**Chapter 9 - Surface Area and Volume**

**Practice Set 9.1**

**Practice Set 9.1**

**1. Length, breadth and height of a cuboid shape box of medicine is 20 cm, 12 cm and 10 cm respectively. Find the surface area of vertical faces and total surface area of this box.**

**Given:**

Length of the box = 20 cm

Breadth of the box = 12 cm

Height of the box = 10 cm**To find:**

Surface area of vertical faces and total surface area of the box**Solution:**

There are 4 vertical faces of a cuboid.

Surface area of vertical faces of the box = 2(lh) + 2(bh)

∴ Surface area of vertical faces of the box = 2(l + b) h

∴ Surface area of vertical faces of the box = 2(20 + 12) × 10

∴ Surface area of vertical faces of the box = 2 × 32 × 10

∴ Surface area of vertical faces of the box = 640 sq.cm

And,

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

∴ Total surface area of the box = 2(20 × 12 + 12 × 10 + 20 × 10)

∴ Total surface area of the box = 2(240 + 120 + 200)

∴ Total surface area of the box = 2 × 560

∴ Total surface area of the box = 1120 sq.cm**Ans:** The surface area of vertical faces is 640 sq.cm and total surface area of the box is 1120 sq.cm.

** **

**2. Total surface area of a box of cuboid shape is 500 sq. unit. Its breadth and height is 6 unit and 5 unit respectively. What is the length of that box ?**

**Given:**

Total surface area of the box = 500 sq. unit

Breadth of the box = 6 unit

Height of the box = 5 unit**To find:**

Length of the box**Solution:**

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

∴ 500 = 2 (6l + 6 × 5 + 5l)

∴ \(\large \frac{500}{2}\) = (11l + 30)

∴ 250 = 11l + 30

∴ 250 – 30 = 11l

∴ 220 = 11l

∴ \( \large \frac{220}{11}\) = l

∴ l = 20 units**Ans:** The length of the box is 20 units.

** **

**3. Side of a cube is 4.5 cm. Find the surface area of all vertical faces and total surface area of the cube.**

**Given:**

Side of the cube = 4.5 cm**To find:**

Surface area of all vertical faces and total surface area of the cube.**Solution:**

There are 4 vertical faces of a cube and all faces of a cube is a square.

∴ Area of vertical faces of cube = 4(l²)

∴ Area of vertical faces of cube = 4 (4.5)²

∴ Area of vertical faces of cube = 4 × 20.25

∴ Area of vertical faces of cube = 81 sq.cm.

And,

Total surface area of the cube = 6(l²)

∴ Total surface area of the cube = 6 (4.5)²

∴ Total surface area of the cube = 6 × 20.25

∴ Total surface area of the cube = 121.5 sq.cm.**Ans:** The surface area of all vertical faces is 81 sq.cm and the total surface area of the cube is 121.5 sq.cm.

**4. Total surface area of a cube is 5400 sq. cm. Find the surface area of all vertical faces of the cube.**

**Given:**

Total surface area of a cube = 5400 sq. cm**To find:**

Surface area of all vertical faces of the cube**Solution:**

Total surface area of cube = 6(l²)

∴ 5400 = 6l²

∴ \(\large \frac{5400}{6}\) = l²

∴ l² = 900 …(i)

Now,

Area of vertical faces of cube = 4(l²)

∴ Area of vertical faces of cube = 4 × 900 …[From (i)]

∴ Area of vertical faces of cube = 3600 sq.cm**Ans:** The surface area of all vertical faces of the cube is 3600 sq.cm.

** **

**5. Volume of a cuboid is 34.50 cubic metre. Breadth and height of the cuboid is 1.5m and 1.15m respectively. Find its length.**

**Given:**

Volume of the cuboid = 34.50 cubic metre

Breadth of the cuboid = 1.5m

Height of the cuboid = 1.15m **To find:**

Length of the cuboid**Solution:**

Volume of cuboid = l × b × h

∴ 34.50 = l × b × h

∴ 34.50 = l × 1.5 × 1.15

∴ \(\large \frac{34.50}{1.5\, ×\, 1.15}\) = l

∴ l = \(\large \frac{34.50}{1.725}\)

∴ l = \(\large \frac{3450\, ×\, 1000}{1725\, ×\, 100}\)

∴ l = \(\large \frac{34500}{1725}\)

∴ l = 20**Ans:** The length of the cuboid is 20 m.

