# Volumes and Areas

Volumes and Areas form a subsection of the chapter we know as mensuration. In this section, we will recall and introduce formulae that will help us obtain the areas of some known geometrical figures. We will also see how these formulae can be applied to check the aptitude and reasoning.

## Volumes And Areas

### 1. Square

Area = SÂ²

Perimeter = 4s

s = length of the sides, d = length of diagonal.

### 2. Rectangle

Area = base x height = b x h

Perimeter = 2 (b + h)

### 3. Triangle

Area = Â½ x base x height

Perimeter = x + y + z ( summation of three sides of a triangle)

### 4. Rhombus

Area = Â½ x product of the diagonals between the sides * sine of the angle between the sides

Perimeter = 4 x side (any side)

Diagonal = 2 x area / diagonal

### 5. Parallelogram

Area = product of any two sides x sine of the included angle

Perimeter = 2 (a + b) (a and b are the two adjacent sides)

### 6. Trapezium

Area = Â½Â Ã— sum of the parallel sides Ã— height.

## Volume

The magnitude or capacity of a solid space like a cube, cylinder, etc is known as the volume of that solid.

Different solids and their volumes:

### 1. Cube

In a cube, length = breadth = height = s

The surface area of a cube = 6sÂ²

The volume of a cube = SÂ³

Diagonal of a cube = âˆš3 s

### 2. Cuboid

Total surface area of a cuboid: 2 (lb + bh + lh)

The volume of a cuboid: lbh

### 3. Cylinder

Curved surface area of a cylinder: 2 Ï€rh( r = radius of the base, h = height)

Total surface area of a right circular cylinder = 2 Ï€rh + 2 Ï€rÂ²

The volume of the right circular cylinder: Ï€rÂ²h

### 4. Cone

Curved surface area of a cone: pirl (l is the slant height)

Total surface area of a cone: Ï€rl + Ï€rÂ²

The volume of a cone: 1/3 Ï€rÂ²h

### 5. Sphere

The surface area of a sphere: 4 Ï€rÂ²

The volume of a sphere: 4/3 Ï€rÂ³

A half sphere is known as a hemisphere.

### 6. Hemisphere

Curved surface area of a hemisphere: 3 Ï€r2

Total surface area of a hemisphere: 3 Ï€r2

The volume of a hemisphere: 2/3 Ï€r3

## Examples Based on Area

Q. What must be the side of an equilateral triangle so that so that its area may be equal to the area of an isosceles triangle with the base and equal sides as 12 m and 10 m respectively?

A. 9mÂ  Â  Â  Â  Â  Â  B. 10mÂ  Â  Â  Â  Â  Â  Â C.10.5mÂ  Â  Â  Â  Â  Â  Â  Â  D.11.5m

Ans: This question can be easily solved if you know the formula for the area of an isosceles triangle.

Area of an isosceles triangle = b/4(4aÂ² â€“ bÂ²)^Â½

= 12/4(4 x 10^2 â€“ 12^2)^1/2

= 3 x 16 = 48m^2.

Now as per the given problem area of equilateral triangle = 48m^2

Therefore âˆš3/4(a2) = 48 => a = 10.5m.

Hence the required side of the equilateral triangle = 10.5m. Thus C is the correct answer.

Q. The length of the floor if a rectangle hall is 10 m more than its breadth. If 34 carpets of size 6 Ã— 4m are required to cover the floor of the hall, then find the length and breadth of the hall.

A. 24, 24mÂ  Â  Â  Â  Â B. 24, 34mÂ  Â  Â  Â  Â C. 22, 32mÂ  Â  Â  Â  Â  Â  D. 34, 34m

Ans: Let breadth = b then length = b + 10m.

Floor area of the rectangle hall = A = lengthÂ Ã— breadth = bÂ Ã— (b + 10)

Also area of each carpet = 6Â Ã— 4m and 34 pieces are required to cover the floor

Therefore area of hall = 34Â Ã— 6Â Ã— 4

bÂ Ã— (b + 10) = 34Â Ã— 24

Therefore b = 24 m and length = b + 10 = 34m. Thus B is the correct answer.

## Examples Based on Volume

Q. Find the volume and the surface area of the sphere of radius 6.3cm.

Volume of the sphere = 4/3Ï€rÂ³ = 4/3 x 22/7 x 6.3 x 6.3 x 6.3 = 1047.82 cmÂ²

Total surface area of the sphere = 4 Ï€rÂ³ = 4 x 22/7 x 6.3 x 6.3 = 498.96 cmÂ²

## Practice Questions

Q. The cost of carpeting a room 15m long with a carpet 75cm wide at 30paise per meter is Rs. 36. Find the width of the room.

A.6mÂ  Â  Â  Â  Â  Â B. 8mÂ  Â  Â  Â  Â  Â  C. 1omÂ  Â  Â  Â  Â  D. 9m

Q. The area of the right-angled triangle is 30 cmÂ² and the length of the hypotenuse is 13 cm. Find the length of the shorter leg.

A. 3cmÂ  Â  Â  Â B. 4cmÂ  Â  Â  C. 5cmÂ  Â  Â  D. 7 cm

Q. The volume of the sphere is 38808 cmÂ³, find the curved surface area of the sphere.

A. 5202Â  cmÂ³ Â  Â  B. 5544Â  cmÂ³ Â  C. 5562Â  cmÂ³ Â  Â D. 5542Â  cmÂ³

Q. The radii of two cylinders are in the ratio of 2:3 and their heights are in the ratio 5:3. Find the ratio of their curved surface area.

A. 10:9Â  Â  Â  B. 10:11Â  Â  Â  C. 9:10Â  Â  Â  Â  D. 9:11

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