and perimeter
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NAME
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SHAPE
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CHARACTERISTICS
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AREA
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PERIMETER
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1
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triangle
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Have 3 sides and3 angles it may be (i)
scalene,(ii)equilateral(iii)isosceles or according to angles,i.e(i) right
angle,(ii) acute angle triangle,(iii) obtuse angle triangle
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(a) ½ base× height
(b) ½ab Sin C
(c )√s(s̶−a)(s−b)(s−c)
Where s=a+b+c
2
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a+b+c
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2
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Square
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D C
A B
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(i)All four sides are equal.(ii)also opposite
sides are parallel (iii). diagonal bisects each other at 900.(iv)all
the angles are right angle
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L X B=L2
Since L=B
i.e.AB=BC=CD=DA
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4L=AB+BC+CD+DA
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3
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Rectangle
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D C
A B
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(i)
opposite sides are equal (ii)opposite
sides are parallel (iii)the diagonal meets each other at 900(iv)
all angles are 900.
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L=length=AB=CD
B=breadth=BC=DA
LX B
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2 (L+B)
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4
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circle
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A round shape of object whose sides is determined by
its radius
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πr2
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2πr = circumference
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5
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trapezium
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D C
A B
D C
A B
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The opposite sides are parallel
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½(sum of parallel sides )X height
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AB+BC+CD+DA
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6
|
rhombus
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D C
h
Ɵ
A B
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(i) all the four sides
Are equal
(ii)2 opposite sides are parallel
(iii) opposite angles are equal.
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L2× SinƟ
L=AB=BC=CD=DA
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4Li=AB+BC+CD+DA
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7
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Parallelogram
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D C
h length
Ɵ
A base B
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(i)opposite sides are parallel
(ii)opposite are equal
(iii)the diagonal bisects each other
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Base×
height
=b× lSin Ɵ
Where l=AD=BC and the base =AB
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AB+BC+CD+DA
i.e 2(L+B)
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9
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Segment of a
cicrle
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Ɵ r
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An arc of a
circle cut off from the sector by a
chord
|
Ɵ
×πr2 ½r2SinƟ
3600 − 1
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2πrƟ +2r
3600
−
2r + the length of the chord)
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|
10
|
kites
|
C
D B
A
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(i) opposite
sides
Are parallel
(ii) adjacent
sided are equal
(iii) the
diagonal intersect at right angle
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AC× BD(the length
of the diagonal are to be used to multiply each other)
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2(L+B)
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THE PROPETIES OF THE SOLID SHAPE
AND THEIR FORMULAR FOR SURFACE
AREA,PERIMETER AND VOLUME
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S/N
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NAMES
|
SHAPE
|
CHARACTERISTIC
|
SURFACE AREA
|
PERIMETER
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VOLUME
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1
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CUBE
|
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(i) it have 6 equal faces
(ii)it have 8 vertices
(iii)it have 12 equal edges
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6 × (area of one of the faces)
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12 ×one of the length
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L×B×H=L3
Since all the length are equal
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2
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CUBOID
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(i) it have 6 faces
(ii)it have 12 edges
(iii)it have 8 vertices
(iv)opposite faces/sides are
equal
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2LB+2BH+2LH i.e
2(LB+BH+LH)
Where L=length
B=breadth
H=height
|
4L+4B+4H
i.e
4(L+B+H)
|
L×B×H
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3
|
CYLINDER
|
|
This is a ring shaped object
that can be closed at both end or at one end. it have 3 faces,(the top layer
the bottom layer and the round sides)
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Closed at both end =
2πr2+2πrh=
2πr(r+h)
Closed at one end=
Πr2+2πrh=
Πr(r+2h)
Open at both end=
2πrh
|
Closed at both end=
2πr+2πr+ h +h+x+x
(where x is the length of the
edges of the curved surfaces)=
4πr+2x +2h
Closed at one end=
2πr+2h+2x
Open at both end=
2h+2x=
2(x+h)
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Πr2h
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4
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SPHERE
|
|
This is
slightly flattened at both end
with curved surfaces
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4πr2
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2πr
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4/3πr3
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5
|
CONE
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|
Triangular shaped object with
curved or circular ends
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Curved surface area=πrl
Total surface area=
πr2+πrl=
πr(r+l)
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2πr+2l
2(πr+l)
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1/3πr2h
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6
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FRUSTUM
|
Conical
frustum=
Pyramidal frustum
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(a)beheaded cone having three
faces,2 edges and no vertex
A beheaded pyramid having 6
faces ,8 edges and 8 vertex
|
Conical frustum=
Surface area of the whole cone
–surface area of the cut off part i.e=
(πR2+Πrl)−
(πr2−πrl)
Pyramidal frustum
Surface area of the bigger
pyramid – the surface area of the smaller cut of
pyramid=
(4× area of one of the triangle of the bigger pyramid +
its base) −
(4×area of the triangle of the
smaller pyramid + its base
|
2πR+2πr +2l=
2(πR+πr+l)
4x+4l+4y=
4(x+y+l)
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1/3πR2H−1/3πr2h
i.e the volume of the bigger cone –the volume of the smaller cut off cone.
