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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. (ii)The cost of sheet for it, if a sheet measuring 1mଶ costs ᲏20.

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 = 5.45 mଶ Hence, the area of the sheet required for making the box is 5.45 m ଶ . (ii)Cost of sheet:

1mଶ

=᲏ 20 5.45m ଶ = 5.45×20 =᲏ͳͲͻ

Hence, the cost of sheet is ᲏109.

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 mଶ .

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 ଶ Area of ceiling of rooms =lb = 5× 4 =20 m ଶ Area to be whitewashed =54+20 =74 m ଶ

Cost of whitewashing 1m

ଶ =᲏ 7.50 Hence , Cost of whitewashing 74 m ଶ =74× 7.5 =᲏ 555 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 ଶ is ᲏15000, find the height of the hall.

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 ଶ

Cost of painting the four walls is ᲏10 per m

ଶ

15000=250× h× 10

Hence , h = ଵହ଴଴଴ ଶହ଴଴ = 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 ଶ . 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

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

Total area that can be painted = 9.375 m

ଶ =93750 cm ଶ Hence ,

Number of bricks

that can be painted = ଽଷ଻ହ଴ ଽଷ଻.ହ =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

ଶ = 4 (10) ଶ =400 cm ଶ 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 ଶ The difference in lateral surface area =400െ360 =40 cm ଶ Hence the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm ଶ (ii) Total surface area of the cubical box = 6a ଶ = 6 (10) ଶ =600 cm ଶ 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 ଶ The difference in Total surface area =610െ600 =10 cm ଶ Hence the total surface area of the cubical box is smaller than the total surface area of the cuboidal box by 10 cm ଶ 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 ଶ = 2 (750+750+625) cm ଶ = 2 (2125) cm ଶ =4250 cm ଶ

Hence, the area of the glass is 4250 cm

ଶ (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 the ca rdboard is ᲏ 4 for 1000 cm ଶ , 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 ଶ = 2 (500+125+100) cm ଶ = 2 (725) cm ଶ =1450 cm ଶ 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 ଶ = 2 (180+75+60) cm ଶ = 2 (215) cm ଶ =630 cm ଶ Total surface area of 250 boxes of each type =(250×1450)+(250×630) =362500+157500 =520000 cm ଶ Extra area required by both types for overlapping =1450× ହ ଵ଴଴ +630×
ହ ଵ଴଴ =72.5+ 31.5 =104

For 250 boxes =104×250

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

Cost of

1000 cm

ଶ cardboard sheet =᲏ 4 Hence , cost of 546000 cm ଶ cardboard sheet required for 250 boxes: =

546000

1000

× 4

=᲏ 2184 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 ଶ

Exercise

: 13.2 1.

Assume ߨ

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

Find the diameter of the base of the cylinder.

Solution:

Given,

Height (h) of cylinder =14 cm

Curved

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

× d× 14 cm=88 cm

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

Assume Ɏ=

ଶଶ ଻ , 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 =

ଵସ଴ ଶ 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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ฺ ൤2 ×22

7× 0.7(0.7+1 )൨ m

ଶ = (4.4×1 .7 ) m ଶ = 7.48 m ଶ Hence , it will require 7.48 m ଶ 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 ଵ = ସ ଶ = 2 cm o uter radius = r ଶ = ସ.ସ ଶ = 2.2 cm

Length of the cylindrical pipe =77 cm

CSA of inner surface of pipe =ʹɎr

ଵ h =൤2 ×22 7

× 2× 77൨cm

ଶ =968 cm ଶ

CSA of outer surface of pipe =ʹɎr

ଶ h =൤2 ×22 7

× 2.2×77൨ cm

ଶ =1064.8 cm ଶ Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe Class- XI-CBSE-Mathematics Surface Area And Volumes

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=968+1064.8+ ʹɎ(r ଶଶ െr ଵଶ ) cm ଶ =968+1064.8+ ʹɎ{2.2 ଶ െ2 ଶ } cm ଶ = (2032.8+ 5. 28) cm ଶ =2038.08 cm ଶ 4.

