Surface Areas and Volumes

Surface Area is the area of the outer part of any 3D figure
Volume is the capacity of the figure i.e. the space inside the solid.

To find the surface areas and volumes of the combination of solids, we must know the surface area and volume of the solids separately. Some of the formulas of solids are -

Cuboid and its Surface Area

The surface area of a cuboid is equal to the sum of the areas of its six rectangular faces.

Consider a cuboid whose dimensions are l × b × h respectively.

The total surface area of the cuboid (TSA) = Sum of the areas of all its six faces

TSA (cuboid)=2(l × b) + 2(b × h) + 2(l × h) = 2(lb + bh + lh)

Lateral surface area (LSA) is the area of all the sides apart from the top and bottom faces.

Consider Cuboid ABCDEFGH,

The lateral surface area of the cuboid = Area of face AEHD + Area of face BFGC + Area of face ABFE + Area of face DHGC

LSA (cuboid) = 2(b × h) + 2(l × h) = 2h(l + b)

Length of diagonal of cuboid

Length of diagonal of a cuboid = √( l²+b²+h²)

Volume of Cuboid

Volume of a cuboid = (base area) × height = (lb)h = lbh

Cube and its Surface Area

For a cube, length = breadth = height

Consider Cube with length l,

Total Surface area of a cube

TSA (cube) =2 × (3l²) = 6l²

Curved Surface Area of a cube

LSA (cube) = 2(l × l + l × l) = 4l²

Length of diagonal of a cube

Diagonal of a cube =√3l

Volume of a cube

Volume of a cube = base area × height. Since all dimensions of a cube are identical,

Volume of cube = l³

Cylinder and its Surface Area

Take a cylinder of base radius r and height h units.

Transformation of a Cylinder into a rectangle.

Curved Surface Area of a Cylinder

The curved surface of this cylinder, if opened along the diameter (d = 2r) of the circular base can be transformed into a rectangle of length 2πr and height h units. Thus,

CSA of a cylinder of base radius r & height h = 2π × r × h

Total Surface Area of a Cylinder

Total Surface Area of a Cylinder of base radius r and height h = 2π × r × h + area of two circular bases

=2π × r × h + 2πr²
=2πr(h + r)

TSA of a cylinder of base radius r & height h=2πr(h + r)

Volume of a cylinder = Base Area * Height = ( πr²) * h = πr²h

Volume of a cylinder = πr²h

Right Circular Cone and its Surface Area

Consider a right circular cone with slant length l, base base radius r and height h.

Curved Surface Area of a Cone

CSA of right circular cone = πrl

Total Surface Area of a Cone

TSA = CSA + area of base = πrl + πr² = πr(l + r)

TSA of right circular cone = πr(l + r)

Volume of a Right Circular Cone
The volume of a Right circular cone is 1/3 times that of a cylinder of same height and base.
^{In other words, 3 cones make a cylinder of the same height and base.}
The volume of a Right circular cone =(1/3) πr²h
Where r is the radius of the base and h is the height of the cone.

Sphere and its Surface Area

For a sphere of radius r,

Surface Area of a Sphere

Curved Surface Area (CSA) = Total Surface Area (TSA) = 4πr²

Volume of a Sphere
The volume of a sphere of radius r = (4/3)πr³

Hemisphere and its Surface Area

Curved Surface Area of Hemisphere
We know that the CSA of a sphere = 4πr². A hemisphere is half of a sphere. Thus ,
CSA of a hemisphere of radius r = 2πr²
Total Surface Area of Hemisphere
Total Surface Area = curved surface area + area of the base circle
TSA of a hemisphere of radius r = 3πr²

Volume of Hemisphere
The volume (V) of a hemisphere will be half of that of a sphere.
The volume of a hemisphere of radius r = (2/3)πr³

Quick Summary of all the formulae

Name	Figure	Lateral or Curved Surface Area	Total Surface Area	Volume	Length of diagonal and nomenclature
Cube		4l²	6l²	l³	√3 l = edge of the cube
Cuboid		2h(l +b)	2(lb + bh + hl)	lbh	l²+b²+h² l = length b = breadth h = height
Cylinder		2πrh	2πr²+ 2πh = 2πr(r + h)	πr²h	r = radius h = height
Hollow cylinder		2πh (R + r)	2πh (R + r) + 2πh (R² - r²)	-	R = outer radius r = inner radius
Cone		$\pi rl = \pi \sqrt{h^{2}+r^{2}$	πr² + πrl = πr(r + l)	1/3 πr²h	r = radius h = height l = slant height
Sphere		4πr²	4πr²	4/3 πr³	r = radius
Hemisphere		2πr²	3πr²	2/3 πr³	r = radius
Spherical shell		4πR² (Surface area of outer)	4πr² (Surface area of outer)	4/3 π(R³ – r³)	R = outer radius r = inner radius
Prism		Perimeter of base × height	Lateteral surface area + 2(Area of the end surface)	Area of base × height	-
pyramid		1/2 (Perimeter of base) × slant height	Lateral surface area + Area of the base	1/3 area of base × height	-

^{Combination of Solids}

^{Surface Area of Combined Figures}

^{Areas
of complex figures can be broken down and analysed as simpler known
shapes. By finding the areas of these known shapes, we can find out the
required area of the unknown figure.}

Example

Find the total surface area of the given figure.

