Volume and Surface Area Exercise 13.1 |

Volume and Surface Area Exercise 13.2 |

Volume and Surface Area Exercise 13.3 |

**Q.1:** **Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:**

**(i) length = 12 cm, breadth = 8 cm and height = 4.5 cm.**

∴ Volume of cuboid = l × b × h

= (12 ×8 ×4.5) cm^{3}

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

= [2(12 + 8) × 4.5] cm^{2}

= ( 2 × 20 × 4.5) cm^{2} = 180 cm^{2}

∴ Total surface area of cuboid = 2(lb + bh + lh)

= 2(12 × 8) = (8 × 4.5) + (4.5 × 12) cm^{2}

= 2(96 + 36 + 54) cm^{2} = 2(186) cm^{2} = 372 cm^{2}

**(ii) Length = 26 m, breadth = 14 m and height = 6.5 m**

∴ Volume of cuboid = l × b × h = (26 × 14 × 6.5) m^{3} = 2366 m^{3}

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

= [2(26 + 14) × 6.5] m^{2} =(2 × 40 × 6.5) m^{2} = 520 m^{2}

∴Total surface area = 2(lb + bh + hl)

= 2((26 × 14) + (14 × 6.5) + (6.5 × 26)) m^{2} = 2(364 + 91 + 169) m^{2} = (2 × 624) m^{2} = 1248 m^{2}

**(iii) Length =**** 15 m, breadth = 6 m and height = 5 dm = 0.5 m.**

∴ Volume of the cuboid = l × b × h = (15 × 6 × 0.5) m^{3}

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

= [2(15 + 6) × 0.5] m^{2} = (2 × 21 × 0.5) m^{2} = 21 m^{2}

∴Total surface area = 2(lb + bh + hl)

= 2((15 × 6) + (6 × 0.5) + (0.5 × 15)) m^{2} = 2(90 + 3 + 7.5) m^{2} = (2 × 100.5) m^{2} = 201 m^{2}

**(iv) Length = 24 m, breadth = 0.25 m and height = 6 m.**

∴ Volume of the cuboid = l × b × h = (24 × 0.25 × 6) m^{3}

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

= [2(24 + 0.25) × 6] m^{2} = (2 × 24.25 × 6) m^{2} =291 m^{2}

∴Total surface area = 2(lb + bh + hl)

= 2((24 × 0.25) + (0.25 × 6) + (24 × 6)) m^{2} = 2(6 + 1.5 + 144) m^{2} = (2 × 151.5) m^{2} = 303 m^{2}

**Q.2:** **Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet requires making the cistern.**

**Solution:**

Length of Cistern = 8 m

Breadth of Cistern = 6 m

Height of Cistern = 2.5 m

Therefore, the capacity of the cistern = Volume of the cistern

Volume of cistern = l × b × h = (8 × 6 × 2.5) m^{3} =120 m^{3}

Area of the iron sheet required = Total surface area of the cistern.

∴Total surface area = 2(lb + bh + hl)

= 2((8 × 6) + (6 × 2.5) + (2.5 × 8)) m^{2}

= 2(48 + 15 + 20) m^{2} = (2 × 83) m^{2} = 166 m^{2}

**Q.3:** **The dimensions of a room are (9 m × 8 m × 6.5 m). It has one door of dimensions (2 m × 1.5 m) and two windows, each of dimensions (1.5 m × 1 m). Find the cost of whitewashing the walls at Rs. 6.40 per square meter.**

**Solution:**

Length of a room = 9 m

Breadth of a room = 8 m

Height of a room = 6.5 m

Therefore, area of 4 walls =Lateral surface area

Lateral surface area of the room = 2(l + b) × h

= [2(9 + 8) × 6.5] m^{2} = (2 × 17 × 6.5) m^{2} = 221 m^{2}

∴ Area not white washed = (area of one door) + (area of 2 windows)

