TS 10th Class Maths Notes Chapter 10 Mensuration

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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 = 4a²
• Total surface area = 6a²
• Volume = a³
• 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 = πr²h

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 = \(\frac{1}{3}\)πr²h
• l² = h² + r²

Sphere :

If ‘r’ is radius of sphere, then

• Lateral surface area = 4πr²
• Total surface area = 4πr²
• Volume = \(\frac{4}{3}\) πr³

Hemisphere :

If Y is the radius of hemi-sphere, then

• Lateral surface area = 2πr²
• Total surface area = 3πr²
• Volume = \(\frac{4}{3}\)πr³

Right Circular Hollow Cylinder:

• Area of each end = π(R² – r²)
• Curved surface area of hollow Cylinder = External area + Internal area
  = 2πrh + 2πRh = 2πh(R + r)
• Total surface area = 2πRh + 2πrh + 2(πR² – πr²)
  = 2πh(R + r) + 2π(R + r)(R – r)
  = 2π(R + r) (R + h – r)
• Volume of the material = External volume – Internal volume
  = πR²h – πr²h = 7th(R² – r²)

Spherical Shell:

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

• Outer surface area = 4πR²
• 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.