**Exercise 16.3**

**Q1.**Â **A bucket has top and bottom diameters of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm.Â Also find the cost of tin sheet for making the bucket at the rate of Rs 1.20 perÂ dm ^{2Â }**

**Soln**:

Given that top part of bucket diameter is 40 cm

So, the radius r_{1}Â =40/2=20cm

Bottom part of bucket diameter is 20 cm

r_{2}Â =30/2=10cm

Depth of the bucket (h) = 12 cm

Therefore, the Volume of the bucket =(Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h

h=(Ï€Â /3)(20^{2}+10^{2}+ 10x 20) 12

Therefore, h = 8800 cm^{3}

Let the slant height of the bucket be “L”

L = \(\sqrt{\left ( r_{1}-r_{2} \right )^{2}+h^{2}}\)

L = \(\sqrt{\left ( 20-10 \right )^{2}+12^{2}}\)

L = 15.620cm

Therefore, The slant height “L”=Â 15.620cm

So, the total surface area of bucket =Ï€Â (r_{1}+ r_{2})L +Â Ï€(r_{1}^{2}+^{Â }r_{2}^{2Â })Square units

= Ï€Â (20+ 10)15.620+Â Ï€Â (20^{2}+10^{2Â })

= \( \frac {1320\sqrt{61}+2200}{7}\)

=17.87 dm^{2Â }

Also, it is given that the cost of tin sheet used for making bucket perÂ dm^{2Â } = Rs 1.20

Therefore, the total cost forÂ 17.87 dm^{2Â }=1.20x 17.87

= Rs 21.40.

Hence, the cost forÂ Â 17.87 dm^{2Â }isÂ Rs 21.40

**Q2. A frustum of a right circular cone has a diameter of base 20cm, of top 12 cm and height 3 cm. find the area of its whole surface and volume.**

**Sol:**

Given that the base diameter of cone d_{1}= 20cm

Therefore, the radius r_{1}= 20/2=10cm

Also, the top diameter of ConeÂ d_{2}Â Â = 12 cm

Therefore, the radius Therefore, the rÂ = 12/2cm=6cm

cone height (h)= 3cm

Therefore the volume of the frustum right circular cone = (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

h=(Ï€Â /3)(10^{2}+6^{2}+ 10x 6) 3

=616 cm^{3}

Let the slant height of the bucket be “L”

=L = \(\sqrt{\left ( r_{1}-r_{2} \right )^{2}+h^{2}}\)

L = \(\sqrt{\left ( 10-6 \right )^{2}+3^{2}}\)

L = \(\sqrt{25}\)

L = 5 cm

Therefore, The slant height “L”= 5 cm

So, the total surface area of bucket =Ï€Â (r_{1}+ r_{2})L +Â Ï€(r_{1}^{2}+^{Â }r_{2}^{2Â })Square units

= Ï€Â (10+ 6)5+Â Ï€Â (10^{2Â }+6^{2Â })

= Ï€*(80+100+36)

= Ï€(216)

= 678.85 cm^{2Â }

Therefore, the total surface area of a cone isÂ 678.85 cm^{2Â }

**Q 3. The slant height of the frustum of a cone is 4 cm and perimeters of its circular ends are 18 cm and 6 cm.Â ****Find the curved surface of the frustum.**

**Soln:**

Given that theÂ slant height of frustum of a cone, L = 4 cm

LetÂ the top and bottom ratio circles beÂ r_{1Â }andÂ r_{2}

Also given that perimeters of its ends are 18 cm and 6 cm

=>2Ï€Â r_{1Â }= 18 cm

2Ï€Â r_{2}= 6 cm

ThereforeÂ Â Ï€Â r_{1Â }= 9 cmÂ — (a)

Ï€Â r_{2}= 3 cm —(b)

Therefore, theÂ Curved surface area of frustum cone =Ï€Â (r_{1}+ r_{2})L

=Â Â (Ï€r_{1}+ Ï€r_{2})L

Now, substitute the values,

=Â (9+ 3)4

=(12)4

= 48Â cm^{2Â }

Hence, the curved surface area of the frustum cone isÂ 48Â cm^{2Â }

**Q4. The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum is 16 cm, find its volume, the slant surface, and the total surface.**

