Basic Mathematics Form Four Notes – Area and Perimeter

Basic Mathematics Form Four Notes – Area and Perimeter

These Basic Mathematics Form Four notes cover Area and Perimeter. The material is arranged in a clear mobile-friendly reading format while preserving the recognized definitions, explanations, examples, activities, calculations and revision material from the source notes.

Formula & Symbol Clarity

  • Circle circumference: C = 2πr
  • Circle area: A = πr²
  • Cylinder volume: V = πr²h
  • Sphere volume: V = 4πr³/3

Area of any Triangle The Formula for the Area of any Triangle

Derive the formula for the area of any triangle Area of triangle is given by'2bh, whereby b is the base of the triangle and h is the height of the

given triangle. Consider the illustrations below:

B

b b

Figure a Figure b

b

Figure c

From the figure above, we see where the base and height are located.

Applying the Formula to find the Area of any Triangle Apply the formula to find the area of any triangle Example 1

The base of a triangle is 12cm long. If the corresponding height is 7cm, find the area of the triangle.

Solution

Consider the figure below:

12 cm The area of a triangle is given by'2bh. Area ="2*12em*7em
Area = 42cm?

Therefore area of a triangle is 42cm"

Example 2

The lengths of two sides of a triangle are 6cm and 8cm. Find the area of a triangle if the included is

Solution:

Consider the triangle below, name it triangle ABC.

A

a 8cm

The area of a triangle above is given by'2bxh So, Area = 2 8cm* 6cm ~ sin 45° =24em'x sin45°
= 16.97em?
Therefore the area of ABC = 16.97cm?

Area of a Rhombus The Formula for Finding the Area of Rhombi in Terms of the Diagonals

Derive the formula for finding the area of rhombi in terms of the diagonals. The area of a rhombus is the same as the area of a parallelogram because rhombus is a special kind of parallelogram. Rhombus is a parallelogram with equal sides. Consider the figure below

of a rhombus with base b and height h.

D Cc

So, Area of rhombus = bh

Another formula for finding the area of rhombus can be obtained by using the diagonals.

Consider the rhombus below:

D Cc

A B

Diagonals of a rhombus bisect each other at right angles (means the diagonal lines are half

equally), so the area of a rhombus ABCD can be found as follows:

Area of a triangle ABC = area of a triangle ADC
= ; x(base (AC)) x height G DB)
x (AC x DB) =-AC x DB Since the triangles are equal, the area of ABCD: (= AC x DB) x2 =-xACxDB

Therefore, the area of a rhombus is equal to half the product of the length of the diagonals.

The Formula to Find the Areas of Quadrilaterals

Apply the formula to find the areas of quadrilaterals

Area of a Trapezium

Consider the trapezium with constructed lines as shown in the figure below:

Af T

In order to find the area of a trapezium, first let us find the area of the triangles ABD and BDF

with the same height h. The base of the triangle ABD is b;

The base of the triangle BDF is bz

The area of the triangle ABD = 2byh The area of the triangle ABE = '2b,h Area of ABDF =2b;h +'2b2h
Area of ABDF=!2 (b; + b2)h

Generally, the area of the trapezium is given by''2 (b;+ b2)hor is the product of half the sum of

the parallel sides (bases) and the perpendicular distance between them (height)

Example 3

Find the height of the trapezium with area 90 square units and bases of 6 units and 14 units

Solution:

Consider the trapezium below:

6 units

14 units The area of the trapezium = . (b; + bz)h
90 = = (6+14)h 90 = 10h 9=h

Therefore the height of the trapezium is 9 units.

Area of a Parallelogram

Consider parallelogram below with constructed lines as shown in the figure.

B

The area of the parallelogram can be formed from the formula for the area of the trapezium. The

important thing to note is that the bases for a parallelogram are equal. Now, the area of the parallelogram ABCD='2 (AB+DC)hsquare units.Since AB = DC Then, the area of parallelogram ABCD ='4 × 2AB x h = ABx h
If AB =b, then Area of the parallelogram ABCD = bh, where b is the base and h is the height of a parallelogram.

Therefore; the area of a parallelogram is equal to the product of the base and the perpendicular

height. Area of a Rectangle

Consider the rectangle below:

D

The rectangle ABCD is divided into two congruent triangles, which are triangle ABD and triangle ACD by the diagonal AD.

