# Class 10 RD Sharma Solutions – Chapter 16 Surface Areas and Volumes – Exercise 16.3 | Set 1

**Question 1. 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. Also, find the cost of tin sheet used for making the bucket at the rate of Rs 1.20 per dm**^{2}.

^{2}.

**Solution:**

Given,

Diameter of the top of the bucket = 40 cm

Therefore,

Radius (r

_{1}) = 40/2 = 20 cmAlso,

Diameter of bottom part of the bucket = 20 cm

Therefore,

Radius (r

_{2}) = 30/2 = 10cmDepth of the bucket (h) = 12 cm

Now,

Volume of the bucket = 1/3 π(r

_{2}^{2}+ r_{1}^{2}+ r_{1}r_{2})h= π/3(20

^{2}+ 10^{2}+ 20 × 10)12= 8800 cm

^{3}Now,

Let us assume L to be the slant height of the bucket.

Also,

Total surface area of bucket =

Since,

The cost of tin sheet used for making bucket per dm

^{2}= Rs 1.20Therefore, the total cost for making bucket 17.87dm

^{2}= 1.20 × 17.87 = Rs 21.40

**Question 2. A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm and height 3 cm. Find the area of its whole surface and volume.**

**Solution:**

Given,

We know,

Base diameter of the cone (d

_{1}) = 20 cmTherefore,

Radius (r

_{1}) = 20/2 cm = 10 cmTop diameter of Cone (d

_{2}) = 12 cmTherefore,

Radius (r

_{2}) = 12/2 cm = 6 cmAlso,

Height of the cone (h) = 3 cm

Volume of the frustum of a right circular cone = 1/3 π(r

_{2}^{2}+ r_{1}^{2 }+ r_{1}r_{2})hSubstituting the values,

= π/3(10

^{2}+ 6^{2}+ 10 × 6)3= 616 cm

^{3}Let us assume ‘L’ to be the slant height of the cone, then we know that

Now,

L = √(r

_{1}– r2_{1})^{2}+ h^{2}L = √(10 – 6)

^{2}+ 3^{2}L = √(16 + 9)

Solving, we get,

L = 5cm

Therefore,

The slant height of cone (L) = 5 cm

Thus,

Total surface area of the frustum = π(r

_{1}+ r_{2}) x L + π r_{1}^{2}+ π r_{2}^{2}Substituting the values,

= π(10 + 6) × 5 + π × 10

^{2}+ π × 6^{2}= π(80 + 100 + 36)

= π(216)

= 678.85 cm

^{2}

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

**Solution:**

Given,

Slant height of frustum of the cone (l) = 4 cm

Let us assume the ratio of the top and bottom circles be r

_{1}and r_{2}respectively.And, we know,

Perimeters of the circular ends are 18 cm and 6 cm respectively.

Substituting the values,

⟹ 2πr

_{1}= 18 cm and 2πr_{2}= 6 cmSolving, we get,

⟹ πr

_{1}= 9 cm and πr_{2}= 3 cmSince, we know,

Curved surface area of frustum of a cone = π(r

_{1}+ r_{2})l= (πr

_{1}+πr_{2})l= (9 + 3) × 4

= (12) × 4 = 48 cm

^{2}Therefore, the curved surface area = 48 cm

^{2}

**Question 4. 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 be 16 cm, find its volume, the slant surface and the total surface.**

**Solution:**

Given values,

Perimeter of the upper end of the circular cone= 44 cm

Now,

2 π r

_{1}= 442(22/7) r

_{1}= 44Solving, we get,

r

_{1}= 7 cmAlso,

Perimeter of the lower end of the circular cone= 33 cm

2 π r

_{2}= 332(22/7) r

_{2}= 33Solving,

r

_{2}= 21/4 cmNow,

Let us assume the slant height of the frustum of a right circular cone to be L

Given,

L = 16.1 cm

We know, the curved surface area of the frustum cone = π(r

_{1 }+ r_{2})lSubstituting values,

= π(7 + 5.25)16.1

Curved surface area of the frustum cone = 619.65 cm

^{3}And, The volume of the frustum cone = 1/3 π(r

_{2}^{2}+ r_{1}^{2}+ r_{1}r_{2})h= 1/3 π(72 + 5.252 + (7) (5.25)) x 16

= 1898.56 cm

^{3}Now, the total surface area of the frustum cone

= π(r

_{1}+ r_{2}) x L + π r_{1}^{2}+ π r_{2}^{2}= π(7 + 5.25) × 16.1 + π7

^{2}+ π5.25^{2}= π(7 + 5.25) × 16.1 + π(7

^{2}+ 5.25^{2}) = 860.27 cm^{2}Therefore, the total surface area of the frustum cone is 860.27 cm

^{2}

**Question 5. If the radii of the circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm, find the capacity of the bucket.**

**Solution:**

Given,

The height of the conical bucket = 45 cm

Also,

The radii of the two ends of the conical bucket are 28 cm and 7 cm respectively.

Now, x = 28 cm and y = 7 cm

Now, Volume of the conical bucket = 1/3 π(x

^{2}+ y^{2}+ xy )hSubstituting the values, we get,

= 1/3 π(28

^{2}+ 7^{2}+ 28 × 7)45 = 15435πTherefore, the volume of the bucket is 48510 cm

^{3}.

**Question 6. The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If its volume be 1/125 of the volume of the original cone, determine at what height above the base the section is made.**

**Solution:**

Let the radius of the small cone and big cone be R and r cm respectively.

Given, height of the big cone = 20 cm

Let us assume the height of section made to be h cm

Now,

Calculating the height of small cone = (20 – h) cm

Considering △OAB and △OCD

∠AOB = ∠COD [common]

And, since both of the angles are 90

^{o}∠OAB = ∠OCD

Now, by AA similarity

Then, OAB ~ △OCD

So, by C.P.S.T we have

OA/ OC = AB/ CD

Substituting values,

(20 – h)/ 20 = r/ R …… (i)

And,

Volume of small cone = 1/125 x volume of big cone, that is,

1/3 π r

^{2}(20 – h) = 1/125 × 1/3 πR^{2}x 20=>r

_{2}/ R^{2}= 1/125 × 20/ (20 – h) [From (i)]=>(20 – h)

^{2}/ 20^{2}= 1/125 × 20/20 – h=>(20 – h)

^{3}= 20^{3}/ 125=>20 – h = 20/5

=>20 – h = 4

Now,

h = 20 – 4 = 16 cm

Therefore, the section was made at a height of 16 cm above the base.

**Question 7. If the radii of the circular ends of a bucket 24 cm high are 5 and 15 cm respectively, find the surface area of the bucket.**

**Solution:**

We know,

Height of the bucket (h) = 24 cm

Radii of the circular ends of the bucket x and y are 5 cm and 15 cm respectively.

Let us assume L to be the slant height of the bucket

Now,

Curved surface area of the bucket is given by= π(x+ y)l + πx

^{2}Substituting the values, we get,

= π(5 + 15)26 + π5

^{2}= π(520 + 25) = 545π cm^{2}

**Question 8. The radii of circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 12 cm. Find the total surface area and volume of frustum.**

**Solution:**

Given,

Height of frustum cone = 12 cm

Now,

The radius of the lower end (x) of the frustum of cone= 12 cm

The radius of the upper end (y) of the frustum of cone= 3 cm

Let us assume L to be the slant height of the frustum cone

Now,

L = 15 cm

Now,

The total surface area of frustum of a cone = π (x + y) x L + π x

^{2 }+ π y^{2}Substituting values,

= π (12 + 3)15 + π12

^{2 }+ π3^{2}Solving, we get,

= 378 π cm

^{2}Also,

Volume of frustum cone = 1/3 π(y

^{2}+ x^{2}+ x y )h= 1/3 π(122 + 3

^{2}+ 1^{2}× 3) × 12= 756π cm

^{3}

**Question 9. A tent consists of a frustum of a cone capped by a cone. If radii of the ends of the frustum be 13 m and 7 m, the height of frustum be 8 m and the slant height of the conical cap be 12 m, find the canvas required for the tent.**

**Solution:**

Given,

Radii of the frustum cone are 13 cm and 7 cm respectively.

Height of the frustum cone, h = 8 m

Now,

The radius of the lower end (x) of the frustum of cone= 13 cm

The radius of the upper end (y) of the frustum of cone= 7 cm

Let us assume L to be the slant height of the frustum cone

Now,

Curved surface area of the frustum (s

_{1}) = π(x + y) × L= π(13 + 7) × 10

= 200 π m

^{2}Also,

Slant height of conical cap = 12 m

And,

Base radius of upper cap cone = 7 m

Now,

The curved surface area of upper cap cone (s

_{2}) = π *r *l= π × 7 × 12

= 264 m

^{2}Therefore, the total canvas required for tent (S)

= Curved surface area of the frustum + Curved surface area of upper cap cone

= s

_{1}+ s_{2}S = 200π + 264

= 892.57 m

^{2}which is the canvas required for the tent

**Question 10. A milk container of height 16 cm is made of metal sheet in the form of frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of Rs. 44 per litre which the container can hold.**

**Solution:**

Given,

The radius of the lower end (x) of the frustum of cone= 8 cm

The radius of the upper end (y) of the frustum of cone= 20 cm

Height, h = 16 cm

Now,

The capacity of the container = Volume of frustum of the cone

which is equivalent to,

= 1/3 π(y

^{2 }+ x^{2}+ xy )hSubstituting the values,

= 1/3 π(20

^{2}+ 8^{2 }+ (20) (8) ) x 16= 10459.42 cm

^{3 }= 10.46 litresCost of 1 litre of milk = Rs 44

Cost of 10.46 litres of milk = Rs (44 x 10.46) = Rs 460.24

Therefore,

The cost of 21.98 litres of milk = Rs (25 x 21.98) = Rs 549.50