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Class 10th Science
Class 10th Maths
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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.

Surface Areas and Volumes

Cuboid with length l, breadth b and height h

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.
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 a cuboid =√(l2 + b2 + h2)

To know more about Surface Area of Cuboid, visit here.

Cube and its Surface Area

For a cube, length = breadth = height

Surface Areas and Volumes

Cube with length l

TSA (cube) =2 × (3l2) = 6l2
Similarly, the Lateral surface area of cube = 2(l × l + l × l) = 4l2
Note: Diagonal of a cube =√3l

To know more about Surface Area of Cube, visit here.

Cylinder and its Surface Area

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

Surface Areas and Volumes

Transformation of a Cylinder into a rectangle.

CSA of a cylinder of base radius r and height h = 2π × r × h
TSA  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)

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.

Surface Areas and Volumes

Cone with base radius r and height h

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

To know more about Surface Area of Right Circular Cone, visit here.

Sphere and its Surface Area

For a sphere of radius r

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

Surface Areas and Volumes

Sphere with radius r

To know more about Surface Area of Sphere, visit here.

Volume of a Cuboid

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

Volume of a Cube

Volume of a cube = base area × height
Since all dimensions of a cube are identical, volume = l3
Where l is the length of the edge of the cube.

To know more about Volume of Cube and Cuboid, visit here.

Volume of a Cylinder

Volume of a cylinder = Base area × height = (πr2) × h = πr2h

Surface Areas and Volumes

Cylinder with height h and base radius r 

To know more about Volume of a Cylinder, visit here.

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.

To know more about Volume of a Right Circular Cone, visit here.

The volume of a Sphere

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

To know more about volume of a Sphere, visit here.

Hemisphere and its Surface Area

Surface Areas and Volumes

Hemisphere of radius r

We know that the CSA of a sphere  = 4πr2.

A hemisphere is half of a sphere.
∴ CSA of a hemisphere of radius r = 2πr2
Total Surface Area = curved surface area + area of the base circle
⇒TSA = 3πr2

To know more about Surface Area of Hemisphere, visit here.

Volume of Hemisphere

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

To know more about Volume of Hemisphere, visit here.

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: 2 cubes each of volume 64 cm3  are joined end to end. Find the surface area of the resulting cuboid.
Length of each cube = 64(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.

Surface Areas and Volumes

Combination of 2 equal cubes

∴ 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 :

Surface Areas and Volumes

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

To know more about Volume of a Combination of Solids, visit here.

Shape Conversion of Solids

Frustum of a Cone

Surface Areas and Volumes

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 Areas and Volumes

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 + r2+ r1r2)

To know more about Frustum, visit here.

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. 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)
⇒4/3πr1= πr22h
⇒4/3π(93) = π(62)h                   (On substituting the values)
⇒h = 27cm

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