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Mensuration concept

Mensuration concept

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CHAPTER 1 : SURFACE AREA AND VOLUME

1.1 Introduction to Solid Geometry

All the geometric shapes discussed in this book till now i.e. polygons, circles etc. are planar shapes. They are called two dimensional shapes i.e. generally speaking they have only length and breadth. In the real world however every object has length breadth and height. Therefore they are called three dimensional objects. No single plane can contain such objects in totality.

Consider the simplest example of a three dimensional shape - a brick.


Figure 1.1

Shown in Figure 8.1 is a brick with length 10 cm, breadth 5 cm and height 4 cm. There cannot be a single plane which can contain the brick.

A brick has six surfaces and eight vertices. Each surface has an area which can be calculated. The sum of the areas of all the six surfaces is called the surface area of the brick.

Apart from surface area a brick has another measurable property. i.e. the space it occupies. This space occupied by the brick is called its volume. Every three dimensional (3-D) object occupies a finite volume. The 3 -D objects or geometric solids dealt within this chapter are

(1) Prism
(2) Cube
(3) Right circular cylinder
(4) Pyramid
(5) Right circular cones and
(6) Sphere

Apart from defining these objects, methods to calculate their surface area and volume are also incorporated in this chapter.

1.2 Prism

Any solid formed by joining the corresponding vertices of two congruent polygons is called a prism.

Figure 1.2

The two congruent polygons are called the two bases of the prism. The lines connecting the corresponding vertices are called lateral edges and they are parallel to each other. The parallelogram formed by the lateral edges are called lateral faces.

(See figure 1.2)Prisms are of two types depending on the angle made by the lateral edges with the base.

If the lateral edges are perpendicular to the base the prism is called a right prism. If the lateral edges are not perpendicular to the base the prism is called an oblique prism.

Consider the right prism shown in Figure 1.3


Figure 1.3

ABCDEF is a prism D ABC & D FED are congruent and seg. AF, seg. CD and seg. BE are perpendicular to the planes containing D ABC & D FED.

Also in DABC m Ð ABC = 900 , l (seg. AB) = 3 cm & l (seg. AC) = 5 cm

The surface area of the prism is the sum of the surface areas of all its surfaces.

base is unknown and height is 3 cm.

1.3 The cuboid and the cube

A book, a match box, a brick are all examples of a cuboid.

The definition of a cuboid is derived from that of the prism.

A cuboid is a solid formed by joining the corresponding vertices of two congruent rectangles such that the lateral edges are perpendicular to the planes containing the congruent rectangles. Figure 1.3 shows a cuboid.

Figure 1.3

As can be seen in Figure 8.3 the cuboid has six surfaces and each one is congruent and parallel to the one opposite to it.

Thus there is a pair of three rectangles which goes to make a cube.

It is already known that the area of a rectangle is the product of its length and breadth, l ´ b = Area.

The surface area of a cuboid is the sum of the areas of all its faces,

Now consider a cuboid made of 6 congruent squares instead of a pair of three rectangles. This structure has l = b = h i.e. the length breadth and height are the same. Such a cuboid is called a cube.

Cube

A cube is a square right prism with the lateral edges of the same length as that of a side of the base. (See figure 1.4)


Figure 1.4

As can be seen in Figure 1.4 in a cube the length, breadth and height are equal. Therefore the surface of a cube is essentially six squares of equal length.

The area of a square is the square of its length. The total surface area of the cube is the sum of the areas of all its 6 surfaces. If length of one side of the cube is l the area of one surface is l 2 \ the surface area of the cube = 6 ´ l 2 = 6 l 2 .

To find volume of a cuboid first a unit cube has to be defined. A unit cube is a cube with length equal to one unit.

This unit is chosen as per convenience. For small volumes 1 cm can be chosen as the unit length . The volume of this cube is 1 cubic cm. It is represented as 1 cm3. For larger volumes 1 meter can be used as the unit length. In this case the volume of the unit cube is 1 cubic meter and is represented by 1m3 .

Depending on the size of the cuboid whose volume is to be measured the appropriate unit cube is chosen.

The volume of a cuboid is the number of unit cubes that are required to fill the cuboid completely.

Figure 8.6 shows a cuboid whose volume is to be measured. This cube has length 10 cm breadth 8 cm and height 4 cm.

The three edges OA, OB and OC are perpendicular to each other. On these sides points LM and N are chosen such that l (seg. O l) = l (seg. ON) = unit length.


Figure 1.5

These sides seg. OL, seg. OM and seg. ON form three sides of a unit cube as shown in Figure 1.6a Let another unit cube be placed on the right side of this unit cube such that one of its edges lies along seg. LA and one of its face covers the right side of the first unit cube. (See Figure 1.6b)

Figure 1.6a                       Figure 1.6b

Since length of seg. OA = 10 and l (seg. OL) = 1, 10 unit cubes can be arranged along OA. Similarly l (seg. OB) = 5. \ 5 unit cubes can be arranged along seg. OB. Thus to cover the face AOBD, 10 ´ 5 = 50 unit cubes will be required. They will form the first layer towards filling the cuboid.

