myhelper: RD Sharma 2020 solution class 9 chapter 18 Surface Area and Volume of a Cuboid and Cube FBQS

RD Sharma 2020 solution class 9 chapter 18 Surface Area and Volume of a Cuboid and Cube FBQS

FBQS

Page-18.36

Question 1:

The lateral surface area of a cube is 256 cm². The volume of the cube is __________.

Answer 1:

Let the edge of the cube be a cm.

Lateral surface area of cube = 256 cm² (Given)

$\Rightarrow 4 a^{2} = 256$

$\Rightarrow a^{2} = 64$

$\Rightarrow a = 8 cm$

∴ Volume of the cube = a³ = (8 cm)³ = 512 cm³

Thus, the volume of the cube is 512 cm³.

The lateral surface area of a cube is 256 cm². The volume of the cube is __512 cm³__.

Question 2:

The length of the longest pole that can be put in a room of dimensions 10 m × 10 m × 5 m is ________.

Answer 2:

The length of the longest pole that can be put in a room is same as the length of the diagonal of the room.

The dimensions of the given room are 10 m × 10 m × 5 m.

Let the length, breadth and height of the room be l, b and h, respectively.

∴ l = 10 m, b = 10 m and h = 5 m

Now,

Length of the longest pole that can be put in the room

= Length of diagonal of the room

$= \sqrt{l^{2} + b^{2} + h^{2}}$

$= \sqrt{10^{2} + 10^{2} + 5^{2}}$

$= \sqrt{100 + 100 + 25}$

$= \sqrt{225}$

$= 15 m$

Thus, the length of the longest pole that can be put in the room of given dimensions is 15 m.

The length of the longest pole that can be put in a room of dimensions 10 m × 10 m × 5 m is ___15 m___.

Question 3:

The number of planks of dimensions 4 m × 50 cm × 20 cm that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is ________.

Answer 3:

Volume of the pit = Length × Breadth × Height = 16 m × 12 m × 4 m = 16 m × 1200 cm × 400 cm (1 m = 100 cm)

Volume of each plank = 4 m × 50 cm × 20 cm

∴ Number of planks that can be stored in the pit

$= \frac{Volume of the pit}{Volume of each plank} = \frac{16 m \times 1200 cm \times 400 cm}{4 m \times 50 cm \times 20 cm} = 1920$

Thus, the number of planks that can be stored in the given pit are 1920.

The number of planks of dimensions 4 m × 50 cm × 20 cm that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is __1920__.

Question 4:

The ratio of the volumes of two cubes is 729 : 1331. The ratio of their total surface areas is __________.

Answer 4:

Let the edges of the two cubes be x units and y units.

Volume of cube 1 : Volume of cube 2 = 729 : 1331            (Given)

$∴ \frac{x^{3}}{y^{3}} = \frac{729}{1331}$                        [Volume of the cube = (Edge)³]

$\Rightarrow {(\frac{x}{y})}^{3} = {(\frac{9}{11})}^{3}$

$\Rightarrow \frac{x}{y} = \frac{9}{11} . . . . . (1)$

$∴ \frac{Total surface area of cube 1}{Total surface area of cube 2} = \frac{6 x^{2}}{6 y^{2}}$

$\Rightarrow \frac{Total surface area of cube 1}{Total surface area of cube 2} = {(\frac{x}{y})}^{2} = {(\frac{9}{11})}^{2} = \frac{81}{121}$                  [Using (1)]

⇒ Total surface area of cube 1 : Total surface area of cube 2 = 81 : 121

Thus, the ratio of their total surface areas is 81 : 121.

The ratio of the volumes of two cubes is 729 : 1331. The ratio of their total surface areas is ___81 : 121___.

Question 5:

The length of a cuboid having breadth = 4 cm, height = 4 cm and total surface area = 148 cm², is __________.

Answer 5:

Let the length of the cuboid be l cm.

Breadth of the cuboid, b = 4 cm

Height of the cuboid, h = 4 cm

Now,

Total surface area of the cuboid = 148 cm² (Given)

$\Rightarrow 2 (l b + b h + h l) = 148$

$\Rightarrow l \times 4 + 4 \times 4 + 4 \times l = 74$

$\Rightarrow 4 l + 4 l = 74 - 16$

$\Rightarrow 8 l = 58$

$\Rightarrow l = \frac{58}{8} = 7.25 cm$

Thus, the length of the cuboid is 7.25 cm.