The area of ABCD = area of triangle ABD + area of triangle ACD

1 1 5 XCD xX AC +> x AB x BD

Since the triangles are equal, the area of ABCD is equals to double the area of one of the triangles. The area of ABCD

= (5x CD x AC) x2 = CD x AC
If CD is the length '/'of the Rectangle and AC is the width 'w' of the Rectangle, then, the area of ABCD = /xw or hw

Therefore, the area of the Rectangle is the product of the length and width. Area of a Square

A Square is a special rectangle with equal sides. Therefore the area of the square is the product

of its lengths. i.e. Area of a square =/ x =I.

We can also find the area of a square by using the length of the diagonals. Consider the square below with diagonals AC and DB:

D Cc

A B

Each of the diagonals of a square bisect at a right angle.Area of a triangle ABC = Area of ADC
; xbase (AC) x height (80) pe 4 = 5 (AC x 380)

x AC x BD

"

But the area of ABCD

= area of AABC x 2
= (5 x AC x BD) x 2
=->xAC xBD
Since the length of the diagonals are equal, then AC = BD.So, the area of ABCD = '2 (AC)? Therefore the area of a Square is equals to the half of the product of the lengths of the diagonals.

Example 4

Find the area of a parallelogram ABCD if AC = 7cm, AB = 9cm and the angle ZWX =. c D
Solution Area of a parallelogram = basexheight

Consider the figure below:

Consider the triangle ACE to find h

opposite hypotenus Sin 58° =

7em

h = 7cm x sin 58°
Recall that; sine =
h = 7cm x 0.8480 h = 5.936cm Now, area of ABCD = 9cm x 5.936cm
= 53.427cm?

Therefore the area of the parallelogram ABCD is 53.427cm'.

Perimeter of a Regular Polygon

The Formula for Finding the Length of a Side of a Regular Polygon

Derive the formula for finding the length of a side of a regular polygon

When we sum up the lengths of the sides of the polygon we obtain what is called perimeter of a polygon. Therefore, perimeter of a regular polygon is the sum of the lengths of the sides of the polygon.

How to find the perimeter of a Regular Polygon inscribed in a circle?

An inscribed polygon is the one whose vertices lie on the circle. If the lengths of the sides of the polygon are the same we say that the polygon is an inscribed Regular Polygon.

A Regular Polygon with number of sides larger than 2 say n sides can be inscribed in a circle as

follows:

For example, if you want to construct an inscribed regular hexagon (6 sides), first draw a circle and locate the center of the circle. Then draw rays that intersect the circle in six points from the center of the circle. Each angle at the center will measure 360°/6 = 60°. Connect the points of intersection on the circle by line segments. The figure formed is an inscribed regular polygon.

See the figure below:

Now, to obtain the formula of finding the perimeter of a regular polygon inscribed in a circle with radius r and center O, let AB be the side of the polygon and OC the perpendicular from O to

AB as shown in the figure below:

ANC B

The angle AOB = 360°/n, since the polygon has n sides.
The angle AOC = the angle AOB and the angle AOB = angle AOC + BOC Therefore, the angle AOC = '4(360°/n) = 180°/n

Let the length of the side of the regular polygon ABbe S.

Then,

ac = +aB =+5 a caer
From the AAOC, sin

Therefore the length of a regular polygon with n sides inscribed in a circle is given by

. 1 S = 2rsin — n

The Perimeter of a Regular Polygon

The formula to determine the perimeter of a regular polygon

If we let 2r=d
180° ThenS = dsin a where d is the diameter of the circle.

From the concept of perimeter that perimeter of a regular polygon is the sum of the lengths of the sides of the polygon , if we have n sides each with length 'S' then the sum of the lengths of these sides will be nS. Therefore, Perimeter P of a regular polygon of n sides each with length S is

given by:

P=nS
180° But S = 2rsin —
So, P = n(2rsin = )
180° Therefore, Pp = 2nrsin — n

If we express 2r as a diameter d, then:

180° P = ndsin — n

Example 5

Find the length of one side of eight-sided regular polygon inscribed in a circle with radius 7em.

