WBBSE Solutions For Class 10 Maths Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid
What is solid object?
Solid object
If any space be surrounded by different planes or curved surfaces, then the space is said to be solids. For examples
Bricks, spheres, cones, cylinders, etc are solid objects. Every solid object covers some space.
If the object is surrounded by planes only, then the objects are known as polyhedrons.
We know that to cover any space at least 4 planes are required.
These planes are called the surface of the solid object and the straight line at which two surfaces intersect each other is known as the edges of the solid object.
WBBSE Solutions for Class 10 Maths
What is rectangular parallelopiped and cuboid?
Rectangular parallelopiped:
The solid objects of which each surface is rectangular and the lengths and breadths of any two opposite surfaces are equal and adjacent surfaces are perpendicular to each other, are called rectangular parallelopiped.
ABCD, AEHD, and BCGF are the surfaces of the parallelopiped, AB, CD, AE, DH, AD, BC,………, etc are the different edges of the parallelopiped.
The surface ABCD is the base of this solid, AB = DC = HG = EF is said to be the length of the rectangular parallelopiped and AD = BC = EH = FG is called the breadth of this solid.
Also, AE = DH = BF = CG is called the height of the rectangular parallelopiped.
The number of surfaces of a rectangular parallelopiped is 6, the number of edges is 12 and the numbers of angular points or vertices is 8.
What is dimensions of a rectangular parallelopiped?
Dimensions of a rectangular parallelopiped
The length, breadth and height of a rectangular parallelopiped are called the dimensions of it.
So, a rectangular parallelopiped has 3 dimensions.
Books, Bricks, etc are the most prominent examples of a rectangular parallelopiped.
Cuboid and Cube:
The rectangular parallelopiped whose length, breadth and height are equal is called a cube.
A cube has also 3 dimensions. Also, it has a number of surfaces 6, the number of edges is 12 and a number of angular points or vertices is 8.
In the given cube beside, ABCD, EFGH, ADHE, BCGF, CDHG, and ABFE are the 6 surfaces of the cube, AB = CD = AD = BC = EF = HG = EH = FG = AE = DH = BF = CG are the 12 edges and A, B, C, D, E, F, G, H are the 8 vertices of the given cube.
Thus we can say that cube is a special kind of rectangular parallelopiped.
Diagonals of a rectangular parallelopiped and a cube:
The straight lines AG, BH, CE and DF are the 4 diagonals of the given rectangular parallelopiped.
Similarly, The straight lines AG, BH, CE and DF are the 4 diagonals of the given cube.
Necessary formulas related to rectangular parallelopiped and cube:
Rectangular parallelopiped:
1. The total surface area of a rectangular parallelopiped = 2 x (Length x breadth + breadth x height + height x length)
Thus, the total surface area of the parallelopiped.
= 2 x (AB x BC + BC x CG + CG x AB), when AB = length, BC = breadth and CG = height.
2. Volume of a rectangular parallelopiped = length x breadth x height.
Thus, the volume of the rectangular parallelopiped.
= AB x BC x CG, when AB = length, BC = breadth and CG = height.
3. Length of the diagonals of a rectangular parallelopiped
=\(\sqrt{(\text { length })^2+(\text { breadth })^2+(\text { height })^2}\)
Thus, the length of the diagonals of the parallelopiped = \(\sqrt{(\mathrm{AB})^2+(\mathrm{BC})^2+(\mathrm{CG})^2},\)
where AB = length, BC = breadth and CG = height.
The total surface area of a cube:
1. Total surface area = 6 x (length of equal sides)2
Thus, the total surface area of the cube = 6 x (AB)2, where AB = length of the equal sides.
2. Volume of a cube = (length of the equal sides)3.
Thus, the volume of the cube = (AB)3, where AB = length of equal sides.
3. Length of the diagonals of a cube = √3 x (length of equal sides)
Thus, the length of the diagonals of the cube = √3 x AB, where AB = length of equal sides.
Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid Multiple Choice Questions
“WBBSE Class 10 cuboid solved examples”
Example 1. The length, breadth and height of a cuboidal hole are 30 m, 10 m and 14 m respectively. The number of planks having a height of 3 m, a breadth of 5 m and the thickness of 7 m can be kept in that hole is
- 20
- 30
- 40
- 50
Solution:
Given:
The length, breadth and height of a cuboidal hole are 30 m, 10 m and 14 m respectively. The number of planks having a height of 3 m, a breadth of 5 m and the thickness of 7 m
The volume of the hole = 30 x 10 x 14 cubic-m.
The volume of each of the planks = (3 x 5 x 7) cubic-m.
Hence the planks can be kept = \(\frac{30 \times 10 \times 14}{3 \times 5 \times 7}\) = 40.
∴ 3. 40 is correct.
The planks can be kept in that hole is 40
Example 2. The inner volume of a cuboidal box is 540 cc and the area of inner base is 108 sq.cm, the inner height of the box is
- 3 cm
- 4 cm
- 5 cm
- 6 cm
Solution:
Given:
The inner volume of a cuboidal box is 540 cc and the area of inner base is 108 sq.cm
Let the inner height of the box be h cm.
∴ the inner volume of the box = area of inner base x h cubic cm. = 108 x h cc.
As per question, 108 x h = 540
⇒ h = \(\frac{540}{108}\) = 5.
Hence the inner height of the box is 5 cm.
∴ 3. 5 cm is correct.
The inner height of the box is 3. 5
“Mensuration problems on rectangular parallelepiped for Class 10”
Example 3. The lateral surface area of a cube is 324 sq-m, the volume of the cube is
- 216 m3
- 343 m3
- 512 m3
- 729 m3
Solution:
Given:
The lateral surface area of a cube is 324 sq-m,
Let the length of equal sides of the cube be a m.
∴ the surface area of the cube = 4a2 sq-m. [∵ the number of lateral surface is 4]
As per question, 4a2 = 324 or, a2 = \(\frac{324}{4}\) or, a2 = 81 or, a = ± 9.
But the value of a can never be negative. ∴ a = 9.
Hence the volume of the cube = 93 cubic-metre = 729 m3.
∴ 4. 729 m3 is correct.
The volume of the cube is 4. 729 m3
Example 4. The ratio of the volumes of two cubes is 1 : 8; the ratio of total surface areas of two cubes is
- 1: 3
- 1: 4
- 1: 6
- 1: 8
Solution:
Given:
The ratio of the volumes of two cubes is 1 : 8
Let