1/3 area of the
cross-section of the bigger pyramid ×its height) – 1/3 area of the
cross-section of the smaller cut off pyramid × its height
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7
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Prism
|
triangular
prism
Rectangular
prism
|
Having many faces with
cross-sectional area
|
Addition of the faces making up
the prism
|
Addition of thee perimeter of the faces of the figure making up the
prism
|
Base area × the perpendicular
height
|
|
8
|
pyramid
|
|
Rectangular base pyramid
It is having 5 faces i.e 4 triangular faces and the base.it have
8 edges and 5 vertex
|
Area of the triangles + area of
the bases
|
(4 × the perimeter of the
triangles) + the perimeter of the base
|
(1/3 base
area) × the height
|
|
|
|
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Triangular base
It is having 4 faces ,i.e 3
triangles and
the base. 4vertex. And 6 edges
|
Area of the triangle that made
up te faces + the area of the base
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3 × perimeter of the triangles
that made up the faces + perimeter of the triangular base
|
(1/3 base
area)× the height
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Examples
1. Find
the surface area and the volume of the cuboids below
Surface
area = 2(lb + lh + bh) (here
l=length,b=breadth and h= height) l = 18, b =
10, h = 15 surface area = 2[(18×10) + (18×15) +
(15×10)] surface area = 2(180 + 270 + 150) =
2(600) = 1200cm2 Volume = l × b × h
=
18 × 10 × 15 =
2700cm3
2.
From the figure above show that the
volume = πr2 (45-2r) cm3 3
(ii) find the surface area
(the
figure is of two parts ,the cone and the cylinder, the cone is like the cover
of the cylinder. so the cylinder can be said to be open at one end while the
bottom end is closed)
From
the figure above, Volume of cone = 1/3 πr2h
=
1/3 πr2× r = 1/3πr3 Volume
of cylinder = πr2h =
πr2(15-r) =
15πr2 – πr3 total
surface area = 1/3πr3 + 15πr2 – πr3
= 1/3πr3 –π r3 + 15πr2 = 15πr2 +πr2- 3πr3
3
= 15πr2 – 2πr3
3
=
45πr2 – 2πr3
3
= 45πr2 – 2πr3
3
=πr2
(45 – 2r) 3
=πr2 (45 – 2r)cm3 3
(ii)
Surface area of a cone = πrl + πr2
AB
= length=l
l2
= OA2 + OB2 l2
= h2 + r2 i.e. l2 = r2 + r2
since h = r
l2 = √2r2
πrl +πr2 = π × r ×√ 2r2×π
r2 =πr√2r2 +π r2
Surface
area of a cylinder = 2πrh + πr2 the
cylinder is closed at one end
= 2πr(15 – r) +πr2
= 30πr – 2πr2 +πr2
= 30πr – πr2
Total
surface area = πr√2r2 +πr2 + 30πr –πr2 = πr √2r2 + 30πr = πr (30 + √2r2)
3.
The figure below shows a cone with the dimension of the frustum. Calculate the
height of the cone.
Height
of the cone = VO
VOB
and
VOA are similar: V VO : VC = OA : CB 12 + x : x = 12:6 x 12 + x = 12 C
6cm B x 6
12cm 12x
= 6(12 + x) (÷6) 2x = 12 + x O 12cm
A x
= 12 VO = 12 +12 =
24cm
(4)
The solid shape in the figure below is a cylinder surmounted by a hemispherical
bowl. (a) Calculate the total
surface area
(b)
Calculate the volume. ( Take π = 22/7 )
The shape is of two parts: the hemisphere and
the cylinder ( the hemisphere was the cover for
The cylinder and if removed, it will make the cylinder to be
Open at one end and closed at one end)
The cylinder is closed at one end, the second
end is the one covered by the hemisphere, so that end would be opened.. Total Surface Area (T.S.A) of the
hemisphere = 4πr2 (note: T.S.A of a Sphere = 4πr2)
2
The radius of the hemisphere
is the same as the radius of the cylinder =7cm since hemisphere = ½ of a Sphere = 2πr2 = 2×22×7×7
7
= 308cm2 The
surface area of the cylinder = πr2 + 2πrh =
22×7×7 + 2×22×7×10 7 7 =
154 + 440 = 594cm2 then
the T.S.A of the solid = surface area of the hemisphere + surface area of the
cylinder
= 308 + 594 =
902cm2
(b)
Volume of the solid = volume of the hemisphere + volume of the cylinder
= 2 πr3 + πr2h (note that volume of sphere =4πr3 but hemisphere
is half 3 3 a
sphere)
=
2 × 22 × 72 + 22 ×72 × 10 3 7 7 = 2156
+ 1540 3 = 718.67 + 1540 = 2258.67cm3.
(5)
The figure below shows a doorway made up of a circular arc ABC which subtends
an angle of 600at the centre O. The upright poles CD and AE of
length 4m each on an horizontal base of width 3m.