Solution

S = 2rsin
_ 180° S=2 × 7cmxsin
S = 14cm x sin 22.5° S = 5.358cm

Therefore the length of one side of eight-sided regular polygon with radius of 7cm is 5.358cem

Area of a Regular Polygon The Formula for Finding the Area of a Regular Polygon

Derive the formula for finding the area of a regular polygon Consider the regular polygon with n sides inscribed in a circle of radius r and center O as shown

below:

A

1 360° The area of the AAOB = 5 x AO x OB x sin cs

But since each vertex of the polygon is connected to O, the polygon region is divided into n

triangles which are equal.

Now

Area of the polygon = n x (area of AAOB)

Therefore the area of a polygon of n sides inscribed in a circle of radius r is given by:

: a 360° A = = nr? sin — r n

The Formula to Calculate the Area of a Regular Polygon

Apply the formula to calculate the area of a regular polygon Example 6

Find the area of twelve-sided regular polygon inscribed in a circle of radius 14 cm

Solution

360°

A=-=nr' sin 2 ir

Circumference and area of a circle

Circumference of a circle is the distance around it. Circumference of a circle can be estimated by

using a regular polygon with many sides inscribed in a circle with radius r. We know perimeter of the regular polygon is given by:

180° P = n(2rsin = )
Or, alternatively we can write it as: P = 2r(nsin = )

180°

180° 30) 7 ar x 3.141

Ifn = 10, P = 2r(10sin —) = 2r x 3.090
Ifn = 90, P = 2r(90sin
= 2 3.14157 sag) = ar x 3.
) = 2r x 3.14158
Ifn = 540, P = 2r(540sin
If P = 720, P = 2r(720sin

Here we see that, as n increases the value of nsin 180°/napproaches the value of.When n is very large the perimeter of a regular polygon approaches the circumference of the circle. The value

ofnsin 180°/ncan be replaced byxbecause it approaches the value ofawhen n is very large.

Therefore, circumference of the circle C, is given by C = 2ar

Area of a Circle

In similar way we can generate the formula of calculating the area of a circle by considering area

of a regular polygon inscribed in a circle of radius r. We know that, area of a regular polygon is given by:

1 .._ 360° P= phv'sin

Alternatively we can write it as:

360°

10 e Ifn =10,P = r(>sin ) =r? x 2.939
90. Ifn = 90,P = r(Fsin
) =r? x 3.139

540 360°

= =r = Ifn = 540,P =r (Ssin aa? r x 3.14152

360°

720. Ifn = 720, P = r?(—sin 2 720
) =r x 3.14159
As n becomes large and larger the value of sin ni approaches the value of * . We can estimate the n
area of the circle by replacing the value of sin ll by = . Therefore the area of a circle 'A' is given by n
A=ar.

Example 7

Find the circumference of a circle of radius 21cm. (takem= 3.14).
Solution Circumference of a circle, C = 2ar
= 2 × 3.14 × 21cm = 131.88cm

Area of Similar Polygons The Ratio of Areas of Similar Polygons

Find the ratio of areas of similar polygons

Let ABC and A' B' C' be two similar triangles:

c

A c

If we find the ratio of their sides we get;

sides. Area of a triangle ABC = absin Cc
Area of a triangle A'B'C' = a'b'sin C' But sin C = sin C' since AABC ~ AA'B'C'

1 :

area of AABC sabsinC ab

Therefore, = <% = area of AA/B/Cr a/brsinCy

Generally, if the ratio of the lengths of the corresponding sides of two similar polygons is k, then

the ratio of their areas is k.

Problems Related to Ratio for Areas of Similar Polygons Solve problems related to ratio for areas of similar polygons

Example 8

We are given two triangles which are similar. The length of one side is 8cm and the length of the corresponding side is 14cm. if the area of a smaller triangle is 24cm7find the area of the other triangle.

Solution

Ratio of the corresponding sides, k = = =
; 4.5 16 Ratio of their area, k? = (=)? = — yy 49 24cm? xem? x = 49 × 24
x = 73.5cm?
k? = (x is represents the area of the larger triangle)

Therefore the area of the other triangle is 73.5em'.

Example 9

The ratio of the areas of two similar polygons is 36:48. The length of a side of the smaller polygon is 10cm. find the length of the corresponding side of the other polygon.

