The document Ex 13.1 NCERT Solutions- Surface Areas and Volumes Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

All you need of Class 9 at this link: Class 9

**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 1 m ^{2 }costs **â‚¹

Solution:

m = 0.65 m

âˆµ It is open from the top.

âˆµ Its surface area = [Lateral surface area] + [Base area]

= [2(l + b)h] + [l Ã— b] = [2(1.50 + 1.25)0.65 m^{2}] + [1.50 x 1.25 m^{2}]

= [2 x 2.75 x 0.65 m^{2}] + [1. 875 m^{2}]

= 3.575 m^{2} + 1.875 m^{2 }= 5.45 m^{2 }

âˆµ The total surface area of the box = 5.45 m^{2}

âˆ´ Area of the sheet required for making the box = 5.45 m^{2}

(ii) âˆµ Rate of sheet = â‚¹ 20 per m^{2}

âˆ´ Cost of 5.45 m^{2 }= â‚¹ 20 x 5.45

â‡’ Cost of the required sheet = â‚¹ 109

**Question 2. The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of **â‚¹** 7.50 per m ^{2}. Solution:** Length of the room (l)= 5m Breadth of the room (b) = 4 m

Height of the room (h) = 3 m

The room is like a cuboid whose four walls (lateral surface) and ceiling are to be white washed.

âˆ´ Area for white washing = [Lateral surface area] + [Area of the ceiling]

= [2(l + b)h] + [l x b] = [2(5 + 4) x 3 m

âˆ´ Cost of white washing for 74 m

= 555

The required cost of white washing is â‚¹555.

**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 **â‚¹**10 per m ^{2} is **â‚¹

find the height of the hall.

Solution:

A rectangular hall means a cubiod.

Let the length and breadth of the hall be â€˜lâ€™ and â€˜bâ€™ respectively.

âˆ´ [Perimeter of the floor] = 2(l + b) â‡’ 2(l + b) = 250 m.

âˆµ Area of four walls = Lateral surface area = [2(l + b)] x h [where â€˜hâ€™ is the height of hall.]

âˆ´ Cost of painting the four walls = â‚¹ 10 x 250 h = â‚¹ 2500 h

â‡’â‚¹2500 h = â‚¹ 15000

â‡’

Thus, the required height of the hall = 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: **Total area that can be painted = 9.375 m

âˆ´ Total surface area of a brick = 2[lb + bh + hl]

= 2[(22.5 x 10) + (10 x 7.5) + (7.5 x 22.5)] cm

= 2[(225) + (75) + (168.75)] cm

= 2[468.75] cm

= 937.5 cm

Let the required number of bricks = n

âˆ´ Total surface area of â€˜nâ€™ bricks = n x

â‡’

â‡’

â‡’ n = 100

Thus, the required number of bricks = 100

**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:** For the cubical box

âˆµ Edge of the cubical box = 10 cm

âˆ´ Lateral surface area = 4a

Total surface area = 6a

âˆµ For the cuboidal box,

l = 12.5 cm, b = 10 cm, h = 8 cm

âˆ´ Lateral surface area = 2[(l + b)] x h = 2[12.5 + 10] x 8 cm

= 2[22.5 x 8] cm

Total surface area = 2[lb + bh + hl]

= 2[(12.5 x 10) + (10 x 8) + (8 x 12.5)] cm

= 2[125 + 80 + 100] cm

= 2[305] cm

Obviously, (i) âˆµ 400 cm^{2} > 360 cm^{2} and 400 â€“ 360 = 40

âˆ´ The cubical box, has greater lateral surface area by 40 m^{2}.

(ii) âˆµ 610 cm^{2} > 600 cm^{2} and 610 â€“ 600 = 10

âˆ´ The cuboidal box has greater total surface area by 10 m^{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:** The herbarium is like a cuboid Here, l = 30 cm, b = 25 cm, h = 25 cm

(i) âˆµ Area of a cuboid = 2[lb + bh + hl]

âˆ´ Surface area of the herbarium (glass) = 2[(30 x 25) + (25 x 25) + (25 x 30)] cm^{2}

= 2[750 + 625 + 750] cm^{2}

= 2[2125] cm^{2 }= 4250 cm^{2}

Thus, the required area of glass is 4250 cm^{2}.

(ii) Total length of 12 edges = 4l + 4b + 4h = 4(l + b + h)

= 4(30 + 25 + 25) cm

= 4 x 80 cm = 320 cm

Thus, length of tape needed = 320 cm

**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 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 **â‚¹ **4 for 1000 cm ^{2}, find the cost of cardboard required for supplying 250 boxes of each kind. Solution:** For bigger box:

Length (l) = 25 cm,

Breadth (b) = 20 cm,

Height (h) = 5 cm

âˆµ The box is like a cuboid and total surface area of a cuboid

= 2(lb + bh + hl)

âˆ´ Area of a box = 2([25 x 20) + (20 x 5) + (5 x 25)] cm

= 2[500 + 100 + 125] cm

= 2[725] cm

â‡’ Total surface area of 250 boxes = 250 x 1450 cm

For smaller box:

l = 15 cm, b = 12 cm, h = 5 cm

Total surface area of a box = 2[lb + bh + hl]

= 2[(15 x 12) + (12 x 5) + (5 x 15)]cm

= 2[180 + 60 + 75] cm

= 2[315] cm

â‡’ Total surface area of 250 boxes = 250 x 630 cm

Now, total surface area of both kinds of boxes = 362500 cm

= 5,20,000 cm

Area for overlaps = 5% of [total surface area]

= (5/100)x 520000 cm

âˆ´ Total area of the cardboard required = [Total area of 250 boxes] + [5% of total surface area]

= 520000 cm

= 546000 cm

Cost of cardboard:

âˆµ Cost of 1000 cm^{2} = â‚¹ 4

âˆ´ Cost of 546000 cm^{2} = (4/546000)/ 1000

= â‚¹ 4*546

= â‚¹ 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:** Here, height (h) = 2.5 m

Base dimension = 4 m x 3 m

â‡’ Length (l) = 4 m and

Breadth (b) = 3 m

âˆµ The structure is like a cuboid.

âˆ´ The surface area of the cuboid, excluding the base. = [Lateral surface area] + [Area of ceiling] = [2 (l + b)h] + [lb]

= [2 (4 + 3) x 2.5] + [4 x 3] m

= [35] + [12] m

Thus, 47 m

**SURFACE AREA OF A RIGHT CIRCULAR CYLINDER**

When axis of a cylinder is perpendicular to the base radius then it is called a right circular cylinder.

If â€˜râ€™ and â€˜hâ€™ be the radius and height of a cylinder respectively, then (i) Base area (= top area) = Ï€r^{2}

(ii) Curved surface area (Lateral surface area) of a cylinder = 2Ï€rh

(iii) Total surface area of a cylinder = 2Ï€rh + 2Ï€r2 = 2Ï€r(h + r)

**Note: **Unless it is mentioned assume Ï€ = (22/7)

190 videos|232 docs|82 tests

### Ex 13.2 NCERT Solutions- Surface Areas and Volumes

- Doc | 6 pages
### Hots Questions- Surface Areas and Volumes

- Doc | 1 pages
### Ex 13.3 NCERT Solutions- Surface Areas and Volumes

- Doc | 3 pages
### Surface Area and Volume of Cube and Cuboid

- Video | 09:41 min
### What is Surface Area and Volume?

- Video | 02:44 min

- Test: Surface Area & Volumes- 1
- Test | 25 ques | 25 min
- NCERT Textbook- Surface Area & Volumes
- Doc | 30 pages