How To Solve Perimeter Area and Volume Questions Quickly

March 7, 2024

Solving Questions related to  Perimeter Area and Volume

From the information provided on this page we will be learning about How to solve  Perimeter Area and Volume Questions of various geometrical shapes and figures. For your knowledge:
How to solve Perimeter, Area and Shapes

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

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