**CBSE Class 10 Maths Surface Area and Volumes Notes:-**Download PDF Here

The concept of surface area and volume for Class 10 is provided here. In this article, we are going to discuss the surface area and volume for different solid shapes such as the cube, cuboid, cone, cylinder, and so on. The surface area can be generally classified into Lateral Surface Area (LSA), Total Surface Area (TSA), and Curved Surface Area (CSA). Here, let us discuss the surface area formulas and volume formulas for different three-dimensional shapes in detail. In this chapter, the combination of different solid shapes can be studied. Also, the procedure to find the volume and its surface area in detail.

### 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 lÂ Ã—Â bÂ Ã—Â h, respectively.

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

TSA (cuboid) = 2(lÂ Ã—Â b)Â +Â 2(bÂ Ã—Â h)Â +Â 2(lÂ Ã—Â h)Â =Â 2(lbÂ +Â bhÂ +Â lh)

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

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(bÂ Ã—Â h)Â +Â 2(lÂ Ã—Â h)Â =Â 2h(lÂ +Â b)

Length of diagonal of a cuboid =âˆš(l2Â + b2Â + h2)

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### Cube and its Surface Area

For a cube, length = breadth = height

TSA (cube) =2Â Ã—Â (3l2)Â =Â 6l2

Similarly, the Lateral surface area of cube =Â 2(lÂ Ã—Â lÂ +Â lÂ Ã—Â l)Â =Â 4l2

Note: Diagonal of a cube =âˆš3l

#### For More Information On Cube And Cuboid, Watch The Below Video.

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### Cylinder and its Surface Area

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

CSA of a cylinder of base radius *r*Â and height hÂ =Â 2Ï€Â Ã—Â rÂ Ã—Â h

TSA Â of a cylinder of base radius *r*Â and height hÂ =Â 2Ï€Â Ã—Â rÂ Ã—Â h + area of two circular bases

=2Ï€Â Ã—Â rÂ Ã—Â hÂ +Â 2Ï€r2

=2Ï€r(hÂ +Â r)

#### For More Information On Cylinder, Watch The Below Video.

To know more about Surface Area of a Cylinder, visit here.

### Right Circular Cone and its Surface Area

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

CSA of right circular cone =Â Ï€rl

TSA = CSA + area of base =Â Ï€rlÂ +Â Ï€r2Â =Â Ï€r(lÂ +Â r)

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### Sphere and its Surface Area

For a sphere of radiusÂ *r*

Curved Surface Area (CSA) = Total Surface Area (TSA) = 4Ï€r2

#### For More Information On Sphere And Hemisphere, Watch The Below Video.

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### Volume of a Cuboid

Volume of a cuboid =Â (baseÂ area)Â Ã—Â heightÂ =Â (lb)hÂ =Â lbh

### Volume of a Cube

Volume of a cube = baseÂ areaÂ Ã—Â height

Since all dimensions of a cube are identical, volume = l3

Where *l* is the length of the edge of the cube.

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### Volume of a Cylinder

Volume of a cylinder = Base area Ã—Â height = (Ï€r2)Â Ã—Â hÂ =Â Ï€r2h

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### Volume of a Right Circular Cone

The volume of a Right circular cone is 1/3Â times that of a cylinder of same height and base.

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

The volume of a Right circular cone =(1/3)Ï€r2h

Where *r* is the radius of the base and *h* is the height of the cone.

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### The volume of a Sphere

The volume of a sphere of radius *r* = (4/3)Ï€r3

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### Hemisphere and its Surface Area

We know that the CSA of a sphereÂ =Â 4Ï€r2.

A hemisphere is half of a sphere.

âˆ´ CSA of a hemisphere of radius *r* =Â 2Ï€r2

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

â‡’TSA =Â 3Ï€r2

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### Volume of Hemisphere

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

âˆ´Â The volume of the hemisphere of radius *r* =Â (2/3)Ï€r3

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## Combination of Solids

#### For More Information On Combination of Solids, Watch The Below Video.

### 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: 2 cubes each of volume 64 cm3Â are joined end to end. Find the surface area of the resulting cuboid.

Length of each cube =Â 64(1/3)Â =Â 4cm

Since these cubes are joined adjacently, Â they form a cuboid whose length *l* = 8 cm. But height and breadth will remain same = 4 cm.

âˆ´ 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 cm^{2}

### 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 :

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

V(solid) = (1/3)Ï€r2h + (2/3)Ï€r3

V(solid) = (1/3)Ï€(9)(5)Â + (2/3)Ï€(27)

V(solid) = 33Ï€ cm3

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## Shape Conversion of Solids

### 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

CSA of frustum =Ï€(r1+r2)l,Â where l=Â âˆš[h2+(r2Â –Â r1)2]

TSA of the frustum is the CSA + the areas of the two circular faces = Ï€(r1Â +Â r2)lÂ +Â Ï€(r_{1}^{2}Â +Â r_{2}^{2})

### Volume of a Frustum

The volume of a frustum of a cone =(1/3)Ï€h(r_{1}^{2}Â +Â r_{2}^{2Â }+Â r1r2)

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### Shape Conversion of Solids

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.Â 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)

â‡’4/3Ï€r1^{3Â }=Â Ï€r2^{2}h

â‡’4/3Ï€(93)Â =Â Ï€(62)hÂ Â Â Â Â Â Â Â Â Â (On substituting the values)

â‡’hÂ =Â 27cm

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