[PDF] CBSE NCERT Solutions for Class 9 Mathematics Chapter 13

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Class- XI-CBSE-Mathematics Surface Area And Volumes

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CBSE NCERT Solutions for Class

9 Mathematics Chapter 13 Back of Chapter Questions

E xercise : 13.1

1.A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is

open ed at the top. Ignoring the thickness of the plastic sheet, determine: (i)The area of the sheet required for making the box.

Solution:

Given, the dimension of the box

is:

Length = 1.5 m

Breadth = 1.25m

Height =65cm= 0.65m

(i)Area of the sheet (cuboid) =lb+2lh+2bh(since top is opened) Area =(1.5×1 .25)+ 2(

1.5×0 .65)

+ 2(

1.25× 0.65)

= 1.95+ 1.875+ 1.625 Hence, the area of the sheet required for making the box is 5.45 m (ii)Cost of sheet: 5.45m = 5.45×20

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

Solution:

Gi ven, Length =5m, Breadth =4m, Height =3m Area to be whitewashed = Area of the walls + Area of ceiling of roomsArea of the walls =2lh+2bh Class- XI-CBSE-Mathematics Surface Area And Volumes

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= 2(5 ×3 )+ 2(4 ×3 ) =30+24 =54 m Area of ceiling of rooms =lb = 5× 4 =20 m Area to be whitewashed =54+20 =74 m

Cost of whitewashing 1m

Hence , Cost of whitewashing 74 m =74× 7.5 3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four

Solution:

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

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

=250 m Area of the four walls = 2(l +b )× h =250 hm

15000=250× h× 10

Hence , h = = 6 m

Hence, the height of the hall is 6 m.

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

Solution:

Given, d

imension of brick is 22.5 cm×10 cm× 7.5 cm

Total surface area of the bricks =2lb+2lh+2bh

Class- XI-CBSE-Mathematics Surface Area And Volumes

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= 2(22.5× 10)+ 2(22.5× 7. 5)+ 2(10× 7.5) =450+337.5+ 150 =937.5 cm

Total area that can be painted = 9.375 m

=93750 cm Hence

Number of bricks

that can be painted = =100 Hence, 100 bricks can be painted out of the container. 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:

(i)

Given, the

edge of the cubical box =10 cm

Lateral surface area of cubical box = 4a

= 4 (10) =400 cm Dimension of the cuboidal box is l =12.5 cm,b= 10 cm,h= 8 cm

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

= 2(12.5+ 10) ×8 =360 cm The difference in lateral surface area =400െ360 =40 cm Hence the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm (ii) Total surface area of the cubical box = 6a = 6 (10) =600 cm Class- XI-CBSE-Mathematics Surface Area And Volumes

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Total surface area of the cuboidal box =2lb+2lh+2bh = 2 (12.5× 10)+ 2(12.5× 8)+ 2(10× 8) =250+200+160 =610 cm The difference in Total surface area =610െ600 =10 cm Hence the total surface area of the cubical box is smaller than the total surface area of the cuboidal box by 10 cm 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:

Given, d

imension of the green house: l =30 cm,b= 25 cm,h= 25cm (i) Total surface area of the green house = 2(lb+lh+bh) = 2 (30×25+30×25+25×25) cm = 2 (750+750+625) cm = 2 (2125) cm =4250 cm

Hence, the area of the glass is 4250 cm

(ii) Length of the tape needed = 4(l +b +h ) = 4 (30+25+25) cm =320 cm Hence , 320 cm tape is needed for all the 12 edges 7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxe s were required. The bigger of dime nsions

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 , find the cost of cardboard required for supplying 250 boxes of each kind.

Solution:

Dimensions of the bigger box: l =25 cm,b= 20 cm,h= 5 cm Class- XI-CBSE-Mathematics Surface Area And Volumes

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Total surface area of the bigger box = 2(lb+lh+bh) = 2 (25×20+25× 5+ 20× 5) cm = 2 (500+125+100) cm = 2 (725) cm =1450 cm Dimension of the smaller box: l =15 cm,b= 12 cm,h= 5 cm Total surface area of the bigger box = 2(lb+lh+bh) = 2 (15×12+15× 5+ 12× 5) cm = 2 (180+75+60) cm = 2 (215) cm =630 cm Total surface area of 250 boxes of each type =(250×1450)+(250×630) =362500+157500 =520000 cm Extra area required by both types for overlapping =1450× +630×
=72.5+ 31.5 =104

For 250 boxes =104×250

=26000 cm Hence total cardboard required =520000+26000 =546000 cm

Cost of

1000 cm

cardboard Hence , cost of 546000 cm cardboard sheet required for 250 boxes:

546000

1000

× 4

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 ? Class- XI-CBSE-Mathematics Surface Area And Volumes

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Solution:

Dimension of the box like car cover: l =4 m,b=3 m,h =2 .5 m Tarpaulin is required only for the four sides and top of the shelter Hence , area of tarpaulin required = 2(l +b )× h+ lb = 2 (4 +3 )× 2.5+(4 ×3 ) = 2 (7 ×2 .5)+12 =35+12 =47 m

Exercise

13.2 1.

Assume ߨ

, unless stated otherwise. The curved surface area of a right circular cylinder of height 14 cm is 88 cm

Find the diameter of the base of the cylinder.

Solution:

Given,

Height (h) of cylinder =14 cm

Curved

surface area of cylinder =88 cm Let us consider the diameter of the cylinder be d. (r is the radius of the base of the cylinder) h =88 cm (d =2r) 22
7

× d× 14 cm=88 cm

Hence the diameter of the base of the cylinder is 2 cm. 2.

Assume Ɏ=

, unless stated otherwise. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?

Solution:

Given,

Height (h) of cylindrical tank = 1 m

Base radius (r) of cylindrical tank =

cm=70 cm= 0.7 m Area of the sheet required = Total surface area of tank =ʹɎr(r +h ) Class- XI-CBSE-Mathematics Surface Area And Volumes

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7× 0.7(0.7+1 )൨ m

(4.4×1 .7 ) m = 7.48 m Hence it will require 7.48 m sheet. 3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see Fig.). Find its (i)

Inner curved surface area,

(ii) Outer curved surface area, (iii) Total surface area.

Solution:

Given, inner radius

= r = 2 cm o uter radius = r = 2.2 cmquotesdbs_dbs4.pdfusesText_8

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