## Mensuration Maths Notes

**Read and understand:**

**List of Important Formulae:**

**Cuboid:**

Cuboid is made up of six surfaces. Each surface is a rectangle.

Examples: Brick, matchbox, fish tank, etc.

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

- lateral (vertical) surface area = 2(1 + b)h
- total surface area = 2 (lb + bh + hl)
- volume = lbh.

**Cube:**

Cube is made up of six surfaces. Each surface is a square. All squares are equal.

Examples : Ice cubes, sugar crystals, die, etc.

If l is the length of each edge of a cube, then

- lateral (vertical) surface area = 4l
^{2} - total surface area = 6l
^{2} - volume = l
^{3}

**Cylinder:**

Cylinder is made of three surfaces. Base surfaces are circular in shape. It also has a curved surface.

Examples : Solid pipe, ball pens refills, road roller, coins, etc.

If r is the base radius and h is the height of a right circular cylinder, then

- area of the base surface = πr
^{2} - curved šurface area = 2πrh
- total surface area = 2πr(r + h)
- volume = πr
^{2}h.

**Sphere:**

Sphere has only one surface which is entirely curved.

Examples: Football, marbles, soap bubbles, lead shots, etc.

If r is the radius of a sphere, then

- surface area = 4πr
^{2} - volume = \(\frac{4}{3}\)πr
^{3}

**Hemisphere:**

Hemisphere has two surfaces. The base surface is circular in shape. It also has a curved surface.

Examples: Bowl, broken coconut, etc.

If r is the radius of a hemisphere, then

- area of the base = πr
^{2} - curved surface area = 2πr
^{2} - total surface area = 3πr
^{2} - volume = \(\frac{2}{3}\)πr
^{3}.

**Right Circular Cone:**

A cone has two surfaces. Base surface is circular in shape. It also has curved surface.

Examples: Conical tent, ice cream cone, tapered end of pencil, etc.

For a right circular cone of height h, slant height l and base radius r, we have,

- area of the base = πr
^{2} - l
^{2}= r^{2}+ h^{2}. - curved surface area = πrl
- total surface area = πr(r + l)
- volume = \(\frac{1}{3}\) πr
^{2}h.

**Frustum of the Cone:**

For a given cone, if we slice or cut a plane parallel to its base and remove the smaller cone, the left over part on other side of plane is called frustum of the cone. It has 3 surfaces. Two faces are circular and one is curved.

Examples : Bucket, water glass, etc.

If h is the height, r_{1}, r_{2} are the radii of the bases (r_{1} > r_{2}) and the l is the slant height of the frustum of

the cone, then

- l = \(\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}}\).
- curved surface area = πr(r
_{1}+ r_{2}) - total surface area = πl(r
_{1}+ r_{2}) + πr_{1}^{2}+ πr_{2}^{2} - volume = \(\frac{1}{3}\)πh (r
_{1}^{2}+ r_{2}^{2}+ r_{1}× r_{2})

[Note: If we consider r_{2} > r_{1} then l = \(\sqrt{h^{2}+\left(r_{2}-r_{1}\right)^{2}}\)]

**Read and Understand:**

**Sector of a Circle:**

In the figure, the central angle divides the circular region in two parts. The sector of a circle is the region bounded by two radii and the arc of the circle. In the figure. arc ACB and arc ADB are the minor and the major arcs respectively.

The region bounded by the two radii and the minor arc is called minor sector and the region bounded by the two radii and the major arc is the major sector.

In the figure, the region O-ACB is minor sector and the region O-ADB is the major sector.

Let m(arc ACB) = ∠AOB = θ

Area of the sector = \(\frac{\theta}{360}\) × πr^{2}.

Circumference of the circle = 2πr or πd

Arc being the part of the circle, its length can be calculated.

Length ofan arc = \(\frac{\theta}{360}\) × 2πr

The relation between the length of the arc and the area of the sector is shown by the following formula :

**Read and Understand:**

**Segment of a circle:**

The region bounded by a chord and its corresponding arc is called the segment of the circle.

If the corresponding arc is minor, then the segment is minor segment and if the corresponding arc is major, then the segment is major segment.

In the figure, the shaded region PRQ is the minor segment and region PMQ is the major segment of the circle.

In the figure, if we draw radii OP and OQ. we also obtain sector O-PRQ.

Thus,

Area of segment PRQ = Area of sector O-PRQ-Area of triangle OPQ.

[Also the following formula can be used to find the area of the segment:]

Area of the segment = r^{2}\(\left[\frac{\pi \theta}{360}-\frac{\sin \theta}{2}\right]\) where r is the radius of the circle and θ is the angular measure of the arc contained by the sector.