This lesson explains the concepts of Surface area and combines important properties
of Cube, Prism, Pyramids, Cylinders, Cones, and Spheres, to apply
the relevant formulas for determining the surface area of the 3D solids. As a first
step, you’ll need to review earlier learning about the solids and use the formulas
for surface area. Moving forward you’ll be able to calculate lateral area and surface
area of these shapes. Further on will solve several problems involving lateral and
total surface areas of prisms. Once mastered the surface area of cubes, prisms etc.
can move to other 3D figures including pyramids, cones, and spheres etc. The presentation
covering such content will be done by the instructor in own handwriting, watching
video and with the help of several examples with solution. This
will help you apply important geometry properties and relationships to solve problems
from day to day life situations and also doing geometry home work.(More
text below video...)
(Continued from above)Cube:
A Cuboid; where its length, breadth and height are equal i.e. where all side lengths
equal and all angles measure 90,
is known as ‘Cube’.
Commonly in geometry, a cuboid is known by a solid
figure bounded by six faces, forming a convex polyhedron (A polyhedron in geometry
is a 3D solid consisting of a group of polygons, generally joining at their edges).
Further the requirement is that these six faces each be a quadrilateral, and that
the graph formed by the vertices and edges of the polyhedron
should be similar to the graph of a cube. Also a ‘cuboid’ may be a shape of the
type in which each of the faces is a rectangle, and where each pair of adjacent
faces meets in a right angle. The later type is also known as a right cuboid,
rectangular box, rectangular hexahedron, right rectangular prism, or rectangular
parallelepiped. E.g. a chalk box, geometrical box, match box and a
book etc. are the examples of cuboids.
Fig: 3, above shows a 3D figure called cuboid. Notice that it has six rectangular
plane surfaces called faces.
Two adjacent faces meet along a line segment called an edge. It
has 12 edges in all.
It has 8 corners called the vertices (A, B, C,
D, E, F, G, and H). At every vertex,
there are three edges meeting one of these edges represents length,
second the breadth and third as the height.
The main diagonal of the cuboid (ds) is defined by joining the
linesegment of vertex D to the vertex F.
Note: If the length, breadth and height of a cuboid are equal it is a cube.
Walk through a few examples now:
Example 1: Given: AB= l, AE
= b and AD = h, find out the surface area.
Step 1: the total surface area of cuboid, can be worked out by adding areas of six
faces i.e. = sum of the areas of six faces ABCD, EFGH, ADGE,
BCHF, ABFE and DCHG
Step 2: = (lh + lh + bh + bh + lb + lb) = 2(lb +
bh + hl)
Step 3: total surface area of a cuboid = 2(lb
+ bh + hl)
Main diagonal of a cuboid=
(l^{2} + b^{2} + h^{2}) Now if it is considered as cube : A cube is a special case of a cuboid when
all the edges are equal i.e. l = b = h
If each one is a, l = b = h = a
Therefore surface area of the cube = 6a^{2} and main diagonal of
cube is =
3
* a
Main Diagonal of a cube =
3
* a, as the final answer.
Example 2: Given if the surface area of a cube is 96
sq cm, find length of each side.
Step 1: the surface area of a cube equals to = 6a^{2}, where a
is the side of cube
Step 2: thus 96 = 6a^{2} i.e. a^{2} = 16
Step 3: a = 4 cm.
Length of each side is 4 cm, as the final answer.
Example 3 Given Karen grows cabbages in her garden square
in shape. Each cabbage takes 1 square feet of area in the garden. This year, she
has increased her output by 211 cabbages as compared to last year. The shape of
the area used for growing the cabbages has remained a square in both these years.
How many cabbages did she produce this year?
Step 1: the shape of the area used for growing cabbages is a square in both the
years. Say the side of the square area used for growing cabbages this year be X
ft, and last year it was Y feet.
Step 2: the ground area this year is X^{2} and last year it was Y^{2}
sq ft.
Step 3: known that increase in cabbages is 211. Therefore, increase in area over
previous year is 211 sq ft (given that each cabbage takes 1 square feet of area
in the garden).
Step 4: thus (X^{2}  Y^{2}) = 211
OR (X + Y) (X  Y) = 211
Step 5: As 211 is a prime number, therefore, its two factors can be represented
by (106 + 105) * (106  105).
Step 6: (X + Y) (X  Y) = (106 + 105) (106  105).
It can thus be deduced that X = 106 and
Y = 105.
Step 7: Number of cabbages produced this year equals to X^{2} =
(106)^{2} = 11236, as the final answer.
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