Surface Area of Combination of Solids
- We saw different solids in our daily life
- These are not exactly the shape of single cube,cubiod,sphere,cylinder ,regular cone. They seems to be combination of these solids
- Let's take an example of solid below
- So solids can be consider as combination of cylinder and Regular cone
- Surface area of the solid will be given
Surface Area of Solid = Curved Surface Area of Cylinder + Curved Surface area of Regular cone+ Surface area of cylinderical Base
$= 2\pi r h + \pi r L + \pi r^2$
- Similarly we can look at the below figure
- Surface area of the solid will be given
Surface Area of Solid = Curved Surface Area of Cylinder + Curved Surface area of hemispshere+ Curved Surface area of hemispshere
$= 2\pi r2 h + 2 \pi r^2 + 2 \pi r^2$
Volume of Combination of Solids
- We discussed about surface area in the above section,let's see the volumen now.
- Let's take an example of solid below
- So solids can be consider as combination of cylinder and Regular cone
- Volume of the solid will be given
Volume of Solid = Volume of Cylinder + Volume of Regular cone
$= \pi r^2 H + \frac {1}{3} \pi r^2H $
- Similarly we can look at the below figure
- Volume of the solid will be given
Volume of Solid = Volume of Cylinder + Volume of hemispshere+ Volume of hemispshere
$= \pi r^2 h + \frac {2}{3} \pi r^3 + \frac {2}{3} \pi r^3$
Formula for Surface Area and Volume of Common Figures
Surface Area and Volume of Cube and Cuboid<
Type
|
Measurement
|
Surface Area of Cuboid of Length L, Breadth B and Height H
|
$2( LB + BH + LH )$
|
|
Lateral surface area of the cuboids
|
$2( L + B ) H$
|
|
Diagonal of the cuboids
|
$\sqrt {L^2+B^2+H^2} $
|
|
Volume of a cuboids
|
$LBH$
|
|
Length of all 12 edges of the cuboids
|
$4 (L+B+H)$
|
|
Surface Area of Cube of side L
|
$6L^2$
|
|
Lateral surface area of the cube
|
$4L^2$
|
|
Diagonal of the cube
|
$L \sqrt{3}$
|
|
Volume of a cube
|
$L^3$
|
Surface Area and Volume of Right circular cylinder
|
Type
|
Measurement
|
|
Curved or lateral Surface Area of cylinder
|
$2 \pi rh$
|
|
Total surface area of cylinder
|
$2 \pi r (h+r)$
|
|
Volume of Cylinder
|
$\pi r^2h$ |
Surface Area and Volume of Right circular cone
|
Type
|
Measurement
|
|
Curved or lateral Surface Area of cone
|
$\pi rL$
|
|
Total surface area of cone
|
$\pi r (L+r)$
|
|
Volume of Cone
|
$\frac {1}{3} \pi r^2h$
|
Surface Area and Volume of sphere and hemisphere
|
Type
|
Measurement
|
|
Surface area of Sphere
|
$4 \pi r^2$
|
|
Volume of Sphere
|
$\frac {4}{3} \pi r^3$
|
|
Curved Surface area of hemisphere
|
$2 \pi r^2$
|
|
Total Surface area of hemisphere
|
$3 \pi r^2$
|
|
Volume of hemisphere
|
$ \frac {2}{3} \pi r^3$
|
|
Volume of the spherical shell whose outer and inner radii and ‘R’ and ‘r’ respectively
|
$\frac {2}{3} \pi (R^3-r^3)$
|
Surface Area and Volume of frustum of cone
- Frustum of cone is created by removing some portion from top as shown above
- Here h = vertical height of the frustum
l = slant height of the frustum
$r_1$ and $r_2$ are radii of the two bases (ends) of the frustum.
|
Type
|
Measurement
|
|
Volume of a frustum of a cone
|
$\frac {1}{3} \pi h(r_1^2 + r_2^2 + r_1r_2)$
|
|
Slant height of frustum of a cone
|
$\sqrt {h^2 + (r_1 -r_2)^2}$
|
|
Curved surface area of a frustum of a cone
|
$ \pi l(r_1 + r_2)$
|
|
Total Surface area of Frustum
|
$ \pi l(r_1+ r_2) + \pi (r_1^2 + r_2^2)$
|
How to solve Surface Area and Volume Problem of Combination of Solids
1) We have explained the surface area and volume of various common sold shapes.
2) We need to divide the given solid shape into known shapes
3) Find out the given quantities like radius,height
4) Apply the formula from the above given tables and get the answer
5) Make sure you use common units across the problem
Question 1
The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume, the curved surface area and the total suface area?
Solutions
Here $h=45 \;cm ,r_1=7 \; cm ,r_2=28 \; cm$
Volume of a frustum of a cone =$\frac {1}{3} \pi h(r_1^2 + r_2^2 + r_1r_2)$
$=48510 cm^3$
Since Curved surface area and Total Surface requires slant height,let's calculate that first
Slant height of frustum of a cone=$\sqrt {h^2 + (r_1 -r_2)^2}$
=49.65 cm
Now Curved surface area of a frustum of a cone=$ \pi l(r_1 + r_2)=5461.5 cm^2$
And Total curved surface area of the frustum =$ \pi l(r_1+ r_2) + \pi (r_1^2 + r_2^2)=8079.5 cm^2$
Question 2
A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 4 cm and the diameter of the base is 8 cm. Determine the volume of the toy and the total surface area of the toy.
Solutions
Let r be the radius of the hemisphere and the cone and h be the height of the cone
r=4 cm and h=4 cm
Volume of the toy = Volume of the hemisphere + Volume of the cone
$= \frac {2}{3} \pi r^3 + \frac {1}{3} \pi r^2h$
$=\frac {1408}{7} cm^3$
total surface area of the toy=Curved surface area of cone + curved surface area of hemisphere
$=\pi rl +2 \pi r^2$
Now $l=\sqrt h^2 + r^2 =4 \sqrt 2$
So total surface area
$=171.68 cm^2$
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