Assume ߨ

ଶଶ ଻ , unless stated otherwise. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground.

Find the area of the playground in m

ଶ ?

Solution:

It is clear that a roller is cylindrical.

Given,

Height (h) of cylindrical roller = Length of roller =120 cm

Radius (r) of the circular end of roller =

଼ସ ଶ =42 cm

CSA of roller =ʹɎrh

=൤2 ×22 7

×42×120൨cm

ଶ =31680 cm ଶ Area of field =500× CSA of roller = (500×31680) cm ଶ =15840000 cm ଶ =1584 m ଶ 5.

Assume ߨ

ଶଶ ଻ , unless stated otherwise. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of ᲏12.50 per m ଶ .

Solution:

Given,

Height (h) of cylindrical pillar = 3.5 m

Radius (r) of the circular end of pillar =

ହ଴ ଶ =25 cm= 0.25 m

CSA of pillar =ʹɎrh

=൤2 ×22 7

× 0.25× 3.5൨m

ଶ = (44× 0.125)m ଶ = 5.5 m ଶ Class- XI-CBSE-Mathematics Surface Area And Volumes

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Cost of painting

1 m ଶ area =᲏ 12.50

Cost of painting

5.5 m ଶ area =᲏ (5.5×12.50) =᲏ 68.75 Hence , the cost of painting the CSA of the pillar is ᲏ 68.75 6.

Assume ߨ

ଶଶ ଻ , unless stated otherwise. Curved surface area of a right circular cylinder is 4.4 m ଶ . If the radius of the base of the cylinder is 0.7 m, find its height.

Solution:

Given, the height of the

circular cylinder to be h.

Radius

(r) of the base of cylinder = 0.7 m

CSA of cylinder = 4.4 m

ଶ

ʹɎrh= 4.4 m

ଶ =൤2 ×22 7

× 0.7×h൨= 4.4 m

ଶ h =1 m Hence , the height of the cylinder is 1 m. 7.

Assume ߨ

ଶଶ ଻ , unless stated otherwise. The inner diameter of a circular well is

3.5 m. It is 10 m deep. Find

(i)

Its inner curved surface area,

(ii) The cost of plastering this curved surface at the rate of ᲏ 40 per m ଶ

Solution:

(i)

Given, inner radius (r) of circular well =

ଷ.ହ ଶ = 1.75 m d epth (h) of circular well =10 m

Inner curved surface area =ʹɎrh

=൤2 ×22 7

× 1.75×10൨m

ଶ = (44× 0.25×10)m ଶ =110 m ଶ (ii) Cost of plastering 1 m ଶ area =᲏ 40

Cost of

plastering 110 m ଶ area =᲏ (110×40) Class- XI-CBSE-Mathematics Surface Area And Volumes

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=᲏ 4400 Hence , the cost of plastering the CSA of this well is ᲏ 4400. 8.

Assume ߨ

ଶଶ ଻ , unless stated otherwise. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Solution:

Since, height (h) of cylindrical pipe = Length of cylindrical pipe =28 m

Given, radius (r) of circular end of pipe =

ହ ଶ = 2.5 cm= 0.025 m

CSA of cylindrical pipe =ʹɎrh

=൤2 ×22 7

× 0.025×28൨m

ଶ = 4.4 m ଶ Hence, the area of the radiating surface of the system is 4.4 m ଶ . 9.

Assume ߨ

ଶଶ ଻ , unless stated otherwise. Find (i) The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high. (ii) How much steel was actually used, if ଵ ଵଶ of the steel actually used was wasted in making the tank.