The length of a cuboid having breadth = 4 cm, height = 4 cm and total surface area = 148 cm², is ___7.25 cm___.

Question 6:

The number of cubes of edge 4 cm that can be cut from a cube of edge 12 cm, is _________.

Answer 6:

Let the edges of the bigger and smaller cubes be x cm and y cm, respectively.

Edge of the bigger cube, x = 12 cm

Edge of each smaller cube, y = 4 cm

∴ Number of smaller cubes that can be cut from the bigger cube

$= \frac{Volume of the bigger cube}{Volume of each smaller cube}$

$= \frac{12 cm \times 12 cm \times 12 cm}{4 cm \times 4 cm \times 4 cm}$ [Volume of cube = (Edge)³]

= 27

Thus, 27 smaller cubes of edge 4 cm that can be cut from a cube of edge 12 cm.

The number of cubes of edge 4 cm that can be cut from a cube of edge 12 cm, is ___27___.

Page-18.37

Question 7:

The volume of a cuboid is 3840 cm³ and the length of the cuboid is 20 cm. If the ratio of its breadth and its height is 4 : 3, then the total surface area of the cuboid is __________.

Answer 7:

Length of the cuboid, l = 20 cm

Let b and h be the breadth and height of the cuboid, respectively.

Breadth of the cuboid, b = 4x

Height of the cuboid, h = 3x

It is given that, the volume of a cuboid is 3840 cm³.

$∴ l \times b \times h = 3840 {cm}^{3}$

$\Rightarrow 20 \times 4 x \times 3 x = 3840$

$\Rightarrow x^{2} = 16$

$\Rightarrow x = 4 cm$

So,

Breadth of the cuboid, b = 4x = 4 × 4 cm = 16 cm

Height of the cuboid, h = 3x = 3 × 4 cm = 12 cm

∴ Total surface area of the cuboid

= 2(lb + bh + hl)

= 2(20 × 16 + 16 × 12 + 12 × 20)

= 2 × 752

= 1504 cm²

Thus, the total surface area of the cuboid is 1504 cm².

The volume of a cuboid is 3840 cm³ and the length of the cuboid is 20 cm. If the ratio of its breadth and its height is 4 : 3, then the total surface area of the cuboid is $1504 {cm}^{2}$ .

Question 8:

The total surface area of a cuboid is 392 cm² and the length of the cuboid is 12 cm. If the ratio of its breadth and height is 8 : 5, then the volume of the cuboid is _________.

Answer 8:

Length of the cuboid, l = 12 cm

Let the breadth and height of the cuboid be b and h, respectively.

Breadth of the cuboid, b = 8x

Height of the cuboid, h = 5x

It is given that, the total surface area of cuboid is 392 cm².

∴ 2(lb + bh + hl) = 392 cm²

$\Rightarrow 2 (12 \times 8 x + 8 x \times 5 x + 5 x \times 12) = 392$

$\Rightarrow 40 x^{2} + 156 x - 196 = 0$

$\Rightarrow 10 x^{2} + 39 x - 49 = 0$

$\Rightarrow 10 x^{2} - 10 x + 49 x - 49 = 0$

$\Rightarrow 10 x (x - 1) + 49 (x - 1) = 0$

$\Rightarrow (x - 1) (10 x + 49) = 0$

$\Rightarrow x - 1 = 0 or 10 x + 49 = 0$

$\Rightarrow x = 1 or x = - \frac{49}{10}$

Now, x cannot be negative. So, x = 1 cm.

So,

Breadth of the cuboid, b = 8x = 8 × 1 = 8 cm

Height of the cuboid, h = 5x = 5 × 1 = 5 cm

∴ Volume of the cuboid = l × b × h = 12 × 8 × 5 = 480 cm³

Thus, the volume of the cuboid is 480 cm³.