Solution

Ratio of the areas = k? So, k? = 38 : 48

To obtain the ratio of the lengths, take square root both sides

360 6 k = _— = let the other side be y 10cm ss 6

y 43

_ 10cm x43 = 6 y = 11. 547cm

y

Three Dimensional Figures

Three Dimensional Figures Sometimes before you make any purchases you may want to know for example, how much cloth

you need to make a pillow cover. What about a cover for your mattress or sofa cushion? How

much oil paint do you need to paint your drinking water tank?

What about the amount of cloth for the pocket covers of your radio, curtain, suit, gown, trousers,

set of table clothes, etc.

Answers to such questions and of the kind leads you to think more carefully about the size of the surfaces (faces) to be covered or coated on the bodies at work. Perhaps you need to take some

measurements on the surfaces.

The knowledge of the surface areas of such bodies will enable you to choose or purchases the

required amount without unnecessary wastage so as to minimize purchases costs too.

Three Dimensional Figures

Classify three dimensional figures Three-dimensional objects are the solid shapes you see every day, like boxes, balls, coffee cups,

and cans.

It is called three-dimensional or 3D because there are three dimensions: width,

depth and height.

wa Oe

-The following table shows examples of some common three dimensional figures

/

ds

pal \

The Characteristics of Each Class

List the characteristics of each class

Here are some helpful vocabulary terms for solids:

Base: Is the bottom surface of a solid object.

Edge: Is the intersection of two faces on a solid object. This is a line. Face: Is a flat side of a 3-dimensional object.

Prism: Is a solid object with two congruent and parallel faces.

Pyramid: Is a solid object with a polygon for a base and triangles for sides.

Construction of Three Dimensional Figures

Three Dimensional Figures

Construct three dimensional figures When drawing a three dimensional object it is important to show that it is not a drawing of a flat object. Are usually drawn on a two dimensional plane by making oblique drawings under certain

tules as follows: Paralled lines are drawn parallel. Vertical lines are drawn up and down the page. Hidden edges are drawn dotted. Construction lines to guide the eyes are drawn thinly.

Construct three dimensional figures

Sketching Three Dimensional Figures

Three Dimensional Figures

Sketch three dimensional figures There are several ways of doing the drawing that corresponds to looking at the cube from

different angles. The figure shows two ways of doing it.

Properties of Three Dimensional Figures

Identify properties of three dimensional figures

Three dimensional shapes have many attributes such as faces, edges andvertices. The flat surfaces of the 3D shapes are called the faces. The line segment where two faces meet is called

an edge. Avertexis a point where 3 edges meet.

———

Edges

The Angle Between a Line and a Plane

Find the angle between a line and a plane

In finding the angle between the line and a plane in a three dimensional geometry, we use the right angled triangle. Joining the line to define the angle between the line and the plane that provides the least possible angle. Also, projection of one line to another on the plane is mostly

used.

Example 1 For the pyramid VABCD with VA=VB=VC=VD=5m, and ABCD a square of side 4cm; find the angle between VA and ABCD.

Solution

Calculated by dropping a perpendicular from V to ABCD. This meets ABCD at X, the centre of

the square.

So the projection of VA on ABCD is AX. AC= square root of (AB2 +BC2) =square of 42+42 =square root of 32. AX =1 / 2 square root of 32. cos (1 / 2 square root 32)/5=0.5657, so VAX is 55.6.

The Angle Between Two Planes

Calculate the angle between two planes

There are infinite possible lines that could be drawn on planes, making different angles with each other. The angle between planes is the angle between lines within those planes, Must be the lines which are at the middle of the plane for non rectangular planes and any other lines for

rectangular planes. Then Right angled triangles are used to find the angles between those planes.

Example 2

Determine the angle between the following planes:

m² 2x-y+z-1-<0

m= (2-11) 7 = (1,0,1)

RB: t+ (-1)-0+1-1| 3

a(n, x. = a(n,,n,) = arc cos = 7 vm)= (PP) rcpt Veeoer Wea
(5) = arc cos = 30°

Surface Area of Three Dimensional Objects The Formulae for Calculating the Surface Area of Prisms, Cylinder and Pyramids and Cone

Derive the formulae for calculating the surface area of prisms, cylinder and pyramids and cone

Surface Area of a Right Circular Cone

A right circular cone is a cone whose vertex is vertically above the centre of the base of the cone.

height radius side length

Area of curved surface = +3 × 2nr = nts So Area of curved surface = mrs
Area of circular base =IIr (it is an area of a circle)
Therefore the total surface area of a right circular cone =IIr +Ilrs = Ir (r + s)
A= qr(r = s) where F is the base radius and S is the slant height (side length).

