NCERT Solutions For Class 9 Maths Chapter 13

NCERT Solutions Class 9 Maths Surface Areas and Volumes

Ncert Solutions For Class 9 Maths Chapter 13 PDF Download

NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes are given here in a detailed way. The class 9 NCERT maths solutions for surface areas and volumes given here are given in steps and are extremely easy to understand. Also, the NCERT Solutions Class 9 Maths Surface Areas and Volumes PDF is available here which the students can download and study offline.

In the NCERT maths book of class 9, several exercise problems are included in each chapter. Many students often face difficulties in certain questions and eventually pile them up. At this time, the NCERT solutions come in handy and help the students to clear all their doubts instantly. Similarly, these NCERT maths solutions for class 9 chapter 13 (surface areas and volumes) are available below for free and any students can refer to these to clear their respective doubts. The NCERT Solutions For Class 9 Maths Chapter 13 PDF can also be downloaded for free and students can refer to the solutions at any time they want.

NCERT Solutions For Class 9 Maths Chapter 13 Exercises

 

QUESTIONS-:

1. A matchbox measures 5cm × 1cm × 3.5cm. Determine the volume of a packet containing 14 such boxes.

Soln.

Given the dimension of the matchbox = 5cm × 1cm × 3.5cm

Let us assume, l = 5cm, b = 1cm, h = 3.5cm

As we know that, Volume of one matchbox = (l × b × h)

= \(\left ( 5\times 1\times 3.5 \right )cm^{3}=17.5cm^{3}\)

\(∴ The\:volume\:of\:a\:packet\:containing\:14\:such\:boxes=\left ( 17.5\times 14 \right )cm^{3}=245cm^{3}\)

 

2. A cuboidal water tank is 10 m long, 2 m wide and 7.5 m deep. How many litres of water can it hold? (\(1m^{3}\) = 1000 l)

Soln.

Dimensions of water tank = 10m × 2m × 7.5m

Let us assume, l = 10m, b = 2m, h = 7.5m

Therefore Volume of the tank = \(\left ( l\times b\times h \right )m^{3}\)

= \(\left ( 10\times 2\times 7.5 \right )m^{3}=150m^{3}\)

Hence, the tank can hold = 150 x 1000 litres = 150000 litres of water.

 

3. A cuboidal vessel is 25 m long and 12 m wide. Determine the height that must be made to hold 400 cubic metres of a liquid.

Soln.

Given, Length = 25 m , Breadth = 12 m and Volume = 400\(m^{3}\)

As we know, Volume of cuboid = Length x Breadth x Height

Therefore, Height = Volume of cuboid/(Length × Breadth)
= \(\frac{400}{25\times 12}m=1.33m\)

 

4.  Find the value of digging a cuboidal pit 15 m long, 5 m broad and 3 m deep at the rate of Rs.50 per \(m^{3}\).

Soln.

Here, length = 15m, breadth = 5m and height = 3m

As we know that, Volume of the pit = \(\left ( l\times b\times h \right )m^{3}\)

= \(\left (15\times 5\times 3 \right )m^{3}=225m^{3}\)

The rate of digging is = Rs.50 per \(m^{3}\)

∴ The total value of digging the pit = Rs.(225 x 50)

= Rs.11250

 

 

5. The capacity of a cuboidal tank is 2,10000 litres of water. Calculate the breadth, given that the length is 3.5m and depth is 20m.

Soln.

Given, length = 3.5m, depth = 15m and volume = 30000 litres

As we know that, \(1m^{3}=1000\:litres\)

\(∴ 210000\:litres=\frac{210000}{1000}m^{3}\)= 210\(1m^{3}\)

Breadth = \(\frac{volume\:of\:cuboid}{length\times depth}\)

= \(\frac{210}{\left ( 3.5\times 20 \right )}m\)

= 3m

 

6. A village, with a population of 6000, requires 200 litres of water per head per day. It has a tank measuring 30m × 25m × 8m. Justify the number of days that will take to empty the water tank.