** **

**6. What will be the volume of a cube having length of edge 7.5 cm ?**

**Given:**

Length of edge of the cube = 7.5 cm**To find:**

Volume of the cube**Solution:**

Volume of a cube = l³

∴ Volume of a cube = (7.5)³

∴ Volume of a cube = 7.5 × 7.5 × 7.5

∴ Volume of a cube = 421.875 ≈ 421.88 cm**Ans:** Volume of the cube is 421.88 cm³

**7. Radius of base of a cylinder is 20cm and its height is 13cm, find its curved surface area and total surface area. ****(π = 3.14)**

**Given:**

Radius of base of the cylinder = 20cm

Height of base of the cylinder = 13 cm

π = 3.14**To find:**

Curved surface area and total surface area of the cylinder**Solution:**

Curved surface area of cylinder = 2πrh

∴ Curved surface area of cylinder = 2 × 3.14 × 20 × 13

∴ Curved surface area of cylinder = 1632.8 cm²

Now,

Total surface area of cylinder = 2πr(r + h)

∴ Total surface area of cylinder = 2 × 3.14 × 20(20 + 13)

∴ Total surface area of cylinder = 2 × 3.14 × 20 × 33

∴ Total surface area of cylinder = 4144.8 cm²**Ans: **The curved surface area of the cylinder is 1632.8 cm² and the total surface area of the cylinder is 4144.8 cm².

**8. Curved surface area of a cylinder is 1980 cm² and radius of its base is 15 cm. Find the height of the cylinder. ****(π = \(\large \frac{22}{7}\))**

**Given:**

Curved surface area of the cylinder = 1980 cm²

Radius of the base of the cylinder = 15 cm

π = \( \large \frac{22}{7}\)

** **

**To find:**

Height of the cylinder

**Solution:**

Curved surface area of cylinder = 2πrh

∴ 1980 = 2 × \( \large \frac {22}{7}\) × 15 × h

∴ h = \( \large \frac{1980 \times 7}{2 \times 22 \times 15}\)

∴ h = 21 cm**Ans:** The height of the cylinder is 21 cm.

**1. Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.**

**Given:**

Perpendicular height of the cone = 12 cm

**Slant height of the cone = 13 cm**

**To find:**Radius of the base of the cone

**Solution:**We know that, l² = r² + h²∴ 132 = r² + 12²∴ 169 = r² + 144∴ 169 – 144 = r²∴ r² = 25∴ \( \sqrt { {r}^{ 2 } }\) = \( \sqrt {25} \) …[By Taking square root of both sides] ∴ r = 5 cm

**Ans:**The radius of base of the cone is 5 cm.

** **

**2. Find the volume of a cone, if its total surface area is 7128 sq.cm and radius of base is 28 cm. ****(π = \( \large \frac{22}{7}\))**

**Given:**

Total surface area of the cone = 7128 cm²

Radius of base of the cone = 28 cm

π = \( \large \frac{22}{7}\)**To find:**

Volume of the cone**Solution:**

Total surface area of cone = πr (l + r)

∴ 7128 = π × 28 × (l + 28)

∴ 7128 = \( \large \frac{22}{7}\) × 4 × (l + 28)

∴ l + 28 = \( \large \frac{7128}{22\, × \, 4}\)

∴ l + 28 = 81

∴ l = 81 – 28

∴ l = 53 cm

Now,

l² = r² + h²

∴ 532 = 282+ h²

∴ 2809 = 784 + h²

∴ 2809 – 784 = h²

∴ h² = 2025

∴ \( \sqrt { {h}^{ 2 } }\) = \( \sqrt {2025} \) …[By Taking square root of both sides]

∴ h = 45 cm

Volume of a cone = \( \large \frac{1}{3}\) πr²h

∴ Volume of a cone = \(\large \frac{1}{3}\) × \( \large \frac{22}{7}\) × 28² × 45

∴ Volume of a cone = \(\large \frac{1}{3}\) × \( \large \frac{22}{7}\) × 28 × 28 × 45

∴ Volume of a cone = 22 × 4 × 28 × 15

∴ Volume of a cone = 36960 cm³**Ans:** The volume of the cone is 36960 cm³.

