# NCERT Solutions For Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 13 |

Exercise | Class 9 Chapter 13 Surface Areas and Volumes Exercise 13.1 |

Number of Questions Solved | 6 |

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

**NCERT TEXTBOOK EXERCISES**

**Question 1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened 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 sheet for it, if a sheet measuring 1m ^{2} costs ₹20.**

**Solution:**

We have a plastic box of

l = length = 15 m

b = width = 1.25 m

h = depth = 65 cm

= 65/100 m= 0.65 m (∵ 1 m = 100cm)

Surface area of the box = 2 (lb + bh + hl)

= 2(1.5 x 1.25 + 1.25 x 0.65 + 0.65 x 1.5)

= 2(1.875 + 0.8125 + 0.975) = 2 (3.6625)

= 7.325 m^{2}

(i) Area of the sheet required for making the box

= 7.325 – l x b (∵ BOX is opened at the top)

= 7.325 – 1.5 x 1.25

= 7.325 – 1.875 = 5.45 m^{2}

(ii) A sheet measuring 1 m^{2} costs = ₹20

∴ Sheet measuring 5.45 m^{2 }costs = ₹20 x 5.45 = ₹109

**Question 2. The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹17.50 per m ^{2}.**

**Solution:**

We have a room of l = 5m

b = 4m

h = 3m

Required area for white washing

= Area of the four walls + Area of ceiling

= 2(l+ b) x h+ (l x b)

= 2(5+4) x 3 +(5 x 4)

= 2x 9x 3 + 20

= 54+20

= 74 m2

White washing 1 m^{2} costs = ₹7.50

White washing 74 m^{2} costs = ₹7.50x 74 = ₹555

**Question 3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹10 per m ^{2} is ₹15000, find the height of the hall.**

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

**Solution:**

Let the rectangular hall of length = l, breadth = b, height = h

Hence, the height of the hall is 6 m.

**Question 4. 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 x 10 cm x 7.5 cm can be painted out of this container.**

**Solution:**

Given, dimensions of a brick

l = 22.5 cm, b = 10 cm

and b = 7.5cm

Total surface area of bricks = 2 ( l x b + b x h + h x l)

= 2(22.5 x 10 + 10 x 75 + 75 x 225)

= 2(225 + 75 + 168.75)

= 2 x 468.75 cm^{2}

= 9375 cm^{2}

Number of bricks that painted out of this container

**Question 5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.**

**(i) Which box has the greater lateral surface area and by how much?****(ii) Which box has the smaller total surface area and by how much?**

**Solution:**

We have l_{1} for cubical box = 10 cm

For cuboidal box l= 12.5 cm

b = 10 cm

h = 8 cm

(i) Lateral surface area of cubical box = 4l^{2} = 4(10)^{2}

= 4 x 100

= 400 cm^{2}

Lateral surface area of cuboidal box = 4 (l + b) x h

= 2 (125 + 10) x 8

= 2 (225) x 8

= 45 x 8 = 360 cm^{2}

(∵ Lateral surface area of cuboidal box) > (Lateral surface area of cuboidal box) (∵ 400 >360)

∴ Required area = (400 – 360) cm2 = 40 cm^{2}

(ii) Total surface area of cubical box = 6l2 = 6(10)2 = 6x 100= 600 cm^{2}

Total surface area of cuboidal box = 2 ( l x b + b x h + h x l)

= 2(125 x 10 + 10 x 8 + 8 x 125)

= 2(125 + 80+ 100)

= 2 x 305

= 610 cm^{2}

∴ (Area of cuboidal box) > (Area of cubical box) (∵ 610 > 600)

Required area = (610 – 600)cm^{2} = 10 cm^{2}

**Question 6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.**

**(i) What is the area of the glass?****(ii) How much of tape is needed for all the 12 edges?**

**Solution:**

Dimension for herbarium are

l = 30 cm, b = 25 cm and h = 25 cm

Area of the glass = 2 (l x b + b x h + h x l)

= 2 ( 30 x 25 + 25 x 25 + 25 x 30)

= 2(750 + 625 + 750) = 2 (2125) = 4250cm^{2}

∴ Length of the tape = 4 (l + b + h) = 4(30 + 25 + 25)

[∵ Herbarium is a shape of cuboid length = 4 (1+ b + h)] = 4×80= 320cm

**Question 7. Shanti Sweets Stalll was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is ₹4 for 1000 cm ^{2}, find the cost of cardboard required for supplying 250 boxes of each kind.**

**Solution:**

Dimension for bigger box, l = 25 cm, b = 20 cm and b = 5 cm

Total surface area of the bigger size box

=2 ( l x b + b x h + h x l)

= 2(25 x 20 + 20 x 5 + 5 x 25)

= 2(500+ 100+ 125)

= 2(725)= 1450 cm^{2}

Dimension for smaller box, l = 15 cm b = 12 cm and h = 5 cm

Total surface area of the smaller size box = 2(15 x 12 + 12 x 5 + 5 x 15)

= 2 (180 + 60 + 75)= 2 (315)= 630 cm^{2}

Area for all the overlaps = 5% x 2080 cm^{2} = 5/100 – x 2080 cm^{2} = 104 cm^{2}

Total surface area of both boxes and area of overlaps = (2080 + 104) cm^{2} = 2184 cm^{2}

Total surface area for 250 boxes = 2184 x 250 cm^{2}

Cost of the cardboard for 1000 cm2 = ₹4

Costs of the cardboard for 1 cm = ₹ 4/1000

Cost of the cardboard for 2184 x 250 cm^{2 }= ₹ $\frac { 4x 2184 x 250 }{ 1000 }$ = ₹ 2184

**Question 8. Parveen wanted to make a temporary shelter, for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?**

**Solution:**

Dimension for shetter, l = 4 m, b = 3 m and h = 25 cm

Required area of tarpaulin to make the shelter

= (Area of 4 sides + Area of the top) of the car

= 2(l + b) x h+ ( l x b)

= 2(4+ 3) x 25 + (4 x 3)

= (2 x 7 x 25) + 12 = 35 + 12

= 47 m^{2}

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1 PDF

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### Other Chapter of Class 9 Maths Chapter 13 Surface Areas and Volumes

**Surface Areas and Volumes Class 9 Ex 13.1****Surface Areas and Volumes Class 9 Ex 13.2****Surface Areas and Volumes Class 9 Ex 13.3****Surface Areas and Volumes Class 9 Ex 13.4****Surface Areas and Volumes Class 9 Ex 13.5****Surface Areas and Volumes Class 9 Ex 13.6****Surface Areas and Volumes Class 9 Ex 13.7****Surface Areas and Volumes Class 9 Ex 13.8****Surface Areas and Volumes Class 9 Ex 13.9**