**Soln:**

Given that, the volume of the frustum of the cone = 16 cm

The frustum of right circular cone perimeters are 44 cm and 33 cm

Perimeter =Â 2Ï€Â r

2Ï€Â r_{1Â }=Â 44Â Â Â Â Â Â ;Â Â Â Â Â Â Â Â Â 2Ï€Â r_{2}Â = 33

Â r_{1}= 7 cmÂ Â Â Â Â Â Â Â Â Â ;Â Â Â Â Â Â Â Â Â Â Â Â r_{2}= 5.25 cm

Let “L” be the slant height of frustum right circular cone

L = \(\sqrt{\left ( r_{1}-r_{2} \right )^{2}+h^{2}}\)

L = \(\sqrt{\left ( 7-5.25 \right )^{2}+16^{2}}\)

L = 16.1 cm

We know that, the curved surface area of the frustum cone =Ï€Â (r_{1}+ r_{2})L

=Ï€Â (7+ 5.25)16.1

Therefore, the Curved surface area of the frustum cone = 619.65 cm^{2Â }

So, the volume of the frustum cone = Â (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

h=(Ï€Â /3)(7^{2}+5.25^{2}+ 7x 5.25) 16

= 1898.56Â cm^{3}

Therefore, theÂ volume of the cone is 1898.56Â cm^{3}

So, the total surface area of bucket =Ï€Â (r_{1}+ r_{2})L +Â Ï€(r_{1}^{2}+^{Â }r_{2}^{2Â })Square units

=Ï€Â (7+5.25)16.1 +Â Ï€( 7^{2}+5.25^{2})

=860.27Â Â cm^{2Â }

Hence, the totka surface area of frustum of a cone isÂ 860.27Â Â cm^{2Â }

**Q.5:Â If the radius of circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm.Â ****Find the capacity of the bucket.**

**Soln:Â **

Given thatÂ the conical bucket height is 45 cm

The 2 circular ends of the conical bucket is of the radius of 28 cm and 7 cm

It means that,

Â r_{1} = 28 cm

r_{2}= 7 cm

We know that the volume of the conical bucket =Â Â (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

V=(Ï€Â /3)(28^{2}+7^{2}+ 28×7) 45

= 15435Â Ï€ = 48510

Therefore, the volume of the conical bucket isÂ 48510 cm^{3}

**Q6. If the radii of circular end of a bucket 24 cm high are 5 and 15 cm.Â ****Find the surface area of the bucket.**

**Soln:**

Given that the height of the bucket, h is 24 cm

The two radii are:

Â r_{1} = 5 cm

r_{2}=15 cm

Let the slant height of the bucket be “L”

L = \(\sqrt ({r_{1}-r_{2}})^{2}+h^{2}\)

L = \(\sqrt ({5-15})^{2}+24^{2}\)

L = \(\sqrt (100+576)\)

L = 26 cm

Therefore, the slant height of the bucket is 26 cm

We know that,

Curved surface area of the bucket =Ï€Â (r_{1}+ r_{2})L +Â Ï€Â r_{2}^{2}

=Â Ï€Â (5+ 15)26+Â Ï€Â 15^{2}

=Â Ï€Â (520 +225)

= 745Ï€Â cm^{2Â }

Therefore, the curved surface area of abucket isÂ 745Ï€Â cm^{2Â }

**Q7. Â The radii of circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 15 cm.Â ****Find the total surface area and volume of a frustum.**

**Soln:**

Given that the height of frustum cone = 12 cm

TheÂ Radii of frustum cone are 12 cm and 3 cm

i.e.,

Â r_{1} = 12 cm

r_{2}=3 cm

Let slant height of the frustum cone be â€˜Lâ€™

L=\(\sqrt{( r_{1}-r_{2} )^{2}+h^{2}}\)

L = \(\sqrt ({12-3})^{2}+12^{2}\)

\(L=\sqrt{81 + 144}\) = 15 cm

L = 15 cm

Therefore, the slant height of the cone is 15 cm

We know that,

Total surface area of cone =Â Ï€Â (r_{1}+ r_{2})L +Â Ï€(r_{1}^{2}+^{Â }r_{2}^{2Â })Square units

=Ï€Â (12+3)15 +Â Ï€( 12^{2}+3^{2})

=378Â Â cm^{2Â }

Therefore, the total surface area is 378Â Â cm^{2Â }

Also, the volume of frustum cone = Â (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

