## Class 10 Mathematics Note

# Mensuration

**Mensuration **

**Area of Triangle**

Area of the Area = $\frac{1}{2}{\rm{\: }}$base*height

If a , b c be the the sides of the traingle , then perimeter of the triangle = a + b+ c

${\rm{\: }}$Semi – perimeter=$\frac{{{\rm{a}} + {\rm{b}} + {\rm{c}}}}{2}$

Area of the triangle = $\sqrt {{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)} $

**Prisms**

A prism is polyhedron formed by two equal parallel regular polygons, end faces connected by side faces which are either rectangles or parallelograms.

**Types of prism **

Lateral surface area of prism(L.S.A)= perimeter of the base* height of prism

Total surfaca area of the prism=L.S.A+ 2A

Volume of prism = l*b*h= Area of the base * height

** Cylinders **

Let `r’ be the radius of the cylinder and`h’ be the height of the cylinder.

Lateral surface area of the cylinder =perimeter of the base* height of prism

= 2πrh

Total surface area of the cylinder=L.S.A+ 2A

= 2πrh + 2π r^{2}

=2πr(r+h)

Volume of the cylinder =Area of the base * height

= πr^{2}h

**Sphere **

A sphere is a perfectly round geometrical object in three-dimensional space that is the surface of a completely round ball.

Surface area of the sphere = 4πr^{2}

Volume of the sphere =$\frac{4}{3}{\rm{\: }}$πr^{3}

**Hemisphere **

Lateral surface area of the hemisphere = 2πr^{2}

Total surface area of the hemisphere = 3πr^{2}

Volume of the hemisphere = $\frac{2}{3}$πr^{2}

**Pyramid**

A pyramid is a three-dimensional shape whose base is a polygon. Each corner of a polygon is attached to a singular apex, which gives the pyramid its distinctive shape. Each base edge and the apex form a triangle.

**Types of pyramid**

Total surface area of the pyramid = Base area + sum of the areas of the all the triangular faces.

= A + $\frac{1}{2}$ P*L

Where P = perimeter of the base and l is slanting height.

Volume of pyramid = $\frac{1}{3}{\rm{\: }}$base area * height of the pyramid

**Cone **

A cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a point called the apex or vertex.

Curved surface area of the cone =πrl

Total surface area of cone =π r (r +l)

Volume of the cone =$\frac{1}{3}$ π r^{2} h

**Examples **

1. Find the volume of the triangular prism.

soln

Area of the base of the triangle = $\frac{1}{2}{\rm{\: }}$base *height${\rm{\: }}$

= $\frac{1}{2}{\rm{\: }}$3*4${\rm{\: }}$= 6cm^{2}

Volume of prism = $\frac{1}{2}{\rm{\: }}$base *height * height of prism

= 6* 10= 60 cm^{2}

2.

Soln

r =$\frac{{42}}{2}$ = 21 cm

Volume of hemisphere (V)=$\frac{2}{3}{\rm{\pi }}{{\rm{r}}^3}$=$\frac{2}{3}{\rm{*}}\frac{{22}}{7}{21^3}$ =19404 cm^{2}

3.

Given solid is made up of cone and the cylinder . The base area of the cylinder is 100 sq. cm and height of the cylinder is 3 cm . If the volume of the whole solid is 600 cubic cm . Find the height of the solid .

Soln

Base area of the cylinder = 100

πr^{2} = 100

r = 5.64 cm

Heightof the cylinder = 3 cm

Volume of the whole solid = 600 sq . cm

Volume of the cylinder + volume of the cone = 600

πr^{2}h + $\frac{1}{3}$π r^{2}h’ = 600 [where h’ is the height of the cone]

100 *3 + $\frac{1}{3}$ *100 *h’ =600

h’= 9 cm

Total height = h+ h’ = 9+3=12cm

4. In the adjoining figure , the solid pyramid having a square base has length of its base30 cmand height 20cm . Find the volume and total surface area of it .

Soln

Volume of the pyramid =${\rm{\: }}\frac{1}{3}$Area of the base * height${\rm{\: }}$

= $\frac{1}{3}$ 30*30 *20 = 6000cm^{3}

Slant height (l) = $\sqrt {{{20}^2} + {{15}^2}} $= 25cm

Surface area of the pyramid = A + $\frac{1}{2}{\rm{\: P*\: L}}$

= 30*30 + $\frac{1}{2}$ *4*30 *25

=2400cm^{2}