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 Shapes

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 breadth = 3b

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 find
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

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

Breadth = 12.5m

Correct option: A

Read Also –  Formulas for Perimeter,  Area and volume questions