**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

(i) The area of the sheet required for making the box.

(ii) The cost of sheet for it, if a sheet measuring 1m

^{2}costs Rs 20.**Answer**

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)×h + (l×b)

= 2[(1.5 + 1.25)×1.25] + (1.5 × 1.25) m

^{2 }

^{ }= (3.575 + 1.875) m

^{2 }

= 5.45

^{ }m

^{2 }

^{}

The sheet required required to make the box is 5.45 m

^{2 }

(ii) Cost of 1 m

^{2 }of sheet

^{ }= Rs 20

∴ Cost of 5.45

^{ }m

^{2 }of sheet = Rs (20 × 5.45) = Rs 109

**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 ₹7.50 per m**

^{2}.**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)×h + (l×b)

= 2(5+4)×3 + (5×4) m

^{2 }

^{ }= (54 + 20) m

^{2 }

= 74

^{ }m

^{2 }Cost of white washing = ₹7.50 per m

^{2}

Total cost = ₹

^{ }(74×7.50) = ₹ 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**

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

^{2}is ₹15000, find the height of the hall.[Hint : Area of the four walls = Lateral surface area.]

**Answer**

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

Total cost of painting = ₹15000

Rate per m

^{2 }=

^{ }₹10

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

^{2 }= (250×h) m

^{2}

A/q,

(250×h)×10 = ₹15000

⇒ 2500×h = ₹15000

⇒ h = 15000/2500 m

⇒ 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 cm×10 cm×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 cm×10 cm×7.5 cm

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

^{2}

= 2(22.5×10 + 10×7.5 + 22.5×7.5) cm

^{2}

= 2(225 + 75 + 168.75) cm

^{2}

= 2×468.75 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) 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?

**Answer**

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

^{2}cm

^{2}= 400 cm

^{2}

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

= 2×(12.5+10)×8 cm

^{2}

= 2×22.5×8 cm

^{2}= 360 cm

^{2}

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

^{2}= 40 cm

^{2}

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

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

= 2(12.5×10 + 10×8 + 8×12.5) cm

^{2}

= 2(125+80+100) cm

^{2}

= (2×305) cm

^{2 }= 610 cm

^{2}

Thus, total surface area of cubical box is smaller by 10 cm

^{2}

^{}

**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) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

**Answer**

(i) Dimensions of greenhouse:

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

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

= 2(30×25 + 25×25 + 25×30) cm

^{2}

= 2(750 + 625 + 750) cm

^{2}

= 4250 cm

^{2}

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

= 4(30 + 25 + 25) cm

= 4×80 cm = 320 cm

**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 × 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 cardboard is ₹4 for 1000 cm**

^{2}, find the cost of cardboard required for supplying 250 boxes of each kind.**Answer**

Dimension of bigger box = 25 cm × 20 cm × 5 cm

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

= 2(25×20 + 20×5 + 25×5) cm

^{2}

= 2(500 + 100 + 125) cm

^{2}

= 1450 cm

^{2}

Dimension of smaller box = 15 cm × 12 cm × 5 cm

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

= 2(15×12 + 12×5 + 15×5) cm

^{2}

= 2(180 + 60 + 75) cm

^{2}

= 630 cm

^{2}

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

^{2}

= 250×2080 cm

^{2 }= 520000

^{ }cm

^{2}

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

^{2 }= 26000

^{ }cm

^{2}

Total Cardboard required = 520000 + 26000 cm

^{2}= 546000

^{ }cm

^{2}

Total cost of cardboard sheet = ₹ (546000 × 4)/1000 = ₹ 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 4m × 3m?**

**Answer**

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

Tarpaulin only required for all the four sides and top.

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

= [2(4+3)×2.5 + 4×3] m

^{2}

= (35×12) m

^{2}

= 47 m

^{2}

**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 = 2πrh = 88 cm

^{2}

⇒ 2 × 22/7 × r × 14 = 88

⇒ r = 88/ (2 × 22/7 × 14)

⇒ r = 1 cm

Thus, the diameter of the base = 2r = 2×1 = 2cm

**2. 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 metres of the sheet are required for the same?**

**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 = 2πr(h + r)

= (2 × 22/7 × 0.7) (1 + 0.7) m

^{2}

= (2 × 22 × 0.1 × 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.

(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 cmh = 77 cm

(i) Inner curved surface = 2πrh cm

^{2}

= 2 × 22/7 × 2 × 77cm

^{2}

= 968 cm

^{2}

(ii) Outer curved surface = 2πRh cm

= 2 × 22/7 × 2.2 × 77 cm^{2}^{2}

= 1064.8 cm

^{2}

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

= 2πrh + 2πRh + 2π(R^{2 -}r

^{2})

= [968 + 1064.8 + (2 × 22/7) (4.84 - 4)] cm

^{2}

= (2032.8 + 44/7 × 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}.**Answer**

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 = 2πrh

= (2 × 22/7 × 0.42 × 1.2) m

^{2}

= 3.168 m

^{2}

Area of the playground = (500 × 3.168) m

^{2 }= 1584 m

^{2}

**5. 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**

^{2}.**Answer**

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

Height of the pillar (h) = 3.5 m.