= (2 × 1.5) m^{2} + (2 × 1.5 × 1) m^{2} = 3 + 3 = 6 m^{2}

∴ Area whitewashed = (221 – 6) m^{2 }= 215 m^{2}

∴ Cost of whitewashing the walls at the rate of Rs.6.40 per square meter = Rs. (6.40 × 215) = Rs. 1376

**Q.4:** **How many planks of dimensions ( 5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?**

**Solution:**

Length of plank = 5 m = 500 cm

Breadth of plank = 25 cm

Height of plank = 10 cm

Hence, volume = l × b × h = (500 × 25 × 10) cm^{3}

Now,

Length of pit = 20 m = 2000 cm

Breadth of pit = 6 m = 600 cm

Height of pit = 80 cm

Hence, volume = l × b × h = ( 2000 × 600 × 80) cm^{3}

∴ Number of planks that can be stored = \( \frac{Volume\, of\, pit}{Volume\, of\, plank}\)

= \(\frac{( 2000 × 600 × 80)}{(500 × 25 × 10)}\)= 768

**Q.5:** **How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?**

**Solution:**

Length of wall = 8 m = 800 cm

Breadth of wall = 6 m = 600 cm

Height of wall = 22.5 cm

Hence, volume = l × b × h = (800 × 600 × 22.5) cm^{3}

Length of brick = 25 cm

Breadth of brick = 11.25 cm

Height of brick = 6 cm

Hence, volume = l × b × h = (25 × 11.25 × 6) cm^{3}

∴ Number of bricks required = \( \frac{Volume\, of\, wall}{Volume\, of\, brick}\)

= \(\frac{(800 × 600 × 22.5)}{(25 × 11.25 × 6)}\)= 6400

**Q.6:** **A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring (22 cm × 12.5 cm × 7.5 cm). If (1/12) of the total volume of the wall consists of the motor, how many bricks are there in the wall?**

**Solution:**

Length of wall = 15 m

Breadth of wall = 0.3 m

Height of wall = 4 m

Hence, volume = l × b × h = (15 × 0.3 × 4) m^{3 }= 18 m^{3}

Volume of motor = \( \frac{1}{12} \times 18 = 1.5 m^{3} \)

Volume of wall = (18 – 1.5) m^{3} = 16.5 = \( \frac{33}{2} m^{3} \)

Length of brick = 22 cm

Breadth of brick = 12.5 cm

Height of brick = 7.5 cm

Hence, volume = l × b × h

= \( [ \frac{22}{100} \times \frac{12.5}{100} \times \frac{7.5}{100} ] m^{3} \) = \( \frac{33}{16000} m^{3} \)

Therefore, the number of bricks = \( \frac{Volume\, of\, bricks}{Volume\, of \, 1\, brick}\) = \( \frac{33}{2} \times \frac{16000}{33} \) = 8000

**Q.7:** **An open rectangular cistern when measured from outside is 1.35 m long, 1.08 m broad and 90 cm deep. It is made up of iron, which is 2.5 cm thick. Find the capacity of the cistern and the volume of the iron used.**

**Solution:**

External Length of Cistern = 1.35 m = 135 cm

External Breadth of Cistern = 1.08 m = 108 cm

External Height of Cistern = 90 cm

External Volume of cistern = l × b × h = (135 × 108 × 90) cm^{3} =1312200 cm^{3}

Internal Length of Cistern = (135 – 2 * 2.5) cm = 130 cm

Internal Breadth of Cistern = (108 – 2 * 2.5) cm = 103 cm

Internal Height of Cistern = (90 – 2.5) cm = 87.5 cm

Therefore, the Internal capacity of the cistern = Volume of the Internal cistern

Volume of cistern = l × b × h

= (130 × 103 × 87.5) cm^{3} =1171625 cm^{3}

Volume of iron used = External Volume of the cistern. – Internal Volume of the citern.

