**Chapter 13 Surface Areas and
Volumes**

**Ex 13.1**

**Ex 13.1
Class 9 Maths 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 1m2 costs ₹20.

Solution:

(i) Here, length (l) = 1.5 m, bread th(b) = 1 .25 m

and height (h) = 65 cm =

**∵**** It is open from the top.
**

**∴**

**Its surface area**

= [Lateral surface area] + [Base area]

= [2(1 + b)h] + [lb]

= [2(1.50 + 1.25)0.65] m

= [2 x 2.75 x 0.65] m

= 3.575 m

= [Lateral surface area] + [Base area]

= [2(1 + b)h] + [lb]

= [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}**∴**

**Area of the sheet required for making the box = 5.45 m**

^{2}

^{}^{}

**(ii) Cost of 1 m ^{2} sheet = Rs. 20
Cost of 5.45 m^{2} sheet = Rs. (20 x 5.45)
= Rs. 109
Hence, cost of the required sheet = Rs. 109**

**Ex 13.1
Class 9 Maths 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:

Length of the room (l) = 5 m

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.

**∴**** Area for white washing
= [Lateral surface area] + [Area of the ceiling]
= [2(l + b)h] + [l x b]
= [2(5 + 4) x 3] m**

^{2}+ [5 x 4] m

^{2}= 54 m

^{2}+ 20 m

^{2}= 74 m

^{2}Cost of white washing for 1 m2 area = Rs. 7.50

**∴**

**Cost of white washing for 74 m**

Thus, the required cost of white washing = Rs. 555

^{2}area = Rs. (7.50 x 74) = Rs. 555Thus, the required cost of white washing = Rs. 555

**Ex 13.1
Class 9 Maths 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:

A rectangular hall means a cuboid.

Let the length and breadth of the hall be l and b respectively.

Solution:

A rectangular hall means a cuboid.

Let the length and breadth of the hall be l and b respectively.

**∴**

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

**⇒**

**2(l + b) = 250 m**

**∵**

**Area of four walls = Lateral surface area = 2(1 + b) x h, where h is the height of the hall = 250 h m**

Cost of painting the four walls

= Rs. (10 x 250 h) = Rs. 2500h

^{2}Cost of painting the four walls

= Rs. (10 x 250 h) = Rs. 2500h

**⇒**

**2500 h = 15000**

**⇒**

**h =**
= 6

Thus, the required height of the hall = 6 m

Thus, the required height of the hall = 6 m

**Ex 13.1
Class 9 Maths 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

Here, Length of a brick (l) = 22.5 cm

Breadth of a brick (b) = 10 cm

Height of a brick (h) = 7.5 cm

Since a brick is like a cuboid, then

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

= 2[(225 x 1(0) + (10 x 7.5) + (7.5 x 22.5)] cm

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

= 2[468.75] cm
m

Let the required number of bricks be n

Solution:

Total area that can be painted = 9.375 m

^{2}Here, Length of a brick (l) = 22.5 cm

Breadth of a brick (b) = 10 cm

Height of a brick (h) = 7.5 cm

Since a brick is like a cuboid, then

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

= 2[(225 x 1(0) + (10 x 7.5) + (7.5 x 22.5)] cm

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

^{2}= 2[468.75] cm

^{2}= 937.5 cm^{2}=^{2}Let the required number of bricks be n

**∴**

**Total surface area of n bricks = n x**
m

Thus, the required number of bricks = 100

^{2}Thus, the required number of bricks = 100

**Ex 13.1
Class 9 Maths 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 with edge (a) = 10 cm

Lateral surface area = 4a

= 400 cm

Total surface area = 6a

= 600 cm

For the cuboidal box with dimensions,

Length (l) = 12.5 cm,

Breadth (b) = 10 cm,

Height (h) = 8 cm

Solution:

For the cubical box with edge (a) = 10 cm

Lateral surface area = 4a

^{2}= 4 x 10^{2}cm^{2}= 400 cm

^{2}Total surface area = 6a

^{2}= 6 x 10^{2}cm^{2}= 600 cm

^{2}For the cuboidal box with dimensions,

Length (l) = 12.5 cm,

Breadth (b) = 10 cm,

Height (h) = 8 cm

**∴**

**Lateral surface area = 2[l + b] x h = 2[12.5 + 10] 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

= 610 cm

(i) A cubical box has the greater lateral surface area by (400 – 360) cm

(ii) Total surface area of a cubical box is smaller than the cuboidal box by (610 – 600) cm

^{2}= 360 cm^{2}Total surface area = 2[lb + bh + hl]

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

^{2}= 2[125 + 80 + 100] cm

^{2}= 2[305] cm

^{2}= 610 cm

^{2}(i) A cubical box has the greater lateral surface area by (400 – 360) cm

^{2}= 40 cm^{2}.(ii) Total surface area of a cubical box is smaller than the cuboidal box by (610 – 600) cm

^{2}= 10 cm^{2}.**Ex 13.1
Class 9 Maths 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, length (l) = 30 cm,

breadth (b) = 25 cm,

height (h) = 25 cm

(i) Surface area of the herbarium (glass)

= 2[lb + bh + hl]

= 2[(30 x 25) + (25 x 25) + (25 x 30)] cm

= 2[2125] cm

= 4250 cm

Thus, the required area of the glass = 4250 cm

Solution:

The herbarium is like a cuboid.

Here, length (l) = 30 cm,

breadth (b) = 25 cm,

height (h) = 25 cm

(i) Surface area of the herbarium (glass)

= 2[lb + bh + hl]

= 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 the glass = 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, the required length of tape = 320 cm**

**Ex 13.1
Class 9 Maths 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 cm², 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

Total surface area of a box = 2(lb + bh + hl)

= 2[(25 x 20) + (20 x 5) + (5 x t25)] cm^{2}

= 2 [500 + 100 + 125] cm^{2}

= 2[725] cm^{2}

= 1450 cm^{2}

Total surface area of 250 boxes = (250 x 1450) cm^{2} = 362500 cm^{2}

**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}= 2[180 + 60 + 75] cm

^{2}= 2[315] cm

^{2}= 630 cm

^{2}

**∴**

**Total surface area of 250 boxes = (250 x 630) cm**

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

=
x 520000 cm

^{2}= 157500 cm^{2}Now, total surface area of both type of boxes = 362500 cm

^{2}+157500 cm^{2}= 520000 cm^{2}Area for overlaps = 5% of [total surface area]=

^{2}= 26000 cm^{2}**∴**

**Total surface area of the cardboard required = [Total surface area of 250 boxes of each type] + [Area for overlaps]**

= 520000 cm

= 520000 cm

^{2}+ 26000 cm^{2}= 546000 cm^{2}**∵**

**Cost of 1000 cm**

^{2}cardboard = Rs. 4**∴**

**Cost of 546000 cm**

= Rs.
= Rs. 2184

^{2}cardboard= Rs.

**Ex 13.1
Class 9 Maths 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, length (l) = 4 m,

breadth (b) = 3m

and height (h) = 2.5 m

The structure is like a cuboid.

**∴**** 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] m**

^{2}+ [4 x 3] m

^{2}= 35 m

^{2}+ 12 m

^{2}= 47 m

^{2}Thus, 47 m

^{2}tarpaulin would be required.

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