Example 3 Find the total surface of right circular cone whose slant light is 10cm and whose base radius is

8cm.Use IIr(r +s)

Solution:

tr=8cm = s=10cm
Surface area = 3.14 × 8 (8 + 10) cm? =3.14 × 8 × 18 = 452.16cm?
=Total surface are = 452.16cm?

Example 4

Find the total surface area of a cone with diameter 8m and slant height of 10m. Use I= 3.14
Solution: A=nr(r +s) d=8, sor=4 A=3.14 × 4(4+10) =3.14 × 4 × 40 = 175.84 m? Therefore the total surface area is 175.84 m*

Surface Area of A Right Cylinder

If you want to know the amount of the covering the surface of a blue band margarine can, then you are finding the surface area of a right cylinder. Total surface area of the can is the sum of the

areas of the top and bottom. Circular surfaces plus the area of the curved surface,

Now, consider a right cylinder of radius r and height h.

If the cylinder is opened up, the curved surface flattens out to form a rectangle. The length of the

rectangle is 2[Ir(the circumference of the circular base) and the width is h (the height of the

cylinder). Total surface area of cylinder:

= Area of curved surface + Area of two bases. Area of curved surface = 2th + 271? = 2nr(r + h)
+. The total surface area of a right cylinder is given by A=2mr(r + h)
Example 5 Find the total surface area of a cylinder with radius of 3m and height of 10m. Use I= 3.14

Solution:

A=a2nr(r + h)
=3, h=10

Substituting:

A=2 × 314 × 3(+10) =6 × 3.14 × 13 = 244.92cm?

«Total surface area is 244.92cm*

Surface Area of a Right Pyramid

A right pyramid is one in which the slant edges joining the vertex to the corner of the base are equal

A right pyramid with a square base.

Total surface area = area of lateral surfaces + area of base.

Example 6

A right rectangular pyramid is such that the rectangle is 12cm by 8cm and each slant edge is 12cm. Find the total surface area of the pyramid.

Solution: By Pythagoras a slant edge from the midpoint of the base length to the common vertex of the pyramid is ¥12? — 6?=6,V3 and that from the midpoint of the base width is ¥12? — 47=8y2 Area of lateral surfaces = 2 (% x 12 × 6V3 ) + 2(% x 8 × 82)

=72V3 + 64y2
= 124. 71+90.51
= 215.22cm?
Area of rectangular base = |x w = 12 × 8cm? = 96cm? Total surface area = area of lateral surfaces + area of the base = (215.22 + 96) cm? = 311.22cm? Area =311.22cm'

Surface Area of a Right Prism

A full brick or concrete block is an example of a right rectangular prism

> 2A

A right prism is a prism in which each of the vertical edges is perpendicular to the plane of the base.

The figure above shows a rectangular right prism in which there are 6 faces though only three of

them can be seen easily. Surface Area = Total or sum of the areas of each face.

Generally for any right prism,

Total surface area = area of lateral surface + area of bases

Example 7 The height of a right prism is 4cm and the perimeter of its base is 30cm. Find the area of its lateral surface.

Solution:

Area of lateral surface = perimeter of base x height =30 × 4cm? = 120cm?

Example 8

Find the total surface area of a rectangular prism 12 by 8 by 6 cm high.

Solution:

Lateral surface area = 6 × 2 (12 + 8) cm? = 240cm?
Area of base = 2 (12 × 8) = 192 cm? Total surface area = (240 + 192) cm?
= 432 cm?
+. Total surface area = 432 cm?

The Formulae to Calculate the Surface Area of Spheres

Apply the formulae to calculate the surface area of spheres

Surface Area of a Sphere

frm

The figure above shows a sphere (ball) with radius "r

The surface area of a sphere is four times the area of circle with the same radius. The area of a

circle is Ilr. Hence, the surface area of sphere is equal to 41'

Surface area of a sphere = 41tr?

Example 9

Find the surface area of a sphere of radius Scm. (II= 3.14)

Solution:

Surface area of sphere = 41°. =4 × 3.14 × 5x Sem?