= Â (Ï€Â /3)(12^{2}+3^{2}+12 x3)12

=756Ï€Â cm^{3}

Hence, the volume of the frustum cone isÂ 756Ï€Â cm^{3}

**Q8. A tent consists of a frustum of a cone capped by a cone. If radii of ends of the frustum are 13 m and 7 m the height of frustum be 8 m and the slant height of the conical cap be 12 m.Â ****Find canvas required for the tent.**

**Soln:**

Given that the height of frustum”h” is 8 m

Also, the radii of the frustum cone are 13 cm and 7 cm

r_{1} = 13 m

r_{2}=7 m

Let slant height of the frustum cone be â€˜Lâ€™

L = \(\sqrt{\left ( r_{1} -r_{2}\right )^{2}+h^{2}}\)

L = \(\sqrt{\left ( 13 -7\right )^{2}+8^{2}}\)

=>L = \(\sqrt{36+64}\)

L = 10 m

Therefore, the slant height of the cone is 10 cm

The curved surface area of frustumÂ S_{1}= Ï€Â (r_{1}+ r_{2})L

= Ï€Â (13+ 7)10

S_{1}=200Ï€ m^{2}

Therefore, the curved surface area of the frustum, S_{1}=200Ï€ m^{2}

Also, given that the slant height of conical cap = 12 m

The upper cap cone base radius is 7 m

Therefore, the curved surface area of upper cap coneÂ S_{2}Â =Ï€Â r L

S_{2}Â =Ï€Â (7)(12)

S_{2}Â = 264Â m^{2}

So, the total canvas required for tent,Â S =Â S_{1}+S_{2}

S =200Ï€ +264

S =892.57Â m^{2}

Hence, the canvas required for the tent isÂ 892.57Â m^{2}

**Q9. A bucket is in the form of a frustum of a cone with a capacity of 12308.8Â cm ^{3Â }of water. Radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making.**

**Soln:**

Given that the radii of the top and bottom circular end of the bucket are 20 cm and 12 cm respectively

Let “h” be the height of the bucket

We know that the volume of frustum cone = (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

V= (Ï€Â /3)(20^{2}+12^{2}+20 x12)h

(784/3)Ï€ hÂ cm^{3Â }—-(a)

Also, given that theÂ capacity or volume of the bucket = 12308.8Â cm^{3}—–(b)

Equate the equations (a) and (b)

(784/3)Ï€ hÂ cm^{3Â }= 12308.8

h = ( 12308.8 x 3)/(784 xÂ Ï€ )

h = 15 cm

Hence, the height of the bucket is 15 cm

Let the slant height of bucket beÂ â€˜Lâ€™

\(L^{2}= \left ( r_{1}-r_{2} \right )^{2}+h^{2}\)

L = \(\sqrt{\left ( r_{1}-r_{2} \right )^{2}+h^{2}}\)

L = 17 cm

So, the slant height of the bucket is 17 cm

Curved surface area of bucket =Â Ï€Â (r_{1}+ r_{2})L +Â Ï€^{Â }r_{2}^{2Â }Square units

=Â Ï€Â (20+ 12)17 +Â Ï€ 12^{2}

=Â Ï€ ( 9248 + 144)

=Â 2160.32 cmÂ²

Hence, the curved surface area isÂ 2160.32 cmÂ²

**Q10. A bucket made of aluminium sheet is of height 20 cm and it’s upper and lower ends are of radius 25 cm and 10 cm.Â Â ****Find cost of making if the aluminum sheet costs Rs 70 per 100 cm ^{2Â }**

**Soln:**

Given that the bucket height (h) = 20 cm

bucket upper radiusÂ Â r_{1} = 25 m

bucket lower radius r_{2}=10 m

Let the slant height of the bucket be â€˜Lâ€™

L = \(\sqrt{\left ( r_{1}-r_{2} \right )^{2}+h^{2}}\)

= \(\sqrt{\left ( 25-10\right )^{2}+20^{2}}\)

L= 25 cm

Therefore, the slant height of bucket (L) = 25 cm

Curved surface area of bucketÂ Â =Â Ï€Â (r_{1}+ r_{2})L +Â Ï€^{Â }r_{2}^{2Â }Square units

