# How To Solve Perimeter Area and Volume Questions Quickly

## How to solve Perimeter Area Volumes

Perimeter can be thought of as the length of the outline of a shape. Surface area is the area of the two-dimensional surface of a three-dimensional object. Volume is the space that an object occupies.

### 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 $\frac{1}{2}$:$\frac{1}{3}$: $\frac{1}{4}$. 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 $\frac{1}{2}$:$\frac{1}{3}$: $\frac{1}{4}$ = 6: 3 : 4

Perimeter of the triangle = 78 cm

Therefore 6x + 4x + 3x = 78

13x = 78

x = 6

The Sides are : 6 x 6 = 36 cm

6 x 4 = 24 cm

6 x 3 = 18 cm

Correct option: D

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 = 2*π*10*120…(1)

Similarly,

x = 2 * π * 20 * a…(2)

From (1) and (2)

10 * 120 = 20 * a

=> a = 60

Correct option: B

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

Correct option: A

### Type 2: How To Solve Perimeter, Area and Volume Questions Quickly.By finding the volume and total surface area

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

Correct option: C

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

$\frac{\text{ Volume of cylinder }}{\text{ Curved Surface area of cylinder }}$

= $\frac{ π r^2 h}{2πrh}$ = $\frac{924}{264}$

r = $\frac{924}{264}\times 2$

r = 7

Curved Surface area of cylinder = 2πrh = 264

$2\times \frac{22}{7}\times 7\times h = 264$

$h = 264\times \frac{7}{22}\times \frac{1}{2}\times \frac{1}{7}$

h = 6

Now, required ratio = 2r/h = 2 *7/6 = 14/6 = 7/3 = 7:3

Correct option: D

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 = $\frac{1}{3}$ πr²h

$\frac{V_{1}}{V_{2}}$= $\frac{1}{10}$

Now the ratio of the radius of both the cone = 1: 2

$\frac{V_{1}}{V_{2}}$=$\frac{(\frac{1}{3} π(r_{1})^2h_{1} )}{(\frac{1}{3} π(r_{2})^2h_{2}}$=

$\frac{(1)^2 h_{1}}{(2)^2 h_{2}}$

On solving we get$\frac{h_{1}}{h_{2}}$ = $\frac{2}{5}$

Therefore, the ratio of the height of the cone = 2:5

Correct option: C

### 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 = $\frac{80}{100}$l
New breadth =$\frac{90}{100}$b

Decrease in the area = lb – $\frac{80}{100}$l * $\frac{90}{100}$b

Decrease in the area = $\frac{7}{25}$lb

Decrease percentage = ($\frac{7}{25}$lb* $\frac{7}{lb}$) * 100
Decrease percentage = $\frac{700}{25}$= 28%

Correct option: C

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 = $\frac{l}{2}$

New area =$\frac{l}{2}$ * 3b = $\frac{3}{2}$lb

Increase in percentage = $\frac{New area – original area}{original area}$ * 100

= $\frac{\frac{3}{2}lb – lb}{lb}$ * 100

Increase in percentage = $\frac{100}{2}$% = 50%

Correct option: B

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 = $\frac{125}{100}$l

New breadth = $\frac{80}{100}$b

New area = $\frac{125}{100}$l * $\frac{80}{100}$b

New area = $\frac{5}{4}$l * $\frac{4}5{}$b = $\frac{20}{20}$lb = lb

Therefore, original and new area are same. It means the area remains unchanged.

Correct option: C

### Type 4: How To Solve Perimeter, Area and Volume Questions Quickly to findfind 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

Correct option: B

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 = a = 361

s = 19

Length of the fence = Perimeter of the plot = 4s = 4 * 19 = 76

Therefore, cost of building the fence = 76 * 50 = Rs. 3800.

Correct option: C

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 =$\frac{2100}{30}$ = 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