Solution

This solid is the combination of three solids i.e.cone, cylinder and hemisphere.

Total surface area of the solid = Curved surface area of cone + Curved surface area of cylinder + Curved surface area of hemisphere

Curved surface area of cone =

Given, h = 5cm, r = 3cm (half of the diameter of hemisphere)

Curved surface area of cylinder = 2πrh

Given, h = 8cm (Total height – height of cone – height of hemisphere), r = 3cm

Curved surface area of hemisphere = 2πr²

Given, r = 3 cm

Total surface area of the solid

Volume of a combination of solids

Find the volume of the given solid.

Solution

The given solid is made up of two solids i.e. Pyramid and cuboid.

Total volume of the solid = Volume of pyramid + Volume of cuboid

Volume of pyramid = 1/3 Area of base x height

Given, height = 6 in. and length of side = 4 in.

Volume of cuboid = lbh

Given, l = 4 in., b = 4 in, h = 5 in.

Total volume of the solid = 1/3 Area of base x height + lbh

= 1/3 x 4 x 4 x 6 + (4) (4) (5)

= 32 + 80

= 112 in³

Conversion of Solid from One Shape to Another

When we convert a solid of any shape into another shape by melting or remoulding then the volume of the solid remains the same even after the conversion of shape.

Example

If we transfer the water from a cuboid-shaped container of 20 m x 22 m into a cylindrical container having a diameter of 2 m and height of 3.5 m. then what will be the height of the water level in the cuboid container if the cylindrical tank gets filled after transferring the water.

Solution

We know that the volume of the cuboid is equal to the volume of the cylinder.

Volume of cuboid = volume of cylinder

l x b x h = πr²h

20 x 22 x h = 22/7 x 1 x 3.5

440 × h =11

H = 2.5 cm

Frustum of a Cone

If we cut the cone with a plane which is parallel to its base and remove the cone then the remaining piece will be the Frustum of a Cone.

Example

Find the lateral surface area of the given frustum of a right circular cone.

Solution

Given, r =1.8 in.

R = 4 in.

l = 4.5 in.

The lateral surface area of the frustum of the cone = πl (R + r)

= π x 4.5 (4 +1.8)

=3.14 x 4.5 x 5.8

= 81.95 sq. in

^Example:

^{2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.}

^{Solution :}

^{Length of each cube = 64sup(1/3) = 4cm
Since
these cubes are joined adjacently, they form a cuboid whose length l =
8 cm. But height and breadth will remain same = 4 cm.}

^{∴ The new surface area, TSA = 2(lb + bh + lh)

TSA = 2 (8 x 4 + 4 x 4 + 8 x 4)

= 2(32 + 16 + 32)

= 2 (80)

TSA = 160 cm2}

^{Volume of Combined Solids}

^{The volume of complex objects can be simplified by visualising it as a combination of shapes of known solids.}

^{Example:
A solid is in the shape of a cone standing on a hemisphere with both
their radii being equal to 3 cm and the height of the cone is equal to 5
cm.
This can be visualised as follows :}

^{Volume of combination of solids
Volume of combined solids

V(solid) = V(Cone) + V(hemisphere)

V(solid) = (1/3)πr2h + (2/3)πr3

V(solid) = (1/3)π(9)(5) + (2/3)π(27)

V(solid) = 33π cm3

Frustum of a cone}

^{If
a right circular cone is sliced by a plane parallel to its base, then
the part with the two circular bases is called a Frustum.
Surface Area of a Frustum

Surface area of frustum of a cone
Frustum with radius r1 and r2 and height h

CSA of frustum =π(r1+r2)l, where l= √[h2+(r2 – r1)2]

TSA of the frustum is the CSA + the areas of the two circular faces = π(r1 + r2)l + π(r12 + r22)
Volume of a Frustum

The volume of a frustum of a cone =(1/3)πh(r12 + r22 + r1r2)

When a solid is converted into another solid of a different shape(by melting or casting), the volume remains constant.

Suppose
a metallic sphere of radius 9 cm is melted and recast into the shape of
a cylinder of radius 6 cm. Since the volume remains the same after a
recast, the volume of the cylinder will be equal to the volume of the
sphere.