** **

**3. Curved surface area of a cone is 251.2 cm² and the radius of its base is 8 cm. Find its slant height and perpendicular height. ****(π = 3.14)**

**Given:**

Curved surface area of the cone = 251.2 cm²

Radius of the base of the cone = 8 cm**To find:**

Slant height of the cone and perpendicular height of the cone**Solution:**

Curved surface area of cone = πrl

∴ 251.2 = 3.14 × 8 × l

∴ \( \large \frac{251.2}{3.14\, × \, 8}\) = l

∴ \( \large \frac{2512\, × \, 100}{314\, × \, 8\, × \, 10}\) = l

∴ l = \( \large \frac{25120}{2512}\)

∴ l = 10 cm

Now,

l² = r² + h²

∴ 10² = 8² + h²

∴ 100 = 64 + h²

∴ 100 – 64 = h²

∴ h² = 36

∴ \( \sqrt { {h}^{ 2 } }\) = \( \sqrt {36} \) …[By Taking square root of both sides]

∴ h = 6 cm**Ans:** The slant height of the cone is 10 cm and the perpendicular height of the cone is 6 cm.

** **

**4. What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is Rs.10 per sq.m ?**

**Given:**

Radius of base of cone = 6 m

Slant height of the cone = 8 m

Rate of making = ₹10 per sq.m**To find:**

Cost of making the cone

**Solution:**

Total surface area of the cone = πr (l + r)

∴ Total surface area of the cone = \( \large \frac{22}{7}\) × 6 × (8 + 6)

∴ Total surface area of the cone = \( \large \frac{22}{7}\) × 6 × 14

∴ Total surface area of the cone = 22 × 6 × 2

∴ Total surface area of the cone = 264 m²

Rate of making the cone per sq.m = ₹10

∴ Total cost of making 264 sq.m. = 264 × Rate of making the cone

∴ Total cost of making 264 sq.m. = 264 × 10

∴ Total cost of making 264 sq.m. = ₹2640**Ans: **The total cost of making the cone of tin sheet is ₹2640.

** **

**5. Volume of a cone is 6280 cubic cm and base radius of the cone is 30 cm. Find its perpendicular height. ****(π = 3.14)**

**Given:**

Volume of the cone = 6280 cubic cm

Radius of the cone = 30 cm

**To find:**

Perpendicular height of the cone

**Solution:**

Volume of a cone = \( \large \frac{1}{3}\) πr²h

∴ 6280 = \( \large \frac{1}{3}\) × 3.14 × 30² × h

∴ 6280 = \( \large \frac{1}{3}\) × 3.14 × 30 × 30 × h

∴ h = \( \large \frac{6280\, × \, 3}{3.14\, × \, 900}\)

∴ h = \( \large \frac{18840}{2826}\)

∴ h = 6.667 cm

**Ans:** The perpendicular height of the cone is 6.667 cm.

** **

**6. Surface area of a cone is 188.4 sq.cm and its slant height is 10 cm. Find its perpendicular height. ****(π = 3.14)**

**Given:**

Surface area of a cone = 188.4 sq.cm

Slant height of the cone = 10 cm

π = 3.14**To find:**

Perpendicular height of the cone**Solution:**

Curved surface area of the cone = πrl

∴ 188.4 = 3.14 × r × 10

∴ \( \large \frac{188.4}{3.14\, × \, 10}\) = r

∴ \( \large \frac{1884\, × \, 100}{314\, × \, 10\, × \, 10}\) = r

∴ r = \( \large \frac{1884}{314}\)

∴ r = 6 cm

Now,

l² = r² + h²

∴ 10² = 6² + h²

∴ 100 = 36 + h²

∴ 100 – 36 = h²

∴ h² = 64

∴ \( \sqrt { {h}^{ 2 } }\) = \( \sqrt {64} \) …[By Taking square root of both sides]

∴ h = 8 cm**Ans:** The perpendicular height of the cone is 8 cm.

** **

**7. Volume of a cone is 1212 cm³ and its height is 24 cm. Find the surface area of the cone. ****(π = \( \large \frac{22}{7}\))**

**Given:**

Volume of a cone is 1212 cm³

Height of the cone is 24 cm.

π = \( \large \frac{22}{7}\)**To find:**

Surface area of the cone**Solution:**

Volume of a cone = \( \large \frac{1}{3}\) πr²h

∴ 1232 = \( \large \frac{1}{3}\) × \( \large \frac{22}{7}\)

× r² × 24

∴ r² = \(\large \frac{1232\, ×\, 3\, ×\,7}{22\, × \, 24}\)

∴ r² = 49

∴ \( \sqrt { {r}^{ 2 } }\) = \( \sqrt {49} \) …[By Taking square root of both sides]

∴ r = 7 cm

Now,

l² = r² + h²

∴ l² = 7² + 24²

∴ l² = 49 + 576

∴ l² = 625

∴ \( \sqrt { {l}^{ 2 } }\) = \( \sqrt {625} \) …[By Taking square root of both sides]

∴ l = 25 cm

Curved surface area of cone = πrl

∴ Curved surface area of cone = \(\large \frac{22}{7}\) × 7 × 25

∴ Curved surface area of cone = 22 × 25

∴ Curved surface area of cone = 550 sq.cm**Ans:** The surface area of the cone is 550 sq.cm.

