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**Exercise 13.1****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 1m**

Length of plastic box = 1.5 m

Width of plastic box = 1.25 m

Depth of plastic box = 1.25 m

(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.

Surface area of the box = Lateral surface area + Area of the base

= 2(l+b)Ã—h + (lÃ—b)

= 2[(1.5 + 1.25)Ã—1.25] + (1.5 Ã— 1.25) m

= (3.575 + 1.875) m

= 5.45

The sheet required required to make the box is 5.45 m

(ii) Cost of 1 m

âˆ´ Cost of 5.45

Answer

length of the room = 5m

breadth of the room = 4m

height of the room = 3m

Area of four walls including the ceiling

= 2(l+b)Ã—h + (lÃ—b)

= 2(5+4)Ã—3 + (5Ã—4) m

= (54 + 20) m

= 74

Total cost = Rs.

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

Perimeter of rectangular hall = 2(l + b) = 250 m

Total cost of painting = Rs.15000

Rate per m

Area of four walls = 2(l + b) h m

A/q,

(250Ã—h)Ã—10 = Rs.15000

â‡’ 2500Ã—h = Rs.15000

â‡’ h = 15000/2500 m

â‡’ h = 6 m

Thus 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 ^{2}. How many bricks of dimensions 22.5 cmÃ—10 cmÃ—7.5 cm can be painted out of this container?**

**Answer**

Volume of paint = 9.375 m^{2 }=^{ }93750 cm^{2}

Dimension of brick = 22.5 cmÃ—10 cmÃ—7.5 cm

Total surface area of a brick = 2(lb + bh + lh) cm^{2}

= 2(22.5Ã—10 + 10Ã—7.5 + 22.5Ã—7.5) cm^{2}

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

= 2Ã—468.75 cm^{2} = 937.5^{ }cm^{2}

Number of bricks can be painted = 93750/937.5 = 100**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?**

(i) Lateral surface area of cubical box of edge 10cm = 4Ã—10

Lateral surface area of cuboid box = 2(l+b)Ã—h

= 2Ã—(12.5+10)Ã—8 cm

= 2Ã—22.5Ã—8 cm

Thus, lateral surface area of the cubical box is greater by (400 â€“ 360) cm

(ii) Total surface area of cubical box of edge 10 cm =6Ã—102cm2=600cm2

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

= 2(12.5Ã—10 + 10Ã—8 + 8Ã—12.5) cm

= 2(125+80+100) cm

= (2Ã—305) cm

Thus, total surface area of cubical box is smaller 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?**

(i) Dimensions of greenhouse:

l = 30 cm, b = 25 cm, h = 25 cm

Total surface area of green house = 2(lb + bh + lh)

= 2(30Ã—25 + 25Ã—25 + 25Ã—30) cm

= 2(750 + 625 + 750) cm

= 4250 cm

(ii) Length of the tape needed = 4(l + b + h)

= 4(30 + 25 + 25) cm

= 4Ã—80 cm = 320 cm

Dimension of bigger box = 25 cm Ã— 20 cm Ã— 5 cm

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

= 2(25Ã—20 + 20Ã—5 + 25Ã—5) cm

= 2(500 + 100 + 125) cm

= 1450 cm

Dimension of smaller box = 15 cm Ã— 12 cm Ã— 5 cm

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

= 2(15Ã—12 + 12Ã—5 + 15Ã—5) cm

= 2(180 + 60 + 75) cm

= 630 cm

Total surface area of 250 boxes of each type = 250(1450 + 630) cm

= 250Ã—2080 cm

Extra area required = 5/100(1450 + 630) Ã— 250 cm

Total Cardboard required = 520000 + 26000 cm

Total cost of cardboard sheet = Rs. (546000 Ã— 4)/1000 = Rs. 2184

Dimensions of the box- like structure = 4m Ã— 3m Ã— 2.5

Tarpaulin only required for all the four sides and top.

Thus, Tarpaulin required = 2(l+b)Ã—h + lb

= [2(4+3)Ã—2.5 + 4Ã—3] m

= (35Ã—12) m

= 47 m

Page No: 216**Exercise 13.2**

**1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. Find the diameter of the base of the cylinder. Answer**

Let r be the radius of the base and h = 14 cm be the height of the cylinder.

Curved surface area of cylinder = 2Ï€rh = 88 cm

â‡’ 2 Ã— 22/7 Ã— r Ã— 14 = 88

â‡’ r = 88/ (2 Ã— 22/7 Ã— 14)

â‡’ r = 1 cm

Thus, the diameter of the base = 2r = 2Ã—1 = 2cm

**Answer**

Let r be the radius of the base and h be the height of the cylinder.