The total surface area of a cuboid is 392 cm² and the length of the cuboid is 12 cm. If the ratio of its breadth and height is 8 : 5, then the volume of the cuboid is $480 {cm}^{3}$ .

Question 9:

If the dimensions of a cuboid decrease by 10% each, then its volume decreases by _________.

Answer 9:

Let the length, breadth and height of the cuboid be l units, b units and h units, respectively.

∴ Volume of the original cuboid = lbh cu. units

If the dimensions of the cuboid are decreased by 10%, then

Length of new cuboid, L = l − 10% of l $= l - \frac{l}{10} = \frac{9 l}{10}$

Breadth of new cuboid, B = b − 10% of b $= b - \frac{b}{10} = \frac{9 b}{10}$

Height of new cuboid, H = h − 10% of h $= h - \frac{h}{10} = \frac{9 h}{10}$

∴ Volume of the new cuboid $= \frac{9 l}{10} \times \frac{9 b}{10} \times \frac{9 h}{10} = \frac{729}{1000} l b h$ cu. units

Now,

Percent decrease in volume of cuboid

$= \frac{Decrease in volume of cuboid}{Original volume of cuboid} \times 100 %$

$= \frac{l b h - \frac{729}{1000} l b h}{l b h} \times 100 %$

$= \frac{1000 - 729}{1000} \times 100 %$

$= \frac{271}{1000} \times 100 %$

$= 27.1 %$

Thus, the volume of cuboid decreases by 27.1%.

If the dimensions of a cuboid decrease by 10% each, then its volume decreases by ___27.1%___.

Question 10:

A cuboid has a total surface area of 96 cm². The sum of the squares of its length, breadth and height (in cm) is 48. The height of the cuboid is ________.

Answer 10:

Let the length, breadth and height of the cuboid be l cm, b cm and h cm, respectively.

Total surface area of the cuboid = 96 cm²         (Given)

⇒ 2(lb + bh + hl) = 96 cm²       .....(1)

Also,

l² + b² + h² = 48 cm²         (Given)

⇒ 2(l² + b² + h²) = 2 × 48 = 96 cm²      .....(2)

Subtracting (1) from (2), we get

2(l² + b² + h²) − 2(lb + bh + hl) = 96 cm² − 96 cm² = 0

⇒ (l² − 2lb + b²) + (b² − 2bh + h²) + (h² − 2hl + l²) = 0

⇒ (l − b)² + (b − h)² + (h − l)² = 0

⇒ l − b = 0, b − h = 0, h − l = 0

⇒ l = b, b = h, h = l

⇒ l = b = h      .....(3)

From (2) and (3), we get

h² + h² + h² = 48 cm²

⇒ 3h² = 48

⇒ h² = 16

⇒ h = 4 cm

Thus, the height of the cuboid is 4 cm.

A cuboid has a total surface area of 96 cm². The sum of the squares of its length, breadth and height (in cm) is 48. The height of the cuboid is ___4 cm___.

Question 11:

The volume of a cube whose surface area is 384 cm², is __________.

Answer 11:

Let the edge of cube be a cm.

Surface area of the cube = 384 cm² (Given)

⇒ 6a² = 384 cm²

⇒ a² = 64

⇒ a = $\sqrt{64}$ = 8 cm

∴ Volume of the cube = a³ = (8 cm)³ = 512 cm³

Thus, the volume of the cube is 512 cm³.

The volume of a cube whose surface area is 384 cm², is ___512 cm³___.

Question 12:

Three cubes of sides 8 cm, 6 cm and 1 cm are melted to form a new cube. The surface area of the cube so formed is __________.

Answer 12:

Let the edge of new cube be a cm.

The sides of the three cubes are 8 cm, 6 cm and 1 cm.

It is given that, the three cubes are melted to form the new cube.

∴ Volume of the new cube = Sum of volumes of the three cubes

⇒ a³ = (8 cm)³ + (6 cm)³ + (1 cm)³ [Volume of cube = (Side)³]

⇒ a³ = 512 + 216 + 1 = 729 cm³

⇒ a³ = (9 cm)³

⇒ a = 9 cm

∴ Surface area of the new cube = 6a² = 6 × (9 cm)² = 6 × 81 cm² = 486 cm²

Thus, the surface area of the new cube is 486 cm².