** **

**8. The curved surface area of a cone is 2200 sq.cm and its slant height is 50 cm. Find the total surface area of cone. ****(π = \(\large \frac{22}{7}\))**

**Given:**

Curved surface area of a cone = 2200 sq.cm

Slant height of the cone = 50 cm

π = \(\large \frac{22}{7}\)**To find:**

Total surface area of cone**Solution:**

Curved surface area of cone = πrl

∴ 2200 = \(\large \frac{22}{7}\) × r × 50

∴ \(\large \frac{2200\, ×\,7}{22\, ×\,50}\) = r

∴ r = 14 cm

Total surface area of cone = πr (l + r)

∴ Total surface area of cone = 227 × 14 × (50 + 14)

∴ Total surface area of cone = 227 × 14 × 64

∴ Total surface area of cone = 22 × 2 × 64

∴ Total surface area of cone = 2816 sq.cm**Ans:** The total surface area of the cone is 2816 sq.cm.

** **

**9. There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq.m. of the ground inside the tent. If the height of the tent is 18m, find the volume of the tent.**

**Given:**

People in the conical tent = 25

Area required by each person = 4 sq.m

Height of the tent = 18 m**To find:**

Volume of the tent**Solution:**

Every person inside the conical tent needs an area of 4 sq.m, of base inside the tent

∴ Surface area of the base of the tent = People in the tent × area required by each person

** **

**Also,**

**Base of the conical tent is a circle∴ Surface area of the base of the tent = Area of circle∴ πr² = 25 × 4∴ πr² = 100 sq.m …(i)Volume of the conical tent = \(\large \frac{1}{3}\) πr²h∴ Volume of the conical tent = \(\large \frac{1}{3}\) × 100 × 18 …[From (i)]∴ Volume of the conical tent = 100 × 6∴ Volume of the conical tent = 600 sq. cm.**

**Ans:**Volume of the conical tent is 600 sq. cm.

** **

**10. In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m. and the diameter of the base is 7.2 m. Find the volume of the fodder if it is to be covered by polythene in the rainy season then how much minimum polythene sheet is needed ? (π = \(\large \frac{22}{7}\) and \( \sqrt{17.37}\))**

**Given:**

**Height of the cone = 2.1 m**

**Diameter of the base = 7.2 m**

**π = \(\large \frac{22}{7}\)**

**\( \sqrt{17.37}\) = 4.17**

** **

**To find:**

**Volume of the fodder **

**Minimum polythene sheet needed to cover the conical heap**

** **

**Solution:**

**Diameter of the base = 7.2 m**

**∴ Radius of the base = \(\large \frac{1}{2}\) × 7.2**

**∴ Radius of the base = 3.6 m …(i)**

** **

**Volume of the fodder = \(\large \frac{1}{3}\) πr²h**

**∴ Volume of the fodder = \(\large \frac{1}{3}\) × \(\large \frac{22}{7}\) × (3.6)² × 2.1 …[From (i)]**

**∴ Volume of the fodder = 22 × 12.96 × 0.1**

**∴ Volume of the fodder = 28.512 m³**

** **

**Now, **

**l² = r² + h²**

**∴ l² = (3.6)² + (2.1)²**

**∴ l² = 12.96 + 4.41**

**∴ l² = 17.37**

**∴ \( \sqrt { {l}^{ 2 } }\) = \( \sqrt {17.37} \) …[By Taking square root of both sides] **

**∴ l = \( \sqrt {17.37} \)**

**∴ l = 4.17 m …[Given]**

** **

**Minimum polythene sheet needed to cover the conical heap = Curved surface area of the conical heap = πrl**

**∴ Minimum polythene sheet needed to cover the conical heap = \(\large \frac{22}{7}\) x 3.6 x 4.17**

**∴ Minimum polythene sheet needed to cover the conical heap = 47.18 sq.m**

** **

**Ans:** The volume of the heap of the fodder is 28.512 m³ and minimum polythene sheet required to cover the conical heap will be 47.18 m².