Solution:

(i)

Given,

Height (h) of cylindrical tank = 4.5 m

Radius (r) of the circular end of cylindrical tank = ସ.ଶ ଶ = 2.1 m Lateral or curved surface area of tank =ʹɎrh =൤2 ×22 7

× 2.1×4. 5൨m

ଶ = (44× 0.3×4. 5)m ଶ =59.4 m ଶ Hence ,

CSA of tank is 59.4 m

ଶ . (ii) Total surface area of tank =ʹɎr (r +h ) Class- XI-CBSE-Mathematics Surface Area And Volumes

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=൤2 ×22

7× 2.1×(2.1+4 .5 )൨m

ଶ = (44× 0.3×6. 6)m ଶ =87.12 m ଶ

Let A m

ଶ steel sheet be actually used in making the tank. ׵ 12 ൨=87.12 m ଶ ֜ 11

×87.12൨m

ଶ ֜ ଶ Hence , 95.04 m ଶ of steel was actually used in making the tank. 10.

Assume ߨ

ଶଶ ଻ , unless stated otherwise. In the given figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and a height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Solution:

Given,

Height (h) of the frame of lampshade =(2.5+30+ 2.5)cm=35 cm Radius (r) of the circular end of the frame of lampshade = ଶ଴ ଶ =10 cm Class- XI-CBSE-Mathematics Surface Area And Volumes

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Cloth required for covering the lampshade =ʹɎrh =൤2 ×22 7

×10×35൨cm

ଶ =2200 cm ଶ

Hence, for covering the lampshade, 2200 cm

ଶ cloth will be required. 11.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Solution:

Given, Radius

(r) of the circular end of cylindrical penholder = 3 cm

Height (h) of penholder =10.5 cm

Surface

area of 1 penholder = CSA of penholder + Area of base of penholder =ʹɎrh+Ɏr ଶ =൤2 ×22 7

× 3× 10.5+ 22

7× 3× 3൨cm

ଶ =൤132× 1.5+198 7 ൨cm ଶ = 1584
7 cm ଶ Area of cardboard sheet used by 1 competitor = ଵହ଼ସ ଻ cm ଶ Area of cardboard sheet used by 35 competitors =൤ 1584 7

×35൨cm

ଶ =7920 cm ଶ Hence , 7920 cm ଶ cardboard sheet will be bought.

Exercise

: 13.3 1.

Assume ߨ

ଶଶ ଻ , unless stated otherwise. Diameter of the base of the cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Solution:

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Given,

diameter of the cone =10.5 cm Hence , radius of the cone, r = ଵ଴.ହ ଶ = 5.25 cm

Slant height of the cone, l =10 cm

Hence, curved surface area of the cone =Ɏrl

= 22
7

× 5.25×10

=165 cm ଶ 2.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:

Given,

diameter of the cone =24 cm Hence , radius of the cone, r = ଶସ ଶ =12 m

Slant height of the cone, l =21 m

Total surface area of the cone =Ɏrl+Ɏr

ଶ = 22
7

×12×(12+21)

= 22
7

×12×33

=1244.57 m ଶ 3.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise.

Curved

surface area of a cone is 308 cm ଶ and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.

Solution:

(i)

Given,

CSA of the cone =308 cm

ଶ ,

Slant height of the cone, l =14 cm

Curved

surface area of the cone =Ɏrl

308=22

7

×14× r

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֜ 44

Hence, radius of the base =7cm

(ii) Total surface area of the cone =Ɏrl+Ɏr ଶ =൤308+22 7

× 7

ଶ ൨ =308+154 =462 cm ଶ

Hence, total surface area of the cone is 462 cm

ଶ . 4.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. A conical tent is 10 m high and the radius of its base is 24 m. Find (i)

Slant height of the tent

(ii) Cost of the canvas required to make the tent, if the cost of 1 m ଶ canvas is ᲏70.