The radius of the cylinder is known however the height is not known. Let h be the height of the cylinder.
r1 and r2 be the radius of the sphere and cylinder respectively. Then,
V(sphere) = V(cylinder)
⇒4/3πr13 = πr22h
⇒4/3π(93) = π(62)h (On substituting the values)
⇒h = 27cm}

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CBSE Grade 10 Maths Chapter 13 - Surface Areas and Volumes

Cuboid and its Surface Area

Cube and its Surface Area

Cylinder and its Surface Area

Right Circular Cone and its Surface Area

Volume of a Right Circular Cone

^{In other words, 3 cones make a cylinder of the same height and base.}

Sphere and its Surface Area

Volume of a Sphere

Hemisphere and its Surface Area

We know that the CSA of a sphere = 4πr². A hemisphere is half of a sphere. Thus ,

CSA of a hemisphere of radius r = 2πr²

Total Surface Area = curved surface area + area of the base circle

TSA of a hemisphere of radius r = 3πr²

Volume of Hemisphere

The volume (V) of a hemisphere will be half of that of a sphere.

The volume of a hemisphere of radius r = (2/3)πr³

Quick Summary of all the formulae

^{Combination of Solids}

^{Surface Area of Combined Figures}

Example

Solution

Volume of a combination of solids

Solution

Conversion of Solid from One Shape to Another

Example

Solution

Frustum of a Cone

Example

Solution

^Example:

^{∴ The new surface area, TSA = 2(lb + bh + lh)

TSA = 2 (8 x 4 + 4 x 4 + 8 x 4)

= 2(32 + 16 + 32)

= 2 (80)

TSA = 160 cm2}

^{Volume of Combined Solids}

^{The volume of complex objects can be simplified by visualising it as a combination of shapes of known solids.}

^{Example:
A solid is in the shape of a cone standing on a hemisphere with both
their radii being equal to 3 cm and the height of the cone is equal to 5
cm.
This can be visualised as follows :}

^{Volume of combination of solids
Volume of combined solids

V(solid) = V(Cone) + V(hemisphere)

V(solid) = (1/3)πr2h + (2/3)πr3

V(solid) = (1/3)π(9)(5) + (2/3)π(27)

V(solid) = 33π cm3

Frustum of a cone}

CBSE Grade 10 Maths Chapter 13 - Surface Areas and Volumes

Cuboid and its Surface Area

Cube and its Surface Area

Cylinder and its Surface Area

Right Circular Cone and its Surface Area

Volume of a Right Circular Cone

In other words, 3 cones make a cylinder of the same height and base.

Sphere and its Surface Area

Volume of a Sphere

Hemisphere and its Surface Area

We know that the CSA of a sphere = 4πr2. A hemisphere is half of a sphere. Thus ,

CSA of a hemisphere of radius r = 2πr2

Total Surface Area = curved surface area + area of the base circle

TSA of a hemisphere of radius r = 3πr2

Volume of Hemisphere

The volume (V) of a hemisphere will be half of that of a sphere.

The volume of a hemisphere of radius r = (2/3)πr3

Quick Summary of all the formulae

Combination of Solids

Surface Area of Combined Figures

Example

Solution

Volume of a combination of solids

Solution

Conversion of Solid from One Shape to Another

Example

Solution

Frustum of a Cone

Example

Solution

Example:

∴ The new surface area, TSA = 2(lb + bh + lh)TSA = 2 (8 x 4 + 4 x 4 + 8 x 4)= 2(32 + 16 + 32)= 2 (80)TSA = 160 cm2

Volume of Combined Solids

The volume of complex objects can be simplified by visualising it as a combination of shapes of known solids.

Example: A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 3 cm and the height of the cone is equal to 5 cm.This can be visualised as follows :

Volume of combination of solidsVolume of combined solidsV(solid) = V(Cone) + V(hemisphere)V(solid) = (1/3)πr2h + (2/3)πr3V(solid) = (1/3)π(9)(5) + (2/3)π(27)V(solid) = 33π cm3Frustum of a cone

^{In other words, 3 cones make a cylinder of the same height and base.}

We know that the CSA of a sphere = 4πr². A hemisphere is half of a sphere. Thus ,

CSA of a hemisphere of radius r = 2πr²

TSA of a hemisphere of radius r = 3πr²

The volume of a hemisphere of radius r = (2/3)πr³

^{Combination of Solids}

^{Surface Area of Combined Figures}

^Example:

^{∴ The new surface area, TSA = 2(lb + bh + lh)

TSA = 2 (8 x 4 + 4 x 4 + 8 x 4)

= 2(32 + 16 + 32)

= 2 (80)

TSA = 160 cm2}

^{Volume of Combined Solids}

^{The volume of complex objects can be simplified by visualising it as a combination of shapes of known solids.}

^{Example:
A solid is in the shape of a cone standing on a hemisphere with both
their radii being equal to 3 cm and the height of the cone is equal to 5
cm.
This can be visualised as follows :}

^{Volume of combination of solids
Volume of combined solids

V(solid) = V(Cone) + V(hemisphere)

V(solid) = (1/3)πr2h + (2/3)πr3

V(solid) = (1/3)π(9)(5) + (2/3)π(27)

V(solid) = 33π cm3

Frustum of a cone}