**1. Find the surface areas and volumes of spheres of the following radii. ****(π = 3.14)**

**(i) 4 cm**

**Given:**

Radius of the sphere = 4 cm

π = 3.14**To find:**

Surface area and volume of the sphere**Solution:**

Surface area of sphere = 4πr²

∴ Surface area of sphere = 4 × 3.14 × 4²

∴ Surface area of sphere = 4 × 3.14 × 16

∴ Surface area of sphere = 200.96 sq.cm

Volume of sphere = \(\large \frac{4}{3}\) πr³

∴ Volume of sphere = \(\large \frac{4}{3}\) × 3.14 × 4³

∴ Volume of sphere = \(\large \frac{4}{3}\) × 3.14 × 4 × 4 × 4

∴ Volume of sphere = \(\large \frac{256\, ×\, 3.14}{3}\)

∴ Volume of sphere = \(\large \frac{803.84}{3}\)

∴ Volume of sphere = 267.944 cubic cm**Ans:** Surface area of sphere is 200.96 sq.cm and volume of sphere is 267.944 cubic cm.

** **

**(ii) 9 cm**

**Given:**

Radius of the sphere = 9 cm

π = 3.14**To find:**

Surface area and volume of the sphere**Solution:**

Surface area of sphere = 4πr²

∴ Surface area of sphere = 4 × 3.14 × 9²

∴ Surface area of sphere = 4 × 3.14 × 81

∴ Surface area of sphere = 1017.36 sq.cm

Volume of sphere = \(\large \frac{4}{3}\) πr³

∴ Volume of sphere = \(\large \frac{4}{3}\) × 3.14 × 9³

∴ Volume of sphere = \(\large \frac{4}{3}\) × 3.14 × 9 × 9 × 9

∴ Volume of sphere = 4 × 3.14 × 3 × 81

∴ Volume of sphere = 3052.08 cubic cm**Ans:** Surface area of sphere is 1017.36 sq.cm and volume of sphere is 3052.08 cubic cm.

** **

**(iii) 3.5 cm.**

**Given:**

Radius of the sphere = 3.5 cm

π = 3.14**To find:**

Surface area and volume of the sphere**Solution:**

Surface area of sphere = 4πr²

∴ Surface area of sphere = 4 × 3.14 × 4²

∴ Surface area of sphere = 4 × 3.14 × 3.5²

∴ Surface area of sphere = 12.56 × 12.25

∴ Surface area of sphere = 153.86 sq.cm

Volume of sphere = \(\large \frac{4}{3}\) πr³

∴ Volume of sphere = \(\large \frac{4}{3}\) × 3.14 × 3.5³

∴ Volume of sphere = \(\large \frac{4}{3}\) × 3.14 × 3.5 × 3.5 × 3.5

∴ Volume of sphere = \(\large \frac{12.56\, ×\, 42.875}{3}\)

∴ Volume of sphere = \(\large \frac{538.51}{3}\)

∴ Volume of sphere = 179.503 cubic cm**Ans:** Surface area of sphere is 153.86 sq.cm and volume of sphere is 179.503 cubic cm.

** **

**2. If the radius of a solid hemisphere is 5 cm, then find its curved surface area and total surface area. ****(π = 3.14)**

**Given:**

Radius of a solid hemisphere = 5 cm

π = 3.14**To find:**

Curved surface area and Total surface area of the solid hemisphere**Solution:**

Curved surface area of hemisphere = 2πr²

∴ Curved surface area of hemisphere = 2 × 3.14 × 5²

∴ Curved surface area of hemisphere = 2 × 3.14 × 25

∴ Curved surface area of hemisphere = 157 sq.cm.

Total surface area of hemisphere = 3πr²

∴ Total surface area of hemisphere = 3 × 3.14 × 5

∴ Total surface area of hemisphere = 3 × 3.14 × 25

∴ Total surface area of hemisphere = 235.5 sq.cm.**Ans:** The curved surface area of the hemisphere is 157 sq.cm and total surface area of the hemisphere is 235.5 sq.cm.

** **

**3. If the surface area of a sphere is 2826 cm² then find its volume. ****(π = 3.14)**

**Given:**

Surface area of sphere = 2826 sq.cm.