= (1312200 – 1171625) cm^{3} = 140575 cm^{3}

**Q.8:** **A river 2 m deep and 45 m wide is flowing t the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.**

**Solution:**

Depth of the river = 2 m

Breadth of the river = 45 m

Length of the river = 3 km/h = \( \frac{3 \times 1000}{60} m/min \) = 50 m/min.

∴ Volume of water running into the sea per minute = (50 × 45 × 2) m^{3} = 4500 m^{3}

**Q.9:** **A box made of sheet metal costs Rs 1620 at Rs. 30 per square metre. If the box is 5 m long and 3 m wide, find its height.**

**Solution:**

Total cost of sheet = Rs. 1620

Cost of metal sheet per square meter = Rs. 30

∴ Area of the sheet required = \( [\frac{Total \, cost}{rate/m^{2}}]\, sq.m. \) = \( [\frac{1620}{30}]\, sq.m.\) = \( 54 sq.m. \)

Length of box = 5 m

Breadth of box = 3 m

Now, let the height of the box be x meters.

∴ Area of the sheet = Total surface area of the box = 2(lb + bh + hl)

54 = 2((5 × 3) + (3 × x) + (x × 5)) m^{2} = 2(15 + 3x + 5x)

54 = (2 × (15 + 18x))

Solving for x, we get, x = 1.5 m

∴ The height of the box = 1.5 m.

**Q.10:** **Find the length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5 m).**

**Solution:**

Length of room = 10 m

Breadth of room = 10 m

Height of room = 5 m

∴ Length of the longest pole = Length of diagonal

= \( \sqrt{l^{2} + b^{2} + h^{2}}\) = \( \sqrt{10^{2} + 10^{2} + 5^{2}}\)

= \( \sqrt{100 + 100 + 25}\) = \( \sqrt{225}\) = 15 m

∴ The length of the longest pole that can be put in a room with given dimensions = 15 m.

**Q.11:** **How many persons can be accommodated in a dining hall of dimensions (20 m × 16 m × 5 m)**

**Solution:**

Length of hall = 20 m

Breadth of hall = 16 m

Height of hall = 4.5 m

∴ Volume of hall = l × b × h = (20 × 16 × 4.5) m^{3}

Volume of air needed per person = 5 m^{3}

∴ Number of persons = \( \frac{Volume\, of\, the\, hall}{Volume\, of\, air\, needed\, per\, person}\)

= \( [\frac{(20 × 16 × 4.5)}{5}] = 288 \)

**Q.12:** **A classroom is 10 m long, 6.4 m wide and 5m high. If each student is given 1.6 m ^{2} of the floor area, how many cubic meters of air would each student get?**

**Solution:**

Length of classroom = 10 m

Breadth of classroom = 6.4 m

Height of classroom = 5 m

Each student is given 1.6 m^{2} of the floor area.

Number of students = \( \frac{Area\, of\, the\, room}{1.6}\) = \( \frac{(10 × 6.4)}{1.6} = \frac{64}{1.6} = 40 \)

Therefore, the number of students = 40

Hence, air required by each student = \(\frac{Volume\, of\, the\, room}{number\, of\, students}\)m^{3}

= \( \frac{10 × 6.4 × 5}{40} = \frac{320}{40} = 8 m^{3} \)

**Q.13:** **The volume of a cuboid is 1536 m ^{3}. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.**

**Solution:**

Volume of a cuboid = 1536 m^{3}

Length of the cuboid = 16 m

Let the breadth and height of the cuboid be 3x and 2x.