=Â Ï€Â (25+ 10)25 +Â Ï€ 10^{2}

=Â 3061.5 cmÂ²

Hence, the curved surface area isÂ 3061.5 cmÂ²

It is given that the cost of making bucket per 100 cmÂ² isÂ Rs 70

Therefore, the cost of making bucket per 3061.5 cmÂ²=(3061.5/100) 70

= Rs 2143.05

Hence, the total cost for 3061.5 cmÂ² isÂ Rs 2143.05

**Q11. Radii of circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Finds its total surface area.**

**Soln**:

Given that the slant height of the frustum cone is 10 cm

The circular ends of frustum cone radii are 33 cm and 27 cm

Â Â r_{1} Â = 33 cm

r_{2}= 27 cm

We know that the total surface area of a solid frustum of cone =Â Ï€Â (r_{1}+ r_{2})L +Â Ï€(r_{1}^{2}+^{Â }r_{2}^{2Â })Square units

=Â Ï€Â (33+ 27)10 +Â Ï€(33^{2}+27^{2Â })

=Â Â Ï€Â (60)10 +Â Ï€(33^{2}+27^{2Â })

=Â Ï€Â (60)10 +Â Ï€(33^{2}+27^{2Â })

=Ï€ ( 600 + 1089 +729)

=Â 2418Â Ï€ cm^{2}

=Â 7599.42 cm^{2}

Therefore, the total surface area of a solid frustum of a cone isÂ 7599.42 cm^{2}

**Q12. A bucket made up of a metal sheet is in from of a frustum of cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm. Find the volume of the bucket. Also find the cost of the bucket if the cost of metal sheet used id Rs 20 per 100 \(cm ^{2Â }.**

**Soln**:

Given that the height of frustum cone is 16 cm

Lower end of bucket diameter Â d_{1} = 16 cm

SoÂ Â Â r_{1} = 16/2 = 8 cm

upper end of bucket diameter Â d_{2} = 40 cm

SoÂ Â Â r_{2} = 40/2 = 20 cm

Let the slant height of frustum of cone beÂ â€˜Lâ€™

L = \(\sqrt{\left ( r_{1}-r_{2} \right )^{2}+h^{2}}\)

L = \(\sqrt{\left ( 20-8 \right )^{2}+16^{2}}\)

L = \(\sqrt{\left (114+256\right )}\)

L = 20 cm

Therefore, the slant height of bucket (L) = 20 cm

We know that

The volume of the frustum cone =Â (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

V= (Ï€Â /3)(8^{2}+20^{2}+8 x20)16

V =10449.92

Hence, the volume of the frustum cone = 10449.92Â cm^{3}

Also, the curved surface area of the frustum cone = Ï€Â (r_{1}+ r_{2})L +Â Ï€^{Â }r_{2}^{2Â }Square units

= Ï€Â (8+ 20)20+Â Ï€8^{2}

= Ï€ (560 +64) =624 Ï€ cm^{2}

Therefore,Â curved surface area of the frustum cone isÂ 624 cm^{2}

It is given that the cost of the metal sheet per 100 cmÂ ^{2}Â is Rs 20

Therefore, the cost of the metal sheet per 624 Ï€ cmÂ =(624 Ï€ x 20)/100 = 391.9

Hence, the cost of the metal sheet is Rs. 391.9

**Q13. A solid is in the shape of a frustum of a cone. The diameters of two circular ends are 60 cm and 36 cm and height is 9 cm. find the area of its whole surface and volume.**

**Soln:**

Given that the height of the frustum cone = 9 cm

Lower end diameter Â d_{1} = 60 cm

SoÂ Â Â r_{1} = 60/2 = 30 cm

upper end diameter Â d_{2} = 36 cm

SoÂ Â Â r_{2} = 36/2 = 18 cm

Let the slant height of frustum of cone beÂ â€˜Lâ€™

L = \(\sqrt{\left ( r_{1}-r_{2} \right )^{2}+h^{2}}\)

L = \(\sqrt{\left ( 18-30 \right )^{2}+9^{2}}\)

L = \(\sqrt{144 + 81}\)

L = 15 cm

Therefore, the slant heightÂ (L) is 20 cm

We know that

The volume of the frustum cone =Â (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

V= (Ï€Â /3)(30^{2}+18^{2}+30 x18)9

V =5292Ï€Â cm^{3}

Hence, the volume of the frustum cone = 5292Ï€Â cm^{3}

Total surface area of frustum cone = Â Ï€Â (r_{1}+ r_{2})L +Â Ï€(r_{1}^{2}+^{Â }r_{2}^{2Â })Square units

= Â Ï€Â (30+ 18)15 +Â Ï€(30^{2}+18^{2Â })

= Â Ï€ ( 720+900+324)

=1944Â Â Ï€ cm^{2}

Therefore, the total surface area of frustum cone isÂ 1944Â Â Ï€ cm^{2}

**Q14. A milk container is made of metal sheet in the shape of frustum cone whose volume is 10459 cm ^{3}. The radii of its lower and upper circular ends are 8 cm and 20 cm. Find the cost of metal sheet used in making container at a rate of Rs 1.40 per cm^{2}.**

**Soln:**

Given, thatÂ radiusÂ r_{1}Â of lower endÂ = 8 cm

upper end radius r_{2}Â ofÂ upper endÂ = 20 cm

LetÂ the height of the container beÂ “h”

We know that,

The volume of the frustum cone =Â (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

V_{1 }= Â =Â (Ï€Â /3)(8^{2}+20^{2}+8 x20)hÂ —- (1)

Volume of the milk container =10459(3/4 )cm^{3}

V_{2} = 73216/7Â cm^{3}

To find height,

V_{1} â€“ V_{2}

(Ï€Â /3)(8^{2}+20^{2}+8 x20)h

h = 10459.42/653.45

h= 16 cm

Hence, the height of frustum cone (h) is 16 cm

Let “L” be the slant height of frustum cone

L = \(\sqrt{\left ( r_{1}-r_{2} \right )^{2}+h^{2}}\)

L = \(\sqrt{\left ( 20-8 \right )^{2}+16^{2}}\)

L = 20 cm

Therefore, the Slant height of frustum cone (L)Â is20 cm

So, Total surface area of frustum cone = Â Ï€Â (r_{1}+ r_{2})L +Â Ï€(r_{1}^{2}+^{Â }r_{2}^{2Â })Square units

= Ï€Â (8+ 20)20 +Â Ï€(8^{2}+20^{2Â })

=Ï€ (560+400 +64)

= 1024Â Ï€

=Â 3216.99 cm^{2}

Hence, the total surface area of the frustum isÂ 3216.99 cm^{2}

**Q.15: A reservoir in the form of a frustum of a right circular cone contains 44×10 ^{7}Â liters of water which fills it completely. The radii of the bottom and top of the reservoir are 50 m and 100 m.Â **

**Find the depth of water and lateral surface area of the reservoir.**

**Â ****Soln**:

Let the frustum cone depth be h

So, the volume of first cone (V) =Â (Ï€Â /3)(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})h cubic units

Â r_{1}=50m

r_{2}=100

V=Â (22/(7×3))(50^{2}+100^{2}+50 x100)h cubic units

V=Â (22/(7×3))(2500^{2}+10000^{2}+5000)hÂ V=Â (22/(7×3))(50^{2}+100^{2}+50 x100)h

Also, given that,

The volume of the reservoir =44x 10^{7Â }liters —- (b)

Equate the equations (a) and (b)

(1/3)Â Ï€ 8500 h =Â 44x 10^{7}

h = 24

Therefore, the depth of the reservoir is 24 m

**Q.16: A metallic right circular cone 20 cm high and whose vertical angle is 90Â°****Â is cut into two parts at the middle point of its axis by a plane parallel to the base.Â ****If frustum so obtained be drawn into a wire of diameter 1/16 cm.Â ****Find the length of the wire.**

**Soln**:

Assume that ABC be the cone

Height of metallic cone AO is 20 cm

It is given that the Cone is cut into two parts at the middle point of its axis

Therefore, the height of the frustum cone AD= 10 cm

Since the angle A is right angles. So each angles B and C =45Â°

Also. the angles E and F =45Â°

Assume that the radii of top and bottom circle of frustum of a Cone be Â r_{1Â }and r_{2}

From â–³Â ADE

DE/AD= Cot 45Â°

=> Â r_{1}/10= 1

=> Â r_{1}= 10cm

Also, From â–³AOBÂ

=>OE/OA= Cot45Â°

=>r_{2}/20= 1

=> r_{2}= 20cm

Hence, the radii of the wire are 10 cm and 20 cm respectively.