Show that the perimeter
is (11 + π)
Note;the
doorway is in form of a rectangle with one side removed and was replaced with a
curve top in form of length of an arc so the perimeter is the addition of the 3
sides of the doorway +the length of an arc of the top cover.
Perimeter
of the rectangle = AE +DE + CD
= 4m + 3m + 3m=11m from
AOC, using cosine rule AC2
= (AO)2 + (CO)2 – 2 (OA)(OC) cos AOC AC
= 3m AC2 = 9 AC2
= r2 + r2 - 2 (r)(r) cos 600 9
= 2r2 – 2r2cos 60 9 = 2r2
– 2r2 × ½ (cos 60 = ½) 9 = 2r2 – r2
9 = r2
r = 3.
Length
of arc = Ɵ × 2πr 360
= 60 × 2 × 22 × 3 = 22
= πm 360 7 7 ___ __
___ perimeter = AE + CD + DE + length of arc
= 11m + πm
= (11 + π)m
(6)
The volume of the figure below is Vcm3. If the radius of the sphere
is 21/2
cm
and height of the cone is 14 cm. Calculate the value of V and if A cm2
is the surface area, what is A?
(the
shape is of two figure. so the shape is broken into two parts the cone and the
hemisphere
The
diameter of the hemisphere is 21cm then the radius is 21/2cm
Volume = volume of cone + volume of hemisphere
= 1/3
πr2h
+ 2/3
πr3
=πr2 h + 2r
3 3
= πr2 2r + h
3
= 22 × 21 × 21 2× 21/2
+
14 7 2 2 3
= 21 × 33 × 35
2 3
= 11×35×21
2
= 4043.66cm3
(ii)
surface area of the shape = curved surface area of the cone + surface area of
the sphere =πrl + 2πr2
to find l
VA
= l
l2
= OA2 + OV2 (VOA is in form of right angle triangle. So
we use Pythagoras thorem)
= 142 + (21/2)2
= 196 + 441 4 l2
= 1225 4 l = 306.27
= 17.5
∙ ∙
∙ Total surface area of the
object = 22 × 21 × 17.5 + 2 × 22 × 21 × 21 7
2 7 2
2
= 11× 3 × 17.5 + 33 × 21
= 33 × 17.5 + 33 × 21
= 33 (21 + 17.5) = 33 (38.5)
= 1270.5 cm2
(7)
The
figure above shows that forms an opened end funnel. Calculate the total surface
area.
Solution:
the figure comprises two kind of shape join together. i.e a cylinder and a
frustum
Total
Surface Area of the solid = total surface area of the cylinder + total surface area
of the frustum
Surface
area of the cylinder = 2πrh. Since it is opened at both ends, = 2 × 22 ×
3 × 7 7 = 44 × 3
= 132cm2 frustum
is a beheaded cone .
So
Surface area of the frustum= Surface area of the bigger cone – surface area of
the smaller cone
From
VOB
VB2
= (VOʹ)2 + (OʹB)2 = 72 + 32
= 49 + 9 = 58 VB = √58 = 7.62 cm ___ ___
___ VA = VB + BA
= 7.62 + 8.7 = 15.82 surface area of the frustum = πRL – πrl
= 22 × 5.5 × 15.82 - 22
× 3 × 7.62 7 7
= 273.46 – 71.85 = 201.61 cm2
Total surface area = 201.61 + 132 = 333.61 cm2
In
the figure below, O is the centre of the concentric circle. OAB is a straight
line such that OA = 5 cm, AB = 4cm. what is the ratio of the area of the outer
circle to that of the inner circle?
Area of small
circle = πr2 If r = 5cm,area
= π × 52 = 25π
Area
of larger circle = πR2 R
= (r + 4cm) = 5 + 4 = 9cm
Area
of larger circle = π × 92 = 81π
Ratio
of area of larger circle to area of smaller circle = 81π꞉25π
(÷π) =
81 ꞉ 25
25 25 = 3.24 ꞉ 1 = 3 ꞉ 1
WORK TO DO
1. The figure below shows a quadrant of a circle of
radius r cm drawn within a square of side 10cm
If the area of the quadrant is ½ of the area of the
square, find r to 2 significant figures.
2. A brass rivet is composed of cylindrical shafts and a
hemispherical cap with dimensions as shown below:
Calculate the mass of 1000 of such rivets if the density
of brass is 8.4g/cm3.
3. The figure below shows the vertical cross-section of a
tent made of tarpaulin in the shape of a
right circular cone placed above a right cylinder. The height of the tent is
6m. the radius and the height of the cylindrical portions are 6m and 3½ m
respectively. Calculate:
(i) the volume of the tent
(ii) the area of the ground covered by the tent
(iii) the area of the tarpaulin used in making the tent
(π = 22/7
and all answers should be in 2 decimal places)
4. Calculate the volume of the solid in the diagram
below:
(NOTE FOR MORE
EXERCISES CHECK NEW CONCEPT
MATHEMATICS SS3 CHAPTER 7 PG
70-73 AND CHAPTER 18 PG 207-212.)
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