Therefore, the volume of the cuboid = l × b × h

\( \Rightarrow \) 1536 = (16 × 3x × 2x)

\( \Rightarrow \) 1536 = 96x^{2}

∴ \( x = \sqrt{16} = 4 m\)

∴ Breadth of the cuboid = 3x = 3 × 4 = 12 m

And height of the cuboid = 2x = 2 × 4 = 8 m

**Q.14:** **The surface area of a cuboid is 758 cm ^{2}. Its length and breadth are 14 cm and 11 cm respectively. Find its height.**

**Solution:**

Surface area of a cuboid = 758 cm^{2}

Length of the cuboid = 14 cm

Breadth of the cuboid = 11 cm

Let the height of the cuboid be h cm

∴ Surface area of cuboid = 2(lb + bh + hl)

758 = 2((14 × 11) + (11 × h) + (h × 14)) cm^{2}

758 = 2(154 + 11h + 14h) cm^{2}

758 = (154 + 25h) cm^{2}

758 = (308 + 50h) cm^{2}

50h = 758 – 308

\( h = \frac{450}{50} = 9 cm \)∴ The height of the cuboid = 9 cm.

**Q.15:** **Find the volume, lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures (a) 9 m, (b) 6.5 cm [Take \( \sqrt{3} = 1.73 \) ]**

**Solution:**

**(a)** Each edge of a cube = 9 m

∴ Volume of the cube = a^{3} = (9)^{3} m^{3} = 729 m^{3}

∴ Lateral surface area of the cube = 4a^{2 }= 4 * (9)^{2 }= (4 * 81) = 324 m^{2}

∴ Total surface area of the cube = 6a^{2} = 486 m^{2}

∴ Diagonal of the cube = \( \sqrt{3} a = \sqrt{3} * 9 = 15.57 m \)

**(b)** Each edge of a cube = 6.5 cm

∴ Volume of the cube = a^{3} = (6.5)^{3} m^{3} = 274.625 cm^{3}

∴ Lateral surface area of the cube = 4a^{2 }= 4 * (6.5)^{2 }= (4 * 42.25) = 169 cm^{2}

∴ Total surface area of the cube = 6a^{2} = 253.5 cm^{2}

∴ Diagonal of the cube = \( \sqrt{3} a = \sqrt{3} * 6.5 = 11.245 cm \)

**Q.16:** **The total surface area of a cube is 1176 cm ^{2}. Find its volume.**

**Solution:**

Let each side of the cube be ‘a’ cm.

Then, the total surface area of the cube = 6a^{2} cm^{2 }

Given,

6a^{2} = 1176

Volume of the cube = a^{3} = (14)^{3} = 2744 cm^{3}

**Q.17:** **The lateral surface area of a cube is 900 cm ^{2}. Find its volume.**

**Solution:**

Let each side of the cube be ‘a’ cm

Then, the lateral surface area of the cube = 4a^{2}

∴ 4a^{2} = 900

Volume of the cube = a^{3} = (15)^{3} = 3375 cm^{3}

**Q.18:** **The volume of a cube is 512 cm ^{3}. Find its surface area.**

**Solution:**

Volume of the cube = 512 cm^{3} [Volume = a^{3}]

∴ Each edge of the cube = \( \sqrt[3]{512} = 8 cm \)

∴ Surface area of the cube = 6a^{2} = 6(8)^{2} cm^{2} = 6(64) cm^{2} = 384 cm^{2}

**Q.19:** **Three cubes of metal with edges 3 cm, 4 cm, 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.**

**Solution:**

Volume of the new cube = [(3)^{3} + (4)^{3} + (5)^{3}] cm^{3} = [27 + 64 + 125] cm^{3 } = 216 cm^{3}

Now, edge of this cube = a cm

And, a^{3 }= 216

Hence, a = 6 cm

Lateral surface area of the new cube = 4a^{2} cm^{2 }= 4 (6)^{2 }cm^{2} = 144 cm^{2}

∴ The lateral surface area of the new cube formed = 144 cm^{2 }

**Q.20:** **In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.**

**Solution:**

1 hectare = 10000 m^{2}

Area = 2 hectares = 2 × 10000 m^{2}

Depth of the ground = 5 cm = \( \frac{5}{100} m\)

Volume of water = (area × depth) = \( (2 \times 10000 \times \frac{5}{100}) m^{3} \) = 1000 m^{3}

∴ Volume of water that falls = 1000 m^{3}

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