# How To Solve Perimeter, Area, and Volume Questions Quickly

## How to Solve Perimeter, Area, and Volume questions:-

Geometry is concerned in calculating the length, perimeter, area and volume of various geometric figures and shapes

## Type 1: Find the area, perimeter, length, breadth and some other sides of the shapes

### Question 1:

The sides of a triangle are in the ratio 1/2:1/3: 1/4. Find the smallest side of the triangle, if the perimeter of the triangle is 78 cm.

Options:

A. 20 cm
B. 24 cm
C. 36 cm
D. 18 cm

#### Solution:

According to the question, ratio of the side of the triangle are 1/2:1/3: 1/4 = 6: 3 : 4
Perimeter of the triangle = 78 cm
Now, side will be 78 * 6/13 = 18 cm
78 * 3/13 = 18
78 * 4/13 = 24
Therefore, the smallest side of the triangle is 18 cm.

### Question 2

A rope makes 120 rounds of cylinder with base radius 10 cm. How many times it can go round a cylinder with base radius 20 cm?

Options:

A. 70
B. 60 cm
C. 45 cm
D. 50 cm

#### Solution:

Le the round be a
If radius is more, then rounds will be less as the length of the ropes remains the same = x

x = 2*π*10*120…(1)

Similarly,

x = 2*π*20*a…(2)

From (1) and (2)

10*120 = 20*a

=> a = 60

### Question 3

Find the area of a parallelogram with base 16 cm and height 7 cm.

Options:

A. 112 cm²
B. 128 cm²
C. 102 cm²
D. 212 cm²

#### Solution:

Area of parallelogram = b* h
Area of parallelogram = 16 * 7 = 112 cm²
style=”font-family: Roboto, sans-serif; color: #3d3d3d;”Type 2: How To Solve Perimeter, Area and Volume Questions Quickly.

## Type 2: How To Solve Perimeter, Area and Volume Questions Quickly.

### Question 1:

A cube of 5 cm was cut into as many 1 cm cubes as possible. Find out the ratio of the surface area of the larger cube to that of the surface areas of the smaller cubes?

Options:

A. 1:2
B. 2:3
C. 1:5
D. 1:3

#### Solution:

Volume of the original cube = 53 = 125 cm3.
Volume of each smaller cubes = 1 cm3. It means there are 125 smaller cubes.
Surface area of the cube = 6a2
Surface area of the larger cube = 6a2 = 6 * 52 = 6 * 25 = 150
Surface area of one smaller cubes = 6 (1²) = 6
Now, surface area of all 125 cubes = 125 * 6 = 750
Therefore,
Required ratio = Surface area of the larger cube: Surface area of smaller cubes
= 150: 750
= 1: 5

### Question 2:

The curved surface area of a cylindrical pillar is 264 m² and its volume is 924m³. Find the ratio of its diameter to its height.

Options:

A. 7:4
B. 3:4
C. 6:5
D. 7:3

#### Solution:

Volume of cylinder = πr2h
Curved Surface area of cylinder = 2 πrh
Volume of cylinder/ Curved Surface area of cylinder = πr2h/2πrh = 924/264
r = 924/264 * 2
r = 7
Curved Surface area of cylinder = 2πrh = 264
2 * 22/ 7 * 7 * h = 264
h = 264 * 7/22* 1/2 * 1/7
h = 6
Now, required ratio = 2r/h = 2 *7/6 = 14/6 = 7/3 = 7:3

### Question 3:

The volumes of two cones are in the ratio 1:10. The radius of the cones are in the ratio of 1: 2. What is the ratio of the height of the cone?

Options:

A. 3:4
B. 3:5
C. 2:5
D. 1:3

#### Solution:

Volume of cone = 1/3 πr²h
V1/V2 = 1/10
Now the ratio of the radius of both the cone = 1: 2
V1/V2 = (1/3 πr1²h1 )/(1/3 πr2²h2) = (1)²h1 / (2)²h2
On solving we get h1/h2= 2/5
Therefore, the ratio of the height of the cone = 2:5

## Type 3: Solve Quickly Perimeter, Area and Volume Questions.Percentage increase or decrease

### Question 1:

A rectangular piece of cloth when soaked in water, was found to have lost 20% of its length and 10% of its breadth. Calculate the total percentage of decrease in the area of rectangular piece of cloth?

Options:

A. 75% decrease
B. 28 % increase
C. 28 % decrease
D. 20% decrease

#### Solution:

Let the original length = l
Let the original breadth = b
Original Area = l * b
New length = 80/100l
Decrease in the area = lb – 80/100l * 90/100b
Decrease in the area = 7/25lb
Decrease percentage = (7/25lb * 1/lb) * 100
Decrease percentage = 700/25 = 28%

### Question 2

The length of a rectangle is decreased to half, while its breadth is increased 3 times. Calculate the percentage change in area of the rectangle?

Options:

A. 25%
B. 50 %
C. 72 %
D. 20%

#### Solution:

Let the original length = l
Let the original breadth = b
Original Area = l * b
New length = l/2
New area = l/2 * 3b = 3/2lb
Increase in percentage = New area – original area/original area * 100 = 3/2lb – lb/lb * 100
Increase in percentage = 100/2% = 50%

### Question 3

If the length of a rectangle is increased by 25% and the width is decreased by 20%, then find the area of the rectangle?

Options:

A. 25% increase
B. 50 % decrease
C. remains unchanged
D. 10 % increase

#### Solution:

Let the original length = l
Let the original breadth = b
Original Area = l * b
New length = 125/100l
New area = 125/100l * 80/100b
New area = 5/4l * 4/5b = 20/20lb = lb
Therefore, original and new area are same. It means the area remains unchanged.

## Type 4: How To Solve Perimeter, Area and Volume Questions Quickly.Find cost

### Question 1:

Calculate the cost of making a garden at one meter boundary around a rectangular plot at the rate of Rs. 20/ sq m? The perimeter of the plot is 340 meters.

Options:

A. Rs. 6810
B. Rs. 6880
C. Rs. 6800
D. Rs. 6600

#### Solution:

Perimeter of the rectangle = 2 (l+b)
So, 340 = 2 (l + b)
Now we have to make garden in one meter boundary
Therefore, we will add 4 to the perimeter
340 + 4 =344
Therefore, required cost = 20 * 344 = 6880

### Question 2

Ajit has a plot of area equal to 361 sq ft. He thought to build a fencing around the four sides of the plot. The cost per foot of building the fence is Rs. 50. Calculate the total cost of building a fencing around the plot?

Options:

A. Rs. 3710
B. Rs. 3890
C. Rs. 3800
D. Rs. 3580

#### Solution:

Area of square = a2 = 361
a = 19
Length of the fence = Perimeter of the plot = 4a = 4 * 19 = 76
Therefore, cost of building the fence = 76 * 50 = Rs. 3800.

### Question 3:

A rectangular wall whose length is 10 m more than its breadth. The cost of painting the wall is at Rs 30 per meter is Rs. 2100, what is the length of the wall in meters?

Options:

A. 22.5m
B. 17.5m
C. 30 m
D. 20 m

#### Solution:

Breadth of the wall = x
Length of the wall = x + 10
Perimeter of the rectangular wall = 2100/30 = 70m
2 (l+b) = 70
2 (x + 10 + x) = 70
2 (2x + 10) = 70
2x + 10 = 35
2x = 25
x = 12.5
Length = x+ 10 = 12.5 + 10 = 22.5m 