class 9 maths chapter 13 exercise 13.1

ncert class 9 maths chapter 13 solution, Surface Areas and Volumes Exercise 13.1 involve complete  answers for each question in the exercise 13.1. The solutions provide students a strategic methods  to prepare for their exam. Class 9 Maths Chapter 13 Surface Areas and Volumes exercise 13.1 questions and answers helps students to perform better in exam and it will  clear doubts definitely. Students will find it extremely easy to understand the questions and learn solving the problems. NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes prepared by www.mathematicsandinformationtechnology.com team in very delicate, easy and creative way. 

Question 1:

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 sheet for it, if a sheet measuring 1 m² costs ₹ 20.

Solution 1:

It is given that, length (l) of box = 1.5 m

Breadth (b) of box = 1.25 m

Depth (h) of box = 0.65 m

(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] m²

= (1.95 + 1.625 + 1.875) m² = 5.45 m²

(ii)Cost of sheet per m² area = ₹ 20

Cost of sheet of 5.45 m² area = ₹ (5.45 × 20) = ₹ 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 ₹ 7.50 per m².

Solution 2:

It is given that

Length (l) of room = 5 m

Breadth (b) of room = 4 m

Height (h) of room = 3 m

It can be observed that four walls and the ceiling of the room are to be whitewashed. The floor of the room is not to be white-washed.

Area to be white-washed = Area of walls + Area of ceiling of room

= 2lh + 2bh + lb

= [2 × 5 × 3 + 2 × 4 × 3 + 5 × 4] m²

= (30 + 24 + 20) m²

= 74 m²

Cost of white-washing per m² area = ₹ 7.50

Cost of white-washing 74 m² area = ₹ (74 × 7.50) = ₹ 555

Question 3:

The floor of a rectangular hall has a perimeter 250 m. If the cost of panting the four walls at the rate of ₹10 per m² is ₹ 15000, find the height of the hall.

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

Solution 3:

Let length, breadth, and height of the rectangular hall be l m, b m, and h m 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 m²

Cost of painting per m² area = ₹ 10

Cost of painting 250h m² area = ₹ (250h × 10) = ₹ 2500h

However, it is given that the cost of paining the walls is ₹ 15000.

∴ 15000 = 2500h

h = 6

Therefore, 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 m². How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

Solution 4:

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

= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)] cm²

= 2(225 + 75 + 168.75) cm²

= (2 × 468.75) cm²

= 937.5 cm²

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

Area of n bricks = (n ×937.5) cm² = 937.5n cm²

Area that can be painted by the paint of the container = 9.375 m² = 93750 m²

∴ 93750 = 937.5n

n = 100

Therefore, 100 bricks can be painted out by the paint of the 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 5:

(i)Edge of cube = 10 cm

Length (l) of box = 12.5 cm

Breadth (b) of box = 10 cm

Height (h) of box = 8 cm

Lateral surface area of cubical box = 4(edge)²

= 4(10 cm)²

= 400 cm²

Lateral surface area of cuboidal box = 2[lh + bh]

= [2(12.5 × 8 + 10 × 8)] cm²

= (2 × 180) cm²

= 360 cm²

Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.

Lateral surface area of cubical box − Lateral surface area of cuboidal box = 400 cm²− 360 cm²= 40 cm².

Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm².

(ii)Total surface area of cubical box = 6(edge)² = 6(10 cm)² = 600 cm²

Total surface area of cuboidal box

= 2[lh + bh + lb]

= [2(12.5 × 8 + 10 × 8 + 12.5 × 100] cm²

= 610 cm²

Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.

Total surface area of cuboidal box − Total surface area of cubical box = 610 cm²−600 cm²= 10 cm²

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm².

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 6:

(i)Length (l) of green house = 30 cm

Breadth (b) of green house = 25 cm

Height (h) of green house = 25 cm

Total surface area of green house

= 2[lb + lh + bh]

= [2(30 × 25 + 30 × 25 + 25 × 25)] cm²

= [2(750 + 750 + 625)] cm²

= (2 × 2125) cm²

= 4250 cm²

Therefore, the area of glass is 4250 cm².

(ii)Total length of tape = 4(l + b + h)

= [4(30 + 25 + 25)] cm

= 320 cm

Therefore, 320 cm tape is required for all the 12 edges.

Question 7:

Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 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 cm², find the cost of cardboard required for supplying 250 boxes of each kind.

Solution 7:

Length (l1) of bigger box = 25 cm

Breadth (b1) of bigger box = 20 cm

Height (h1) of bigger box = 5 cm

Total surface area of bigger box = 2(lb + lh + bh)

= [2(25 × 20 + 25 × 5 + 20 × 5)] cm²

= [2(500 + 125 + 100)] cm²

= 1450 cm²

Extra area required for overlapping = (1450× 5/100) cm²

= 72.5 cm²

While considering all overlaps, total surface area of 1 bigger box

= (1450 + 72.5) cm²=1522.5 cm²

Area of cardboard sheet required for 250 such bigger boxes

= (1522.5 × 250) cm² = 380625 cm²

Similarly, total surface area of smaller box = [2(15 ×12 + 15 × 5 + 12 × 5] cm²

= [2(180 + 75 + 60)] cm²

= (2 × 315) cm²

= 630 cm²

Therefore, extra area required for overlapping =(630×5/100 ) cm²= 31.5 cm²   

Total surface area of 1 smaller box while considering all overlaps

= (630 + 31.5) cm2= 661.5 cm²

Areaof cardboard sheet required for 250 smaller boxes = (250 × 661.5) cm²

= 165375 cm²

Total card board sheet required = (380625 + 165375) cm²= 546000 cm²

Cost of 1000 cm2 cardboard sheet= ₹ 4

Cost of 546000 cm2 cardboard sheet = ₹ 4 × 546000/1000= ₹ 2184

Therefore,the cost of cardboard sheet required for 250 such boxes of each kind will be ₹ 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 × 3 m?

Solution 8:

Length (l) of shelter = 4 m

Breadth (b) of shelter = 3 m

Height (h) of shelter = 2.5 m

Tarpaulin will be required for the top and four wall sides of the shelter.

Area of Tarpaulin required = 2(lh + bh) + lb

= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m²

= [2(10 + 7.5) + 12] m²

= 47 m²

Therefore, 47 m² tarpaulin will be required.