π = 3.14**To find:**

Volume of the sphere**Solution:**

Surface area of sphere = 4πr²

∴ 2826 = 4 × 3.14 × r²

∴ 2826 = 12.56 × r²

∴ r² = \(\large \frac{2826}{12.56}\)

∴ r² = 225

∴ \( \sqrt { {r}^{ 2 } }\) = \( \sqrt {225} \) …[By Taking square root of both sides]

∴ r = \( \sqrt {225} \) cm

∴ r = 15 cm

Volume of sphere = \(\large \frac{4}{3}\) πr³

∴ Volume of sphere = \(\large \frac{4}{3}\) × 3.14 × 15³

∴ Volume of sphere = \(\large \frac{4}{3}\) × 3.14 × 15 × 15 × 15

∴ Volume of sphere = 4 × 3.14 × 5 × 225

∴ Volume of sphere = 14130 cubic cm**Ans: **The volume of the sphere is 14130 cubic cm.

** **

**4. Find the surface area of a sphere, if its volume is 38808 cubic cm. ****(π = \(\frac{22}{7}\))**

**Given:**

Volume of sphere = 38808 cubic cm

π = \(\large \frac{22}{7}\)**To find:**

Surface area of the sphere**Solution:**

Volume of sphere = \(\large \frac{4}{3}\) πr³

∴ 38808 = \(\large \frac{4}{3}\) × \(\large \frac{22}{7}\) × r³

∴ \(\large \frac{38808\, ×\, 3\, ×\, 7}{4\, ×\, 22}\) = r³

∴ r³ = \(\large \frac{9702\, ×\, 21}{22}\)

∴ r³ = 441 × 21

∴ r³ = 21 × 21 × 21

∴ \( \sqrt [ 3 ]{ r }\) = \( \sqrt [ 3 ]{ 21\, ×\, 21\, ×\, 21 }\) …[By Taking cube root of both sides]

∴ r = 21 cm …[By Taking cube root of both sides]

Surface area of sphere = 4πr²

∴ Surface area of sphere = 4 × \(\large \frac{22}{7}\) × 21

∴ Surface area of sphere = 4 × \(\large \frac{22}{7}\) × 21 × 21

∴ Surface area of sphere = 4 × 22 × 3 × 21

∴ Surface area of sphere = 5544 sq.cm.**Ans:** The surface area of the sphere is 5544 sq.cm.

** **

**5. Volume of a hemisphere is 18000π cubic cm. Find its diameter.**

**Given:**

Volume of the hemisphere is 18000π cubic cm**To find:**

Diameter of the hemisphere**Solution:**

Volume of hemisphere = \( \large \frac{2}{3}\) πr³

∴ 18000π = \( \large \frac{2}{3} \) x π x r³

∴ \( \large \frac{18000π\, ×\, 3}{2\, ×\, π} \) = r³

∴ r³ = 9000 × 3

∴ r³ = 27000

∴ r³ = 27 × 1000

∴ \( \sqrt [ 3 ]{ r }\) = \( \sqrt [ 3 ]{ 27000 }\) …[By Taking cube root of both sides]

∴ r = 30 cm

Diameter of the hemisphere = 2 × radius of the hemisphere

∴ Diameter of the hemisphere = 2 × 30

∴ Diameter of the hemisphere = 60 cm**Ans:** Diameter of the hemisphere = 60 cm

**Problem Set 9**

**Problem Set 9**

**1. If the diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations?**

**Given:**

Diameter of the road roller = 0.9 m

length of the road roller = 1.4 m**To find:**

Area of the field pressed in 500 rotations**Solution:**

Area of field pressed in 1 rotation of road roller = Curved surface area of road roller

∴ Curved surface area of the road roller = 2πrh

∴ Curved surface area of the road roller = πdh …[∵ diameter = 2 × radius]

∴ Curved surface area of the road roller = \( \large \frac{22}{7}\) × 0.9 × 1.4

∴ Curved surface area of the road roller = 22 × 0.9 × 0.2

∴ Curved surface area of the road roller = 3.96 sq.m.

Now,

Area of field pressed in 1 rotation = 3.96 sq.m.

∴ Area of field pressed in 500 rotations = 500 × 3.96

∴ Area of field pressed in 500 rotations = 1980 sq.m.**Ans:** Area of field pressed in 500 rotations is 1980 sq.m.