Base diameter = 140 cm and Height (h) = 1m

Radius of base (r) = 140/2 = 70 cm = 0.7 m

Metal sheet required to make a closed cylindrical tank = 2Ï€r(h + r)

= (2 Ã— 22/7 Ã— 0.7) (1 + 0.7) m^{2}

= (2 Ã— 22 Ã— 0.1 Ã— 1.7) m^{2}

=7.48 m^{2}**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. 13.11). Find its (i) inner curved surface area, (ii) outer curved surface area, (iii) total surface area.**

** **

**Answer**

Let R be external radius and r be the internal radius h be the length of the pipe.

R = 4.4/2 cm = 2.2 cm

r = 4/2 cm = 2 cm

h = 77 cm

(i) Inner curved surface = 2Ï€rh cm^{2}

= 2 Ã— 22/7 Ã— 2 Ã— 77cm^{2}

= 968 cm^{2}

(ii) Outer curved surface = 2Ï€Rh cm^{2}

= 2 Ã— 22/7 Ã— 2.2 Ã— 77 cm^{2}

= 1064.8 cm^{2}

(iii) Total surface area of a pipe = Inner curved surface area + outer curved surface area + areas of two bases

= 2Ï€rh + 2Ï€Rh + 2Ï€(R^{2 -} r^{2})

= [968 + 1064.8 + (2 Ã— 22/7) (4.84 - 4)] cm^{2}

= (2032.8 + 44/7 Ã— 0.84) cm^{2}

= (2032.8 + 5.28) cm^{2 }= 2038.08 cm^{2}**4. 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 ^{2}.**

Length of the roller (h) = 120 cm = 1.2 m

Radius of the cylinder = 84/2 cm = 42 cm = 0.42 m

Total no. of revolutions = 500

Distance covered by roller in one revolution = Curved surface area = 2Ï€rh

= (2 Ã— 22/7 Ã— 0.42 Ã— 1.2) m

= 3.168 m

Area of the playground = (500 Ã— 3.168) m

Radius of the pillar (r) = 50/2 cm = 25 cm = 0.25 m

Height of the pillar (h) = 3.5 m.

Rate of painting = Rs.12.50 per m

Curved surface = 2Ï€rh

= (2 Ã— 22/7 Ã— 0.25 Ã— 3.5) m

Total cost of painting = (5.5 Ã— 12.5) = Rs.68.75

Answer

Let r be the radius of the base and h be the height of the cylinder.

Curved surface area = 2Ï€rh = 4.4 m

â‡’ 2 Ã— 22/7 Ã— 0.7 Ã— h = 4.4

â‡’ h = 4.4/(2 Ã— 22/7 Ã— 0.7) = 1m

â‡’ h = 1m

(i) its inner curved surface area,

(ii) the cost of plastering this curved surface at the rate of Rs.40 per m

Answer

Radius of circular well (r) = 3.5/2 m = 1.75 m

Depth of the well (h) = 10 m

Rate of plastering = Rs.40 per m

(i) Curved surface = 2Ï€rh

= (2 Ã— 22/7 Ã— 1.75 Ã— 10) m

= 110 m

(ii) Cost of plastering = Rs.(110 Ã— 40) = Rs.4400

Answer

Radius of the pipe (r) = 5/2 cm = 2.5 cm = 0.025 m

Length of the pipe (h) = 28/2 m = 14 m

Total radiating surface = Curved surface area of the pipe = 2Ï€rh

= (2 Ã— 22/7 Ã— 0.025 Ã— 28) m

(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 1/12 of the steel actually used was wasted in making the tank.

Answer

(i) Radius of the tank (r) = 4.2/2 m = 2.1 m

Height of the tank (h) = 4.5 m

Curved surface area = 2Ï€rh m

= (2 Ã— 22/7 Ã— 2.1 Ã— 4.5) m

= 59.4 m

(ii) Total surface area of the tank = 2Ï€r(r + h) m^{2}

= [2 Ã— 22/7 Ã— 2.1 (2.1 + 4.5)] m^{2 }= 87.12 m^{2}

Let x be the actual steel used in making tank.

âˆ´ (1 - 1/12) Ã— x = 87.12

â‡’ x = 87.12 Ã— 12/11

â‡’ x = 95.04 m^{2}**10. In Fig. 13.12, 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 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.**

**Answer**

Radius of the frame (r) = 20/2 cm = 10 cm

Height of the frame (h) = 30 cm + 2Ã—2.5 cm = 35 cm

2.5 cm of margin will be added both side in the height.