The surface area is 314cm?.

Example 10

Find the surface area of a tennis ball, given that its radius is 3.3cm. Usell= 3.14 Express your

answer to the nearest tenth.

Solution: A=4nr SoA=4 × 3.14x(3.3)2 = (12.56) (10.89) = 136.7784 «The surface area is 136.8cm?.

Exercise 1

Do the exercise to check your understanding. Use x= 3.14 throughout the exercise.

1.The altitude of a rectangular prism is 4cm and the width and lengths of its base are 2cm and

3cm respectively calculate the total surface area of the prism.

2. The following diagram shows a cylinder of diameter 20 units and height 9 units. What is its

curved surface area?

—_

3. The diagram below shows a cone of height 24 cm and base diameter 14 cm. what is its total

surface area?

4. The base of a right pyramid is a rectangle 6cm by 8cm and the slant edges are each 9cm long.

Calculate its lateral surface area. 5. Find the total surface area of a circular cylinder of diameter 8cm and height 6cm.

6. Taking the earth to be a sphere with radius 6400km, find its surface area.

Volume of Three Dimensional Objects

The Formulae for Calculating Volume of Prisms, Cylinders and Pyramids

Derive the formulae for calculating volume of prisms, cylinders and pyramids

Volums Of Some Three — Dimensional Figures

We have seen some formulas for calculating the surface areas of some three dimensional figures. Let us see as well formulas for calculating the volumes of such figures.

-The amount of space that is enclosed by a space figure is called the volume.

The Volume is measured in cubic units, cubic meters (m*), Cubic centimeters (cm*) etc.

When we find (calculate) the volume of a space figure or solid, we are finding the number of

cubic units enclosed by the given spaces figure.

(a) Volume of a Right Prism

co) % AS

The figure above shows a right rectangular prism. Let / be height, w width and | the length of the

prism. Then the Volume of the prism is given by: V = Base area x height= [xWxh
Generally, volume of any right prism is equal to the product of the area of the base and the height V = Base area x height.
Volume (V) = Base area x height

Or

Y= B, < H| Where V = volume , B, = base area and H = height

(b) Volume of a Right Cylinder

Consider a right circular cylinder with radius" r "and height h as shown below.

The volume of a right circular cylinder is equal to the product of the area of the base and the height.

If V is volume, A is area of the base and h is the height,

Then Volume = Area of base x height
Or V = 2r2x h where nr? is the base area (a circular bases area) and h is the height of the cylinder.
V=ar7h

(c) Volume of a Pyramid

Generally, the volume of a pyramid is one — third the product of its altitude (height) and its base

area.

If h is the perpendicular distance from the vertex of the pyramid to its base then,

Volume of the pyramid = {base area x height)
-= +(base area x height)

(d) Volume of a Cone

raed)

Consider a cone of radius "r" and altitude h as shown below.

height radius side length

A base of a circular cone can be considered to be a regular polygon with very many (infinitive) numbers of sides.

Like a pyramid, Volume of a circular cone is one — third the product of its altitude and its base area.

If the base is of radius "r', then the base area is mr' and the volume becomes =nrh
The volume of the cone( V) = <arh

(e) Volume of a Sphere

The figure above shows a sphere of radius r, if the sphere can be put inside a cylinder of the

>

same radius" r', then the height h = 2r.
It follows that Volume of Sphere is given by Volume = mre
Volume of sphere = $ mr

The Formulae to Calculate the Volume of Cylinders, Pyramids and Cones Apply the formulae to calculate the volume of cylinders, pyramids and cones

Example 11

Find the volume of the prism shown below, given that the dimensions are in meters (m)

Solution:

The base of the prism is a rectangle. Area of base = | X w = 11 × 6= 66 m?
h= 6m v= Base area x height (h) v = 66m2 × 6m=396m?
Volume of prism = 396m

Example 12

Calculate the volume of a rectangular prism whose base is 8cm by Scm and whose height is

10cm.

Solution:

Volume = Area of Base x height =(8 × 5)x 10cm? = 400cm3

«The volume is 400cm.

Example 13

Calculate the radius of a right circular cylinder of volume 1570m* and height 20m. Use 2=3.14
Solution: V=nr'xh So 1570 =3.14xrx 20

1570 3.14 × 20

~ r=
r=25 r=5m :.The radius is 5m.