Rate of painting = ₹12.50 per m

^{2}

Curved surface = 2πrh

= (2 × 22/7 × 0.25 × 3.5) m

^{2}

^{ }=5.5 m

^{2}

Total cost of painting = (5.5 × 12.5) = ₹68.75

**6. Curved surface area of a right circular cylinder is 4.4 m**

^{2}. If the radius of the base of the cylinder is 0.7 m, find its height.**Answer**

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

Curved surface area = 2πrh = 4.4 m

^{2}

⇒ 2 × 22/7 × 0.7 × h = 4.4

⇒ h = 4.4/(2 × 22/7 × 0.7) = 1m

⇒ h = 1m

**7. 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

(i) its inner curved surface area,

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

^{2}.**Answer**

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

Depth of the well (h) = 10 m

Rate of plastering = ₹40 per m

^{2}

(i) Curved surface = 2πrh

= (2 × 22/7 × 1.75 × 10) m

^{2}

= 110 m

^{2}

(ii) Cost of plastering = ₹(110 × 40) = ₹4400

**8. 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.**

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 = 2πrh

= (2 × 22/7 × 0.025 × 28) m

^{2}

^{ }= 4.4 m

^{2}

**9. 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 1/12 of the steel actually used was wasted in making the tank.

(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 = 2πrh m

^{2}

= (2 × 22/7 × 2.1 × 4.5) m

^{2}

= 59.4 m

^{2}

(ii) Total surface area of the tank = 2πr(r + h) m

^{2}

= [2 × 22/7 × 2.1 (2.1 + 4.5)] m

^{2}

= 87.12 m

^{2}

Let x be the actual steel used in making tank.

∴ (1 - 1/12) × x = 87.12

⇒ x = 87.12 × 12/11

⇒ 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 + 2×2.5 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 = 2πrh

= (2 × 22/7 × 10 × 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

= 2πrh + πr

^{2}

= [(2 × 22/7 × 3 × 10.5) + 22/7 × 3

^{2}] cm

^{2}

= (198 + 198/7) cm

^{2}

= 1584/7 cm

^{2}

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

^{2}

= 7920 cm

^{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 = (πrl) cm

^{2}

=(22/7 × 5.25 × 10) cm

^{2}

=165 cm

^{2}

**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 = πr (l + r) m

^{2}

= 22/7 × 12 × (21 + 12) m

^{2}

= (22/7 × 12 × 33) m

^{2}

= 1244.57 m

^{2}

**3. Curved surface area of a cone is 308 cm**

(i) radius of the base and (ii) total surface area of the cone.

^{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

^{2}

Slant height (l) = 14cm

Let r be the radius of the base

∴ πrℓ = 308

⇒ 22/7 × r × 14 = 308

⇒ r =308/(22/7 × 14) = 7 cm

(ii) TSA of the cone = πr(l + r) cm

^{2}

= 22/7 × 7 ×(14 + 7) cm

^{2}

= (22 × 21) cm

^{2}

= 462 cm

^{2}

**4. 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

(i) slant height of the tent.

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

^{2}canvas is ₹70**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.

∴ l

^{2 }= h

^{2 }+ r

^{2}

⇒ l = √h

^{2 }+ r

^{2}

⇒ l = √10

^{2 }+ 24

^{2}

⇒ l = √100

^{ }+ 576

^{}

⇒ l = 26 m

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

Cost of 1 m

^{2}canvas = ₹70

∴ πrl = (22/7 × 24 × 26) m

^{2}= 13728/7 m

^{2}

∴ Cost of canvas = ₹ 13728/7 × 70 = ₹137280

**5. What length of tarpaulin 3m wide will be required to make conical tent of height 8m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20cm (Use π = 3.14).**

**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.

∴ l = √h

^{2 }+ r

^{2}

⇒ l = √8

^{2 }+ 6

^{2}

⇒ l = √100

⇒ l = 10 m

CSA of conical tent = πrl

= (3.14 × 6 × 10) m

^{2}= 188.4 m

^{2}

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) × 3] m = 188.4 m

^{2}

⇒ x - 0.2 m = 62.8 m

⇒ x = 63 m

**6. 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**

^{2}.**Answer**

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

Slant height tomb (l) = 25 m

Curved surface area = πrl m

^{2}

=(227×25×7) m

^{2}

=550 m

^{2}

Rate of white- washing = ₹210 per 100 m

^{2}

Total cost of white-washing the tomb = ₹(550 × 210/100) = ₹1155

**7. 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.**

**Answer**

Radius of the cone (r) = 7 cm

Height of the cone (h) = 24 cm

Let l be the slant height

∴ l = √h

^{2 }+ r

^{2}

⇒ l = √24

^{2 }+ 7

^{2}

⇒ l = √625

⇒ l = 25 m

Sheet required for one cap = Curved surface of the cone

= πrl cm

^{2}

= (22/7 × 7 × 25) cm

^{2}

= 550 cm

^{2}

Sheet required for 10 caps = 550 × 10 cm

^{2 }= 5500 cm

^{2}

**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**

^{2}, what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04 = 1.02)**Answer**

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.

∴ l = √h

^{2 }+ r

^{2}

⇒ l = √1

^{2 }+ 0.2

^{2}

⇒ l = √1.04

⇒ l = 1.02 m

Rate of painting = ₹12 per m

^{2}

Curved surface of 1 cone = πrl m

^{2}

= (3.14 × 0.2 × 1.02) m

^{2}

= 0.64056 m

^{2}

Curved surface of such 50 cones = (50 × 0.64056) m

^{2}

^{ }= 32.028 m

^{2}

Cost of painting all these cones = ₹(32.028 × 12)

= ₹384.34

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