## Surface Area And Volume Class 9 PDF Summary

Greetings to all, today we are going to upload the Surface Area And volume Class 9 PDF to assist you all.** NCERT Solutions for Class 9 Maths Chapter 13** Surface Areas and Volumes include the accurately prepared wide range of solved exercise questions for an outstanding understanding. These solutions in Maths for Class 9 are prepared considering the latest CBSE syllabus 2021-22 update of the Term II examination. NCERT for Class 9 Maths Solutions is specially created to be a guiding solution and good reference to assist students to clear doubts spontaneously and effectively. NCERT Solutions for Class 9 Maths is conditioned by the teaching faculties having vast teaching experience along with subject matter experts to serve the purpose.

Class 9 NCERT Solutions are formulated keeping in mind the concept-based approach along with the precise answering method for Term-II examinations. Guide to **NCERT Solutions for Class 9** for a better score. It is an exact and well-structured solution for a good understanding of concept-based knowledge.

### Surface Area And volume Class 9 PDF

**A plastic box 1.5 m long, 1.25 m wide, and 65 cm deep, is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:**

**(i)The area of the sheet required for making the box.**

**(ii)The cost of the sheet for it, if a sheet measuring 1m ^{2} costs Rs. 20.**

**Solution:**

Given: length (l) of box = 1.5m

Breadth (b) of box = 1.25 m

Depth (h) of box = 0.65m

(i) Box is to be open at top

Area of sheet required.

= 2lh+2bh+lb

= [2×1.5×0.65+2×1.25×0.65+1.5×1.25]m

^{2}

= (1.95+1.625+1.875) m

^{2}= 5.45 m

^{2}

(ii) Cost of sheet per m

^{2}area = Rs.20.

Cost of sheet of 5.45 m

^{2}area = Rs (5.45×20)

= Rs.109.

**The length, breadth, and height of a room are 5 m, 4 m, and 3 m respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs 7.50 per m**^{2}.

**Solution:**

Length (l) of room = 5m

Breadth (b) of room = 4m

Height (h) of room = 3m

It can be observed that four walls and the ceiling of the room are to be whitewashed.

Total area to be whitewashed = Area of walls + Area of the ceiling of the room

= 2lh+2bh+lb

= [2×5×3+2×4×3+5×4]

= (30+24+20)

= 74

Area = 74 m^{2}

Also,

Cost of whitewash per m^{2 }area = Rs.7.50 (Given)

Cost of whitewashing 74 m^{2 }area = Rs. (74×7.50)

= Rs. 555

**The floor of a rectangular hall has a perimeter of 250 m. If the cost of painting the four walls at the rate of Rs.10 per m**^{2}is Rs.15000, find the height of the hall.

**[Hint: Area of the four walls = Lateral surface area.]**

**Solution:**

Let length, breadth, and height of the rectangular hall be l, b, and h respectively.

Area of four walls = 2lh+2bh

= 2(l+b)h

Perimeter of the floor of hall = 2(l+b)

= 250 m

Area of four walls = 2(l+b) h = 250h m^{2}

Cost of painting per square meter area = Rs.10

Cost of painting 250h square meter area = Rs (250h×10) = Rs.2500h

However, it is given that the cost of painting the walls is Rs. 15000.

15000 = 2500h

Or h = 6

Therefore, the height of the hall is 6 m.

**The paint in a certain container is sufficient to paint an area equal to 9.375 m**^{2}. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

**Solution:**

Total surface area of one brick = 2(lb +bh+lb)

= [2(22.5×10+10×7.5+22.5×7.5)] cm^{2}

= 2(225+75+168.75) cm^{2}

= (2×468.75) cm^{2}

= 937.5 cm^{2}

Let n bricks can be painted out by the paint of the container

Area of n bricks = (n×937.5) cm^{2 }= 937.5n cm^{2}

As per the given instructions, an area that can be painted by the paint of the container = 9.375 m^{2 }= 93750 cm^{2}

So, we have, 93750 = 937.5n

n = 100

Therefore, 100 bricks can be painted out by the paint of the container.

**You can download the Surface Area And volume Class 9 PDF by clicking on the link given below.**