Example 14

A pipe made of metal lcm thick, has an external (outside) radius of 6cm. Find the volume of metal used in making 4m of pipe. Use 1=3.14
Solution: Let R be the external radius and r the inner radius, and h the height, Volume of pipe = area of cross section x height
=(xR? —ar?)h
=_R*h —ar*h Verh =a(R+j(R-nh R=6cm, r= Sem, h= 4000cm V=3.14 × 11 × 1 × 400m*
= 13816 cm? «Volume of metal is 13816cm>

Example 15

Find the volume of a pyramid with rectangular base with length 6m and width 4m if the height of

the pyramid is 10m.

Solution Base area = length x width = 6 × 4m? = 24m? From V = (base area x height), h= 10m V== 324m? x 10m) = 80m? +The volume is 80m

Example 16

Calculate the volume of a square pyramid whose altitude is 10cm and length of side of base is 6cm.

Solution:

V= (base area x height) = 36 × 6 × 10) cm*= 12 × 10 cm?
= 120cm? «The volume is 120cm'

Example 17

Calculate the volume of a cone having base radius 10cm and altitude 12 cm Use 1=3.14 Solution: Volume = neh =£x3.14 × 102 × 12cm'
= 1256cm3 «The volume is 1256cm'.

Example 18

Find the volume of a sphere whose radius is 10cm. (Take m= 3.14). Solution: Volume =a =2 × 3.14 × 10 × 10 × 10
= 4.1867 × 10° = 4186.7cm? 2. The volume is 4186.7cm?

Example 19

The volume of a spherical tank is 268m'. Calculate the radius of the tank. (m= 3.14) Solution: Volume = mm
p= 2y 4n

3 3 × 268 3 × 3.14

= 64.0
r= V64 =4m

«. The radius of the tank is 4m.

Example 20

Find the volume of rubber in a hollow spherical ball with inner diameter 14cm and outer diameter 16cm. (Take a = 3.14)

Solution: Let inner radius be r and outer radius be R.

Then R = 8cm and r= 7cm Volume of spherical shell = Volume of outer sphere—Volume of inner sphere. Vv =*nrR?— tar

3 3

=n ®-P) But R=8andr=7
V= +nR?- f)= <n(8?- 7) =: x 3.14 × 169
= 707.55em3 -. The volume of rubber is 707.55cm?
Answer the following questions and (use a = 3.14)
1. What is the volume of a right prism whose base is a regular hexagon ( n = 6) with a side of the base 4cm long and the height of the prism.

2. Find the volume of a cylinder whose diameter is 28cm and whose height is 12cm. 3. Find the volume of a square pyramid whose height is 24cm and slant edge 25cm each.

4. The slant height of a cone is 20cm and the radius of its base is 12cm. Find its volume in terms

of z.

5. The volume of a sphere is 827cm'. Find its radius.

6. A cylinder and sphere have the same volume. If the radius of the sphere is Scm and radius of

the cylinder is 3cm, Calculate height of the cylinder.

Find the surface area of this rectangular prism (cuboid)

8.The diagram shows a barn. What is the volume of the barn? (The length of the hypotenuse in

the right triangle is rounded to the nearest foot.)

9.What is the volume of this prism?

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The diagram shows a prism whose cross-section is a right triangle. What is the volume of the prism?

Summary of the topic

Here are the important formulas you have covered under the section on surface areas

summarized.

Surface area of:

1. Right circular cone. Curved surface = ar]
Circular base = ax?
Total surface area, A= mal + a
Or A=a(r+)) 2. Right cylinder
Curved surface = 2arh
Two bases =2ar
Total surface area, A = 2ar? + 2arh
Or A= 2ar[r +h]
3. Right Pyramid Total surface area = area of lateral surface + area of base.
4. Right Prism Area of lateral surface = perimeter of base x height Total surface area = area of lateral surface + area of base
5. Sphere Surface area = 4 × 1.
You can now have a summary of the important formulas for calculating volume of some three dimensional figures as follows:- 1. Prism: V = base area x height
2. Cylinder: V =area of circular base x height = ar*h 3. Cone: V= + (area of circular base x height) = imrh
4. Pyramid: V = = (base area x height)

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