We are offering TS 10th Class Maths Notes Chapter 10 Mensuration to learn maths more effectively.

## TS 10th Class Maths Notes Chapter 10 Mensuration

**Cuboid :**

l, b and h denote respectively the length, breadth and height of a cuboid then

- Lateral surface area = 2h(l + b)
- Total surface area = 2(lb + bh + hl)
- Volume = lbh
- Diagonal of the cuboid = \(\sqrt{l^2+b^2+h^2}\)

**Cube :**

If the length of each edge of cube is “a” units then

- Lateral surface area = 4a
^{2} - Total surface area = 6a
^{2} - Volume = a
^{3} - Diagonal of the cube = √3 × a = a√3

**Right Prism :**

- Lateral surface area = Perimeter of base × height
- Total Surface area = Lateral Surface area + 2(Area of end Surface)
- Volume = Area of base × height

**Right Circular Cylinder:**

If r is the radius of the base and ‘h’ is the height, then

- Lateral surface area = 2πrh
- Total surface area = 2πr(h + r)
- Volume = πr
^{2}h

**Right Pyramid :**

- Lateral surface area = \(\frac{1}{2}\) × perimeter of base × slant height
- Total Surface area = Lateral surface area + area of base
- Volume = \(\frac{1}{3}\) × area of base × height

**Right Circular Cone :**

If ‘r’ is the radius of the base, ‘h’ is the height and is slant height, then

- Lateral surface area = πrl.
- Total Surface area = πr(l + r)
- Volume = \(\frac{1}{3}\)πr
^{2}h - l
^{2}= h^{2}+ r^{2}

**Sphere :**

If ‘r’ is radius of sphere, then

- Lateral surface area = 4πr
^{2} - Total surface area = 4πr
^{2} - Volume = \(\frac{4}{3}\) πr
^{3}

**Hemisphere :**

If Y is the radius of hemi-sphere, then

- Lateral surface area = 2πr
^{2} - Total surface area = 3πr
^{2} - Volume = \(\frac{4}{3}\)πr
^{3}

**Right Circular Hollow Cylinder:**

- Area of each end = π(R
^{2}– r^{2}) - Curved surface area of hollow Cylinder = External area + Internal area

= 2πrh + 2πRh = 2πh(R + r) - Total surface area = 2πRh + 2πrh + 2(πR
^{2}– πr^{2})

= 2πh(R + r) + 2π(R + r)(R – r)

= 2π(R + r) (R + h – r) - Volume of the material = External volume – Internal volume

= πR^{2}h – πr^{2}h = 7th(R^{2}– r^{2})

**Spherical Shell:**

If R and r are the outer and inner radii of a spherical shell, then

- Outer surface area = 4πR
^{2} - Volume of material = \(\frac{4}{3}\) π

→ The volume of the solid-formed by joining two basic solids is the sum of the volumes of the constituents.

→ In calculating the S.A of a combination of solids, we cannot add the surface area of the two constituents, because some part of the surface area disappears in the process of joining them.

Flow Chat:

Brahmagupta (598 – 668):

- Brahmagupta was born in the state of Rajasthan.
- He worked in the great astronomical centre of ancient India – Ujjain.
- He made significant contributions to Trigonometry.