** **

**2. To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it ?**

**Given:**

Thickness of the glass = 2 mm

Outer length of the tank = 60.4 cm

Outer breadth of the tank = 40.4 cm

Outer height of the tank = 40.2 cm

**To find:**

Maximum volume of water the tank will contain**Solution:**

Thickness of the glass = 2 mm = 0.2 cm

Outer length of the tank = 60.4 cm

∴ Inner length of the tank (l) = Outer length – thickness of the glass on both sides

∴ Inner length of the tank (l) = 60.4 – 0.2 – 0.2

∴ Inner length of the tank (l) = 60cm

Outer breadth of the tank = 40.4 cm

∴ Inner breadth of the tank (b) = Outer breadth – thickness of the glass on both sides

∴ Inner breadth of the tank (b) = 40.4 – 0.2 – 0.2

∴ Inner breadth of the tank (b) = 40 cm

Outer height of the tank = 40.2 cm

∴ Inner height of the tank (h) = Outer height of the tank – thickness of the glass on one side …(∵The tank is open on one side)

∴ Inner height of the tank (h) = 40.2 – 0.2

∴ Inner height of the tank (h) = 40 cm

Now,

Maximum volume of water that can be contained in the tank = Volume of the tank

∴ Volume of the tank = l × b × h

∴ Volume of the tank = 60 × 40 × 40

∴ Volume of the tank = 96000 cubic cm.**Ans:** The tank can contain a maximum of 96000 cubic cm. water in it.

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**3. If the ratio of radius of base and height of a cone is 5:12 and its volume is 314 cubic metre. Find its perpendicular height and slant height. ****(π = 3.14)**

**Given:**

Ratio of radius of base and height of a cone is 5:12

Volume of the cone is 314 cubic metre

π = 3.14**To find:**

Perpendicular height and slant height of the cone**Solution:**

Ratio of radius and height of cone is 5 : 12

Let the common multiple be x.

∴ Radius of base = 5x

and Perpendicular height = 12x

Volume of the cone = \( \large \frac{1}{3}\) πr²h

∴ 314 = \( \large \frac{1}{3}\) × 3.14 × (5x)² × 12x

∴ 314 = \( \large \frac{1}{3}\) × 3.14 × 25x² × 12x

∴ 314 = 3.14 × 100x³

∴ 314 = 314x³

∴ x³ = \( \large \frac{314}{314}\)

∴ x³ = 1

∴ \( \sqrt [ 3 ]{ x }\) = \( \sqrt [ 3 ]{ 1 }\) …[By Taking cube root of both sides]

∴ x = 1

∴ Radius of base = 5x = 5 × 1 = 5 m

and Perpendicular height = 12x = 12 × 1 = 12 m

Now,

l² = r² + h²

∴ l² = 5² + 12²

∴ l² = 25 + 144

∴ l² = 169

∴ \( \sqrt { l }\) = \( \sqrt { 169 }\) …[Taking square root on both sides]

∴ l = 13 m**Ans:** The perpendicular height is 12 m and slant height of the cone is 13 m.

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**4. Find the radius of a sphere if its volume is 904.32 cubic cm. ****(π = 3.14)**

**Given:**

Volume of the sphere is 904.32 cubic cm

π = 3.14**To find:**

Radius of the sphere**Solution:**

Volume of sphere = \( \large \frac{4}{3}\) πr³

∴ 904.32 = \( \large \frac{4}{3}\) × 3.14 × r³

∴ \( \large \frac{904.32\,×\,3}{4\,×\,3.14}\) = r³

∴ r³ = \( \large \frac{288\,×\,3}{4}\)

∴ r³ = 72 × 3

∴ r³ = 216

∴ \( \sqrt [ 3 ]{ r }\) = \( \sqrt [ 3 ]{ 216 }\) …[By Taking cube root of both sides]

∴ r = 6 cm**Ans:** The radius of the sphere is 6 cm.

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**5. Total surface area of a cube is 864 sq.cm. Find its volume.**

**Given:**

Total surface area of a cube = 864 sq.cm**To find:**

Volume of the sphere**Solution:**

Total surface area of cube = 6l²

∴ 864 = 6l²

∴ l² = \( \large \frac{864}{6}\)

∴ l² = 144

∴ \( \sqrt { l }\) = \( \sqrt { 144 }\) …[Taking square root on both sides]

∴ l = 12 cm

Now,

Volume of cube = l³

∴ Volume of cube = 12³

∴ Volume of cube = 12 × 12 × 12

∴ Volume of cube = 1728 cubic cm**Ans:** The volume of the cube is 1728 cubic cm.