Soln.

Given, the dimension of the tank = 30m × 25m × 8m

So, l = 30m, b = 25m and h = 8m

As we know that, the total capacity of the tank = \(\left ( 30\times25\times 8 \right )m^{3}=6000m^{3}\)

Water required for a single person per day = 200 litres

The requirement of water for 6000 person in a single day = (6000 x 200) litres

= \(\frac{\left ( 6000\times 200 \right )}{1000}=1200m^{3}\)

Hence, the number of days the water will last = (the capacity of the tank /water required per day) = \(\left ( \frac{6000}{1200} \right )=5\)

∴ The water lasts for 5 days.

 

7. A warehouse measures 50m × 35m × 25m. Calculate the maximum number of wooden boxes each measuring 2.5m × 1.5m × 1m that can be stored in the warehouse.

Soln.

Given the dimensions of the warehouse = 50m x 35m x 25m

As we know that, the volume of the warehouse will be = \(\left ( lbh \right ) m^{3}\)

= \(\left (50\times 35\times 25 \right ) m^{3}=43750m^{3}\)

Now, the dimension of box = 2.5m x 1.5m x 1m

Similarly, volume of 1 box = \(\left (2.5\times 1.5\times 1 \right ) m^{3}=3.75m^{3}\)

Hence, Number of box that can be stored =  volume of warehouse / volume of 1 box = \(\frac{43750}{3.75}=11666.666=11666\)

 

8. A solid cuboid having side 20 cm is cut into 16 cubes of equal volume. Calculate the side of the new cuboid and also calculate the ratio between their surface areas.

Soln.

Here the edge of the cube = 20cm

So, Volume of the cuboid = \(\left ( edge \right )^{3}cm^{3}\)

= \(\left ( 20\times20\times 20 \right )cm^{3}=8000cm^{3}\)

Now, The number of smaller cube = 16

So, the volume of 1 small cube = \(\frac{8000}{16}cm^{3}=500cm^{3}\)

Let us assume the side of small cube as ‘p’

\(p^{3}=500\:\:\Rightarrow p=7.937\:\left ( approx \right )\)

Hence, the surface area the cube = \(7.937\left ( side \right )^{2}\)

Therefore, the ratio of their surface area

= (7.937 x 20 x 20)/(7.937 x 7.937 x 7.937)

= \(\frac{40}{1.585}\) = 40 :1.585

 

9. A river 5 m deep and 60 m wide is flowing at a rate of 6 km per hour. Estimate the amount of water that will fall into the sea in a minute.

Soln.

Given, Depth (h) = 5m

Width (b) = 60m

So, the rate of flow of water (l) = 6km per hour= \(\left ( \frac{6000}{60} \right )m\:per\:minute=100m\:per\:minute\)

Therefore, the volume of water flowing into the sea in a minute = \(lbh\:m^{3}\)

\(\left ( 100\times 60\times 5 \right )m^{3}=30000m^{3}\)

 

Surface areas and volumes chapter of class 9 introduces the concepts of solid shapes and finding the parameters. These concepts are extremely crucial as the formulas are extensively used in several other maths and science concepts in the later grades.

At first, this chapter introduces the concept of surface area of a cuboid and a cube and then to the surface area of a right circular cylinder, cone, sphere. The detailed lateral surface area (or curved surface area) and total surface area of all these figures are explained for all the solids. The volume of cuboid, cylinder, right circular cone, and sphere are explained in detail with diagrams.

In this chapter, several examples are given to help the students get completely acquainted with the terminologies and the methods of solving different questions. Students are suggested to solve all the questions from the exercises to be able to develop a deeper understanding of the subject. Students can always refer to these NCERT Solutions Class 9 Maths Surface Areas and Volumes (chapter 13) to clear all their respective doubts.

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