Three cubes of sides 8 cm, 6 cm and 1 cm are melted to form a new cube. The surface area of the cube so formed is ___486 cm²____.

Question 13:

The perimeter of one face of a cube is 40 cm. The volume of the cube is _________.

Answer 13:

Let each side of the cube be a cm.

Each face of a cube is a square.

∴ Perimeter of one face of cube = 4a

⇒ 4a = 40 cm

⇒ a = 10 cm

∴ Volume of the cube = a³ = (10 cm)³ = 1000 cm³

Thus, the volume of the given cube is 1000 cm³.

The perimeter of one face of a cube is 40 cm. The volume of the cube is ___1000 cm³____.

Question 14:

The volume of a cube is 2744 cm³. Its surface area is __________.

Answer 14:

Let the edge of the cube be a cm.

Volume of the cube = 2744 cm³ (Given)

∴ a³ = 2744 cm³

⇒ a³ = (14 cm)³ (2744 = 2 × 2 × 2 × 7 × 7 × 7)

⇒ a = 14 cm

∴ Surface area of the cube = 6a² = 6 × (14 cm)² = 6 × 196 cm² = 1176 cm²

Thus, the surface area of the cube is 1176 cm².

The volume of a cube is 2744 cm³. Its surface area is ____1176 cm²____.

Question 15:

If the areas of three adjacent faces of a cuboid are 6 cm², 8 cm²and 27 cm², then its volume is ________.

Answer 15:

Let the length, breadth and height of the cuboid be l cm, b cm and h cm, respectively.

Therefore, the areas of the adjacent faces of the cuboid are lb, bh and hl, respectively.

Now,

lb = 6 cm² .....(1)

bh = 8 cm² .....(2)

hl = 27 cm² .....(3)

Multiplying (1), (2) and (3), we get

lb × bh × hl = 6 cm² × 8 cm² × 27 cm²

⇒ l²b²h² = 1296 cm⁶

⇒ (lbh)² = (36 cm³)²

⇒ lbh = 36 cm³

Thus, the volume of the cuboid is 36 cm³.

If the areas of three adjacent faces of a cuboid are 6 cm², 8 cm²and 27 cm², then its volume is ___36 cm³___.

Question 16:

The length of a hall is 15 m and width is 12 m. The sum of the areas of the floor and flat root is equal to the sum of the areas of the four walls. The capacity of the hall is ________.

Answer 16:

Let the height of hall be h m.

Length of the hall, l = 15 m

Width of the hall, b = 12 m

Now,

Area of the floor + Area of the flat roof = Sum of the areas of the four walls

⇒ lb + lb = 2h(l + b)

⇒ 2lb = 2h(l + b)

⇒ 15 m × 12 m = h(15 m + 12 m)

⇒ h = $\frac{15 \times 12}{27}$ = $\frac{20}{3}$ m

∴ Capacity of the hall = lbh = $15 m \times 12 m \times \frac{20}{3} m$ = 1200 m³

Thus, the capacity of the hall is 1200 m³.

The length of a hall is 15 m and width is 12 m. The sum of the areas of the floor and flat root is equal to the sum of the areas of the four walls. The capacity of the hall is ___1200 m³___.

Question 17:

The area of the cardboard needed to make a box of size 25 cm × 15 cm × 8 cm is _________.

Answer 17:

The area of the cardboard needed to make the box of given size is same as the total surface area of the box.

Length of the box, l = 25 cm

Breadth of the box, b = 15 cm

Height of the box, h = 8 cm

∴ Area of the cardboard needed to make the box

= Total surface area of the box

= 2(lb + bh + hl)

= 2(25 cm × 15 cm + 15 cm × 8 cm + 8 cm × 25 cm)

= 2(375 cm² + 120 cm² + 200 cm²)

= 2 × 695 cm²

= 1390 cm²

Thus, the area of the cardboard needed to make the box of given size is 1390 cm².

The area of the cardboard needed to make a box of size 25 cm × 15 cm × 8 cm is ___1390 cm²___.

Question 18:

If the length of the diagonal of a cube is $6 \sqrt{3}$ cm, then its volume is _________.