Cloth required for covering the lampshade = curved surface area = 2Ï€rh

= (2 Ã— 22/7 Ã— 10 Ã— 35)cm^{2}

= 2200 cm^{2}**11. 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? Answer**

Radius of the penholder (r) = 3cm

Height of the penholder (h) = 10.5cm

Cardboard required by 1 competitor = CSA of one penholder + area of the base

= 2Ï€rh + Ï€r

= [(2 Ã— 22/7 Ã— 3 Ã— 10.5) + 22/7 Ã— 3

= (198 + 198/7) cm

= 1584/7 cm

Cardboard required for 35 competitors = (35 Ã— 1584/7) cm

= 7920 cm

**Exercise 13.2****1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area. Answer**

Radius (r) = 10.5/2 cm = 5.25 cm

Slant height (l) = 10 cm

Curved surface area of the cone = (Ï€rl) cm

=(22/7 Ã— 5.25 Ã— 10) cm

=165 cm

**2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.**

**Answer**

Radius (r) = 24/2 m = 12 m

Slant height (l) = 21 m

Total surface area of the cone = Ï€r (l + r) m^{2}

= 22/7 Ã— 12 Ã— (21 + 12) m^{2}

= (22/7 Ã— 12 Ã— 33) m^{2}

= 1244.57 m^{2}**3. Curved surface area of a cone is 308 cm ^{2} and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone. Answer**

(i) Curved surface of a cone = 308 cm

Slant height (l) = 14cm

Let r be the radius of the base

âˆ´ Ï€râ„“ = 308

â‡’ 22/7 Ã— r Ã— 14 = 308

â‡’ r =308/(22/7 Ã— 14) = 7 cm

(ii) TSA of the cone = Ï€r(l + r) cm

= 22/7 Ã— 7 Ã—(14 + 7) cm

= (22 Ã— 21) cm

= 462 cm

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m

Answer

Radius of the base (r) = 24 m

Height of the conical tent (h) = 10 m

Let l be the slant height of the cone.

âˆ´ l

â‡’ l = âˆšh

â‡’ l = âˆš10

â‡’ l = âˆš100

â‡’ l = 26 m

(ii) Canvas required to make the conical tent = Curved surface of the cone

Cost of 1 m

âˆ´ Ï€rl = (22/7 Ã— 24 Ã— 26) m

âˆ´ Cost of canvas = Rs. 13728/7 Ã— 70 = Rs.137280

Answer

Radius of the base (r) = 6 m

Height of the conical tent (h) = 8 m

Let l be the slant height of the cone.

âˆ´ l = âˆšh

â‡’ l = âˆš8

â‡’ l = âˆš100

â‡’ l = 10 m

CSA of conical tent = Ï€rl = (3.14 Ã— 6 Ã— 10) m

Breadth of tarpaulin = 3 m

Let length of tarpaulin sheet required be x.

20 cm will be wasted in cutting.

So, the length will be (x - 0.2 m)

Breadth of tarpaulin = 3 m

Area of sheet = CSA of tent

[(x - 0.2 m) Ã— 3] m = 188.4 m

â‡’ x - 0.2 m = 62.8 m

â‡’ x = 63 m

Radius (r) = 14/2 m = 7 m

Slant height tomb (l) = 25 m

Curved surface area = Ï€rl m

=(227Ã—25Ã—7) m

=550 m

Rate of white- washing = Rs.210 per 100 m

Total cost of white-washing the tomb = Rs.(550 Ã— 210/100) = Rs.1155

Radius of the cone (r) = 7 cm

Height of the cone (h) = 24 cm

Let l be the slant height

âˆ´ l = âˆšh

â‡’ l = âˆš24

â‡’ l = âˆš625

â‡’ l = 25 m

Sheet required for one cap = Curved surface of the cone

= Ï€rl cm

= (22/7 Ã— 7 Ã— 25) cm

= 550 cm

Sheet required for 10 caps = 550 Ã— 10 cm

Radius of the cone (r) = 40/2 cm = 20 cm = 0.2 m

Height of the cone (h) = 1 m

Let l be the slant height of a cone.

âˆ´ l = âˆšh

â‡’ l = âˆš1

â‡’ l = âˆš1.04

â‡’ l = 1.02 m

Rate of painting = Rs.12 per m

Curved surface of 1 cone = Ï€rl m

= (3.14 Ã— 0.2 Ã— 1.02) m

= 0.64056 m

Curved surface of such 50 cones = (50 Ã— 0.64056) m

= 32.028 m

Cost of painting all these cones = Rs.(32.028 Ã— 12)

= Rs.384.34