Solution:

(i)

Given, h

eight of the cone, h =10 m

Radius of the cone, r =24 m

Hence, slant height of the cone, l =ξr

ଶ + h ଶ = ඥ24 ଶ +10 ଶ =26 m (ii) Curved surface area of the cone =Ɏrl = 22
7

×24×26

= 13728
7 m ଶ

Cost of

1 canvas is =᲏ 70

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Hence , cost of canvas required to make the tent = ଵଷ଻ଶ଼ ଻

×70

=᲏ 137280 Hence , the cost of the canvas required to make such a tent is ᲏ 137280. 5. What length of tarpaulin 3 m wide will be required to make conical tent of height

8 m and base radius 6 m? Assume that the extra length of material that will be

required for stitching margins and wastage in cutting is approximately 20 cm. (Use Ɏ= 3.14).

Solution:

Given, h

eight of the conical tent, h =8 m

Radius of the

conical tent, r =6 m

Slant height of the conical tent, l =ξr

ଶ + h ଶ = ඥ6 ଶ + 8 ଶ =10 m

Curved

surface area of the cone =Ɏrl = 3.14× 6× 10 =188.4 m ଶ Area of the tent = Area of the tarpaulin sheet

Length × Breadth =188.4

Length × 3= 188.4

Length =

ଵ଼଼.ସ ଷ =62.8 m

Now, given that wastage of margin is 20 cm= 0.2 m

Hence , actual length of tarpaulin sheet required =62.8+ 0. 2 =63 m Hence , the length of the required tarpaulin sheet will be 63 m. 6.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ᲏

210 per 100 m

ଶ .

Solution:

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Given, radius of the conical tomb r =

ଵସ ଶ = 7 m slant height of the conical tomb l =25 m

Curved

surface area of the tomb =Ɏrl = 22
7

× 7× 25

=550 m ଶ

Cost of

white washing per 100 m ଶ =᲏ 210 Hence , cost for 550 m ଶ = ଶଵ଴×ହହ଴ ଵ଴଴ =᲏ͳͳͷͷ Hence , it will cost ᲏ 1155 while white-washing such a conical tomb. 7.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. A joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:

Given, h

eight of the cone, h =24 cm

Radius of the cone, r =7 cm

Hence, slant height of the cone, l =ξr

ଶ + h ଶ = ඥ7 ଶ +24
ଶ =25 cm

Curved

surface area of the cone =Ɏrl = 22
7

× 7× 25

=550 cm ଶ Hence , area of the sheet required to make 10 caps =550×10 =5500 cm ଶ 8.

A bus stop

is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. if the outer side of each of the cones is to be painted and the cost of painting is ᲏ 12 per m ଶ , what will be the cost of painting all these cones? (Use

Ɏ= 3.14 and take ξ1.04

= 1.02).

Solution:

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Given, h

eight of the cone, h =1 m radius of the cone , r = ସ଴ ଶ =20 cm= 0.2 m

Slant height of the cone, l =ξr

ଶ + h ଶ = ඥ0.2 ଶ + 1 ଶ =

ξ1.04

= 1.02 m

Curved

surface area of the cone =Ɏrl = 3.14× 0.2×1. 02 = 0.64056 m ଶ

Curved

surface area of 50 cones = 0.64056×50 =32.028 m ଶ

The cost of painting 1m

ଶ =᲏ 12 Hence , the cost of painting 50 cones of 32.028 m ଶ =12×32.028 =᲏ 384.336

Exercise

: 13.4 1.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise.

Find the surface area of a sphere of radius:

(i)

10.5 cm

(ii) 5.6 cm (iii) 14 cm

Solution:

(i)

Given, radius of the

sphere, r =10.5 cm

Hence, surface area of the sphere = 4Ɏr

ଶ = 4× 22
7

×10.5× 10.5

=1386 cm ଶ (ii) Given, radius of the sphere, r =5 .6cm

Hence, surface area of the sphere =ͶɎr

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

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= 4× 22

7× 5.6×5. 6

=394.24 cm ଶ (iii) Given, radius of the sphere, r =14 cm

Hence, surface area of the sphere =ͶɎr

ଶ = 4× 22
7

×14×14

=2464 cm ଶ 2.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m

Solution:

(i)

Given,

diameter of the sphere =14 cm

Hence, radius of the sphere, r =

ଵସ ଶ = 7 cm

Hence, surface area of the sphere =ͶɎr

ଶ = 4× 22
7

× 7× 7

=616 cm ଶ (ii) Given, diameter of the sphere =21 cm

Hence, radius of the sphere, r =

ଶଵ ଶ =10.5 cm

Hence, surface area of the sphere =ͶɎr

ଶ = 4× 22
7

×10.52

=1386 cm ଶ (iii) Given, diameter of the sphere = 3.5 m

Hence, radius of the sphere, r =

ଷ.ହ ଶ = 1.75 m

Hence, surface area of the sphere =ͶɎr

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

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= 4× 22

7× 1.752

=38.5 m ଶ 3. Find the total surface area of a hemisphere of radius 10 cm. (Use Ɏ= 3.14)

Solution:

Given, radius of the

hemisphere , r =10 cm Hence, total surface area of the hemisphere =͵Ɏr ଶ = 3× 3. 14×10×10 =942 cm ଶ 4.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Solution:

Given, radius

(r ଵ ) of spherical balloon = 7 cm, and radius (r ଶ ) of spherical balloon, when air is pumped into it =14 cm

Required ratio =

୍୬୧୲୧ୟ୪ ୱ୳୰୤ୟୡୣ ୟ୰ୣୟ

ୗ୳୰୤ୟୡୣ

ୟ୰ୣୟ ୟ୤୲ୣ୰ ୮୳୫୮୧୬୥ ୟ୧୰ ୧୬୲୭ ୠୟ୪୪୭୭୬

=

ͶɎr

ଵଶ

ͶɎr

ଶଶ =r ଵଶ r ଶଶ =ቀ ଻ ଵସ A2 = ଵ ସ Hence , the required ratio is 1:4 5.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ᲏16 per 100 cm ଶ

Solution:

According to the question,

Inner radius (r) of hemispherical bowl =

ଵ଴.ହ ଶ = 5.25 cm

Surface

area of hemispherical bowl =ʹɎr2 =൤2 ×22 7

×(5.25)2൨

Class- XI-CBSE-Mathematics Surface Area And Volumes

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=173.25 cm ଶ

Cost of

tin-plating 100 cm ଶ area =᲏ 16

Cost of

tin-plating 173.25 cm ଶ area =᲏ ଵ଺×ଵ଻ଷ.ଶହ ଵ଴଴ =᲏ 27.72 Hence , the cost of tin-plating the inner side of the hemispherical bowl is ᲏ 27.72. 6.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. Find the radius of a sphere whose surface area is 154 cm ଶ

Solution:

Let the radius of the sphere be r.

Surface

area of sphere =154 cm ଶ ׵ ଶ =154 cm ଶ ֜ ଶ = ଵହସ×଻ ସ×ଶଶ cm ଶ ֜ ଻ ଶ = 3.5 cm Hence , the radius of the sphere whose surface area is 154 cm ଶ is 3.5 cm. 7.

Assume Ɏ=

ଶଶ ଻ , unless stated otherwise. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution:

Let the diameter of the earth be d

Hence , the diameter of the moon will be ୢ ସ

Radius of the

earth is ୢ ଶ

Radius of the moon

ୢ ସ × ଵ ଶ = ୢ ଼

Required ratio =

ୗ୳୰୤ୟୡୣ ୟ୰ୣୟ ୭୤ ୲୦ୣ ୫୭୭୬

ୗ୳୰୤ୟୡୣ ୟ୰ୣୟ ୭୤ ୲୦ୣ ୣୟ୰୲୦

= ସ஠ቀ ౚ ఴ A .

8quotesdbs_dbs2.pdfusesText_2

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