Answer 18:

Let the side of the cube be a cm.

We know

Length of the diagonal of cube = $\sqrt{3} \times Side$

$∴ \sqrt{3} a = 6 \sqrt{3} cm$ (Given)

⇒ a = 6 cm

∴ Volume of the cube = a³ = (6 cm)³ = 216 cm³

Thus, the volume of the cube is 216 cm³.

If the length of the diagonal of a cube is $6 \sqrt{3}$ cm, then its volume is ___216 cm³___.

Question 19:

If the length of each edge of a cube increases by 20%, then the volume of the cube increases by _________.

Answer 19:

Let the length of each edge of the original cube be a units.

∴ Volume of the original cube = a³

If the length of each edge of a cube increases by 20%, then

Length of each edge of new cube = a + 20% of a = $a + \frac{20}{100} a = a + \frac{1}{5} a = \frac{6 a}{5}$

∴ Volume of the new cube $= {(\frac{6 a}{5})}^{3} = \frac{216 a^{3}}{125}$

Percent increase in volume of cube

$= \frac{Increase in volume of cube}{Original volume of cube} \times 100 %$

$= \frac{\frac{216}{125} a^{3} - a^{3}}{a^{3}} \times 100 %$

$= \frac{216 - 125}{125} \times 100 %$

$= \frac{91}{125} \times 100 %$

$= 72.8 %$

Thus, the volume of the cube increases by 72.8%.

If the length of each edge of a cube increases by 20%, then the volume of the cube increases by ___72.8%___.

Question 20:

From a solid cube of side 6 feet, a square hole of side 2 feet is punched through between a pair of opposite faces. The volume of the remaining solid in cubic feet is _________.

Answer 20:

When a square hole of side 2 feet is punched through between a pair of opposite faces of a cube of side 6 feet, then the empty space (or hole) so formed in the cube is in the form of a cuboid. The length, breadth and height of the hole are 6 feet, 2 feet and 2 feet, respectively.

Now,

Volume of the cube = (Side)³ = (6)³ = 216 cubic feet

Volume of the hole = Length × Breadth × Height = 6 × 2 × 2 = 24 cubic feet

∴ Volume of the remaining solid

= Volume of the solid cube − Volume of the hole

= 216 − 24

= 192 cubic feet

Thus, the volume of the remaining solid is 192 cubic feet.

From a solid cube of side 6 feet, a square hole of side 2 feet is punched through between a pair of opposite faces. The volume of the remaining solid in cubic feet is ___192___.

Question 21:

Three small metallic cubes whose edges are in the ratio 3 : 4 : 5 are melted to form a big cube. If the diagonal of the cube so formed is 18 cm, then the total surface area of the smallest cube is __________.

Answer 21:

Let the edge of three small metallic cubes be 3x, 4x and 5x, respectively.

Suppose the edge of the big cube is a cm.

It is given that the three small metallic cubes are melted to form a big cube.

∴ Volume of big cube = Sum of the volumes of three small cubes

⇒ a³ = (3x)³ + (4x)³ + (5x)³

⇒ a³ = 27x³ + 64x³ + 125x³ = 216x³

⇒ a³ = (6x)³

⇒ a = 6x .....(1)

It is given that the length of diagonal to the big cube is 18 cm.

$∴ \sqrt{3} a = 18 cm$ (Length of the diagonal of the cube = $\sqrt{3}$ × Side)

$\Rightarrow 6 \sqrt{3} x = 18$ [Using (1)]

$\Rightarrow x = \frac{18}{6 \sqrt{3}} = \sqrt{3} cm$

Now,

Edge of the smallest cube = 3x = $3 \sqrt{3}$ cm

Surface area of the cube = 6 × (Side)²

∴ Total surface area of the smallest cube = $6 \times {(3 \sqrt{3})}^{2}$ = 6 × 27 = 162 cm²

Thus, the the total surface area of the smallest cube is 162 cm².

Three small metallic cubes whose edges are in the ratio 3 : 4 : 5 are melted to form a big cube. If the diagonal of the cube so formed is 18 cm, then the total surface area of the smallest cube is $162 {cm}^{2}$ .