Since l (seg. OC) = 6 it is obvious that 6 such layers are required to fill the cuboid.

\ Number of unit cubes required = 50 ´ 6

 = 300

Thus the space occupied by the cuboid or its volume
= 300 cubic units

= 300 unit3

In general it may be stated that for a cuboid with length l , breadth b and height h the volume = l ´ b ´ h cubic units.

For a cuboid,

Surface area = 2 (l b + b h + l h)

Volume = (l ´ b ´ h) cubic units.

Since a cube has l = b = c

its surface area = 6 l 2 &

Volume = l 3


Example 1

Draw a right hexagonal prism with height 5 cm and length of one side of the hexagon = 2.5 cm.

Solution :

Example 2

When is a cuboid a cube ?

Solution :

If the length, breadth and height of a cuboid are equal it is a cube.

Example 3

What is the difference between a right prism and an oblique prism ?

Solution :

In a right prism, the angle formed by the lateral edges with the planes containing the bases is 900. In oblique prism the same angle is not 900 . In oblique prism the same angle is not 900 .

Example 4

What is the altitude of a prism ?

Solution :

A prism has two congruent polygons joined together at their corresponding vertices. The perpendicular distance between the two polygons is the height of the prism. It is hence obvious that in a right prism the length of the lateral side is the altitude of the prism.

Example 5

If the volume of a cube is 27. Find the surface area.

Solution :

Volume of a cube    = cube of its length

= l3

= 27

Since   l 3     = 27

                

                  l = 3

Surface area of a cube      =    6 l 2

= 6 (3)2

= 6 ´ 9

= 54 sq. units

Example 6

If the height of a cuboid is zero it becomes a (a) prism (b) cube (c) rectangle.

Solution :

A cuboid is two congruent rectangles that are separated from each other by a distance = height. If the height is zero these two will collapse on each other to form one rectangle.

The other two options i.e. prism and cube are obviously wrong because both cannot have height = zero.

1.5 Pyramids

A pyramid is a polygon with all the vertices joined to a point outside the plane of the polygon.

If the polygon is regular then the pyramid is called a regular pyramid and is named by the polygon which forms its base.

If the base is a square the pyramid is called a regular square pyramid, if it is a pentagon the pyramid is called a regular pentagonal pyramid and so on and so forth.

The parts of the pyramid are named analogous to the geometric solids mentioned earlier in the chapter. It is a base, lateral faces, lateral edges and an altitude.

The terminology varies only for the vertex. In all the geometric solids seen so far all corners are called vertices. In a pyramid however the apex to which all corners of the polygon are joined is called the vertex of the pyramid.

A regular pyramid has a property called slant height which is the perpendicular distance between the vertex and any side of the polygon.

The lateral area of a regular pyramid is defined using this parameter.

Lateral area of a regular pyramid = square units

where ’p’ is the perimeter and ’ l’ is the slant height.

Figure 1.8

In pyramid PABC, P is the vertex and PR is the slant height.

     Perimeter      =   l (seg. AB) + l (seg. BC) + l (seg. CA)

     = p

1.6 Right circular cone

A right circular cone is a circle with all points on its circumference joined to a point equidistant from all of them and outside the plane of the circle.

The lateral area of the right circular cone

The total area TA = (LA + BA) square units.

BA = base area = pr2

\ TA = p r l + pr2

= p r ( l + r) square units

1.7 Sphere

The simplest example of a sphere is a ball. One can call here that a circle is a set of points in a plane that are equidistant from one point in the plane. If this is extended to the third dimension, we have all points in space equidistant from one particular point forming a sphere.

The sphere has only one surfaces and its surface area = 4p r2

The volume of a sphere =

Example 1

If the lateral area of a right circular cylinder is 24p and its radius is 2 . What is its height.

Solution :

Lateral area of a right circular cylinder = 2 p r h

Since r = 2 , LA = 4 p h.

Given that LA = 24 p

\ 24 p = 4 p h

\ h = 6 units

Example 2

Find the total area of the right circular cylinder with a radius of 10 units and a height of 5 units.

Solution :

Total area of a right circular cylinder is 2prh + 2p r2

If r = 10 and h = 5

Total area = 100p + 200 p = 300 p

Example 3

What is the radius of the right circular cylinder with volume 18p cubic units and height = 2.

Solution :

Volume of right circular cylinder = pr2h

where r = radius and h = height

Given that V = 18p and h = 2

\ 18p = pr2 ´ 2

or r2 = 9

r = 3 units

Example 4

If the perimeter and slant height of a regular pyramid are 10 and 3 respectively. Find its lateral area.

Solution :

For a regular pyramid lateral area, LA =´ p ´ l

where p = perimeter and l = slant height

Given that p = 10 and l = 3

\ LA = ´ 10 ´ 3 = 15 square units.

Example 5

If the volume of a sphere is 36p. Find its radius and surface area.

Solution :