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# 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 \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

Now, side will be 78 * **\frac{6}{12}** = 18 cm

78 * **\frac{3}{13}** = 18

78 * **\frac{4}{13}** = 24

Therefore, the smallest side of the triangle is 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{Volume of cylinder}{ Curved Surface area of cylinder}

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

r = \frac{924}{264} * 2

r = 7

Curved Surface area of cylinder = 2πrh = 264

2 * \frac{22}{7}* 7 * h = 264

h = 264 * \frac{7}{22}* \frac{1}{2} * \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 **

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