Mensuration
Solid Geometry
A solid
is a figure bounded by one or more surfaces. Hence a solid has length, breadth
and height. The plane surfaces that bind it are called its faces and the solid so generated
is known as a polyhedron.
A solid has edges, vertices and faces which are shown in the figure given
below.
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.
Shown in Figure above 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 lecture are
(1)
Prism (2) Cube (3) Right circular cylinder
(4)
Pyramid
(5) Right circular cones (6) Sphere
Apart
from defining these objects, methods to calculate their surface area and
volume are also incorporated in this lecture.
Prism
Any solid formed by joining the corresponding vertices of two congruent
polygons is called a prism.
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 parallelograms formed by the lateral
edges are called lateral faces.
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 of the prism, it is called a right prism. If the lateral edges are not
perpendicular to the base of the
prism it is called an oblique prism
Consider the right
prism shown in figure.
ABCDEF
is a prism ΔABC & ΔFED are congruent and seg. AF, seg. CD and seg. BE are
perpendicular to the planes containing Δ ABC and
Δ FED.
Also in
ΔABC; m L ABC = 900,
I (seg. AB) = 3 cm & I (seg. AC) = 5 cm.
The surface area of the prism is
the sum of the surface areas of all its surfaces.
Surface area = A(ΔABC) + A(ΔFED) + A(□ABEF) + A(□ACDF)
+ A(□ BCDE)
1
A(ΔABC)
= A(ΔFED) = ½ x base
x height
base is unknown and height is 3 cm. Since
ΔABC is right angle triangle;
If h = Height, Sc =
Lateral surface area; ST = Total surface area; V = volume; B
= Area of base, then
ü V=Bxh
ü Sc = perimeter of base x h
ü S = Sc + (2 x B).
Difference between a right prism and an oblique prism
In a right prism, the angle
formed by the lateral edges with the planes containing the bases is 90°. In
oblique prism the same angle is not 90°.
Altitude of a prism
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.
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 shows a cuboid.
As can
be seen in figure 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, a x b = Area.
Important
If the height of a cuboid is zero it
becomes a rectangle.
El. The measurement of a luggage box are 48 cm, 36 cm and 28 cm. Find the volume of the box. How many sq. cm of cloth
is required to make a cover for the box?
Sol. Given: length a = 48 cm, breadth b =
36 cm and height c
= 28 cm.
(i) The
required volume
=
48 x
36
x
28
cm3 = 48, 384 cm3.
(ii) The
quantity of cloth required to make the cover of the box
=
Total surface area of the box =2[48x36+36x28+28x48]cm2
=2x48x [lx 36+3x7+ 28 x 1] cm2
=
96 x
85
cm2 = 8160 cm2.
Cube
A six-faced solid figure with all faces
equal and adjacent faces mutually perpendicular is a cube. Also we can say that
a cube is a square right prism with the lateral edges of the same length as
that of a side of the base.
If "a" be the edge of a cube,
then
ü The
longest diagonal = a√3
ü Volume = a3
ü Total surface area = 6 a2.
E2.
If
the volume of a cube is 27. Find the surface area.
Sol. Volume of a cube = cube of its lengthy
l3 = 27 Since I3 =
27
(l3 )1/3= (27 )1/3
l = 3
Surface area of a cube = 6 l2 = 6(3)2 = 6 x 9
= 54sq. units
Right Circular Cylinder
Pillars, pipes etc. are examples of circular
cylinders encountered daily. A circular cylinder is two circles with the same radius at a finite distance from each
other with their circumferences joined.
The definition of a
circular cylinder is a prism with circular bases. The
line joining the centers of the two circles is called the axis.
If the axis is perpendicular to the circles it is a right circular cylinder otherwise it is an oblique circular cylinder.
This cylinder is generated by rotating a rectangle by fixing one of its sides.
E3. If the lateral area of a right circular cylinder is 24π and its radius is 2. What is its height.
Since r = 2,
Given
that Lateral Area = 24 π
24π = 4πh
h=6units.
E6.
A rectangular sheet of aluminium foil is 44 cm long and 20 cm wide. A cylinder is made
out of it, by rolling the foil as in figure.
Find the volume of the cylinder. (Take π = 22/7)
Sol.
The
cylinder obtained from the foil has height h = 20 cm (see
figure).
Its base circle has circumference 44 cm. If r is the radius of the
base,
E7. The radius
of the base and the height of a cone are respectively
35 cm and 72 cm. Find the volume, curved
surface area and total surface area of
the cone.
Sol. Volume =1/3πr2h = 1/3 x 22/7 x 35 x35 x72 cm3
try Curved and Total
surface area yourself.
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.
r = Radius
Volume = 4/3
π r3ü Surface Area = 4 π r2
E8. Find the volume and surface area of a sphere of radius 4.2 cm.
Sol. Given radius = 4.2 cm.
Volume = 4/3 x 22/7 x 4.2 x 4.2 x 4.2 cm3 = 310 cm3 approx.
And
surface area = 4 x 22/7 x 4.2 x 4.2 cm2 =222 cm2 approx.
E9. If
the volume of a sphere is 36rr. Find its radius and surface Area.
Hemisphere
r = Radius
ü
Curved
surface Area = 2 π
r2
ü Total surface Area = Area of Circle + Curved Surface
Area = π
r2 + 2π r2 = 3 π r2
Pyramid
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.
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.
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.
A regular pyramid has a 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 = ½ p x I square units where 'p' is the perimeter and 'I' is the
slant height.
Let
height of side faces = slant height = l
ü Volume = 1
x Base Area x height (H)
ü Sc
(lateral surface area) = 1/2 x perimeter of base x
I
ü ST (Total surface area) =
Base area + Sc.
E10. For a regular
square pyramid if the length of a side of base = 4 and the height = 6. Find the
volume.
Sol. Since length of
the base = 4
Base
Area = (4)2 = 16
Volume = 1/3 x 16 x 6
= 32 cube units.
E11.If the perimeter and slant height of a regular pyramid
are 10 and 3 respectively. Find its lateral area.
Sol. For a
regular pyramid Lateral Area = 1/2 p x l .
where
p = perimeter and I = slant height
Given
that p = 10 and I = 3
.•.
Lateral Area = 1/2 x 10 x 3 = 15 square units.
Frustum of Pyramid
A pyramid with some of its top portion cut off is called the frustum of that original
pyramid.
If Al
and A2 be the areas of top and bottom faces of a frustum, I and h be its slant height and height
respectively, then
ü Sc (lateral surface area) = 1/2 (P1 + P2) x l where P1 & P2 are perimeters of top &
bottom faces.
ü ST
= (total
surface area) = Sc + A1 + A2
ü
V = h/3 (A1 + A2
+ √A1 A2)
Frustum of
Cone
A cone with some
of its top portion cut off is called the frustum of that original cone.
The
formulae for this can be derived from above as cone.
Some
Important Results
ü
For any regular solid
Number of faces + Number of
vertices = No. of edges + 2
ü
In a prism with a base of n sides.
Number of vertices = 2n,
Number of faces = n + 2
ü
In a pyramid with a base of n sides.
Number of vertices = n + 1, Number of
faces = n + 1
ü
Distance travelled by a wheel in n
revolutions (without slippage) = n x
circumference
ü The rise or fall of liquid level in a
container =Total volume
of object submerged or taken out ÷ cross sectional area of container
Imp.
If the given rectangular sheet of paper
is rolled across its length to form a
Cylinder,
having a height b, then the
ü Volume of the
cylinder = l2b/4π
If the given
rectangular sheet of paper is rolled across its breadth to form a cylinder,
having a height I, then
2
Solids
inscribed/circumscribing other solids
1.
If a largest possible sphere is
circumscribed by a cube of a edge
'a' cm, then the radius of the sphere = a/2 .
2. If a largest possible cube is inscribed in a sphere of radius 'a' cm, then
ü The
edge of the cube = 2a/√3.
3. If a largest possible sphere is inscribed in a cylinder of radius 'a' cm and
height 'h' cm, then for h > a,
the
radius of the sphere = a and
ü the
radius = h/2(fora > h)
4. If a largest possible sphere is inscribed in a cone of radius 'a' cm and slant height equal to the diameter of the base, then
ü
the radius of the sphere
= a/√3.
a
5.
If a largest possible cone is inscribed in a cylinder of radius
'a' cm and height 'h' cm, then the radius
of the cone = a and height = h.
6.
If a largest possible cube is inscribed in
a hemisphere of radius 'a' cm, then
the edge of the cube = a √2/3
No comments:
Post a Comment