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**6. Find the volume of a sphere, if its surface area is 154 sq.cm.**

**Given:**

Surface area of sphere = 154 sq.cm.**To find:**

Volume of the sphere**Solution:**

Surface area of sphere = 4πr²

∴ 154 = 4 × \( \large \frac{22}{7}\) × r²

∴ \( \large \frac{154\, ×\, 7}{4\, ×\, 22}\) = r²

∴ r² = \( \large \frac{7\, ×\, 7}{4}\)

∴ r² = \( \large \frac{49}{4}\)

∴ \( \sqrt { {r}^{ 2 } }\) = \( \sqrt {\large \frac{49}{4}}\) …[By Taking square root of both sides]

∴ r = \( \large \frac{7}{2}\) cm

Volume of sphere = \( \large \frac{4}{3}\) πr³

∴ Volume of sphere = \( \large \frac{4}{3}\) × \( \large \frac{22}{7}\) × \( (\large \frac{7}{2})^3 \)

∴ Volume of sphere = \( \large \frac{4}{3}\) × \( \large \frac{22}{7}\) × \( (\large \frac{7}{2}) \) × \( (\large \frac{7}{2}) \) × \( (\large \frac{7}{2}) \)

∴ Volume of sphere = \( \large \frac{11\,×\,49}{3}\)

∴ Volume of sphere = 179.667 cu.cm.**Ans:** The volume of the sphere is 179.667 cu.cm.

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**7. Total surface area of a cone is 616 sq.cm. If the slant height of the cone is three times the radius of its base, find its slant height.**

**Given:**

Total surface area of a cone is 616 sq.cm

Slant height of the cone is three times the radius of its base; i.e. l = 3r**To find:**

Slant height of the cone**Solution:**

Total surface area of cone = πr (l + r)

∴ 616 = πr(l + r)

∴ 616 = \( (\large\frac{22}{7}) \) × r × (3r + r)

∴ 616 = \( (\large\frac{22}{7}) \) × r × (4r)

∴ 616 = \( (\large\frac{22}{7}) \) × 4r²

∴ \( (\large\frac{616\,×\,7}{22\,×\,4}) \) = r²

∴ r² = \( (\large\frac{28\,×\,7}{4}) \)

∴ r² = 7 × 7

∴ \( \sqrt { {r}^{ 2 } }\) = \( \sqrt {7\, ×\, 7}\) …[By Taking square root of both sides]

∴ r = 7 cm

Now,

l = 3r …[Given]

∴ l = 3 × 7

∴ l = 21 cm**Ans:** The slant height of the cone is 21 cm.

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**8. The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate Rs. 52 per sq.m.**

**Given:**

Inner diameter of the well = 4.2 m

Depth of the well = 10 m

Rate of plastering = Rs. 52 per sq.m.**To find:**

Inner surface area of the well

Cost of plastering the well from inside**Solution:**

Inner curved surface area of the well = 2πrh

∴ Inner curved surface area of the well = πdh …[∵ diameter = 2 × radius]

∴ Inner curved surface area of the well = \( (\large\frac{22}{7}) \) × 4.2 × 10

∴ Inner curved surface area of the well = 22 × 0.6 × 10

∴ Inner curved surface area of the well = 22 × 6

∴ Inner curved surface area of the well = 132 sq.m.

Rate of plastering = ₹52 per sq.m.

∴ Total cost of plastering = Curved surface area of the well × Rate of plastering

∴ Total cost of plastering = 132 × 52

∴ Total cost of plastering = ₹6864**Ans:** The cost of plastering the well from inside is ₹6864.

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**9. The length of a road roller is 2.1m and its diameter is 1.4m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of Rs. 7 per sq. m.**

**Given:**

Length of the road roller = 2.1 m

Diameter of the road roller = 1.4 m

Number of rotations required for levelling the ground = 500 rotations

Rate of levelling = ₹7 per sq.m.**To find:**

Area of ground levelled by the road roller and cost of levelling**Solution:**

Area of ground levelled in 1 rotation of road roller = curved surface area of road roller

∴ Curved surface area of the road roller = 2πrh

∴ Curved surface area of the road roller = πdh …[∵ diameter = 2 × radius]

∴ Curved surface area of the road roller = \( (\large \frac{22}{7}) \) × 1.4 × 2.1

∴ Curved surface area of the road roller = 22 × 0.2 × 2.1

∴ Curved surface area of the road roller = 9.24 sq.m.

Area of ground levelled in 1 rotation = 9.24 sq.m.

∴ Area of ground levelled in 500 rotations = 9.24 × 500

∴ Area of ground levelled in 500 rotations = 4620 sq.m.

Now,

Rate of levelling = ₹ 7 per sq.m.

∴ Total cost of levelling = Area of ground levelled × Rate of levelling

∴ Total cost of levelling = 4620 × 7

∴ Total cost of levelling = ₹32340**Ans:** The road roller levels 4620 sq.m. land in 500 rotations, and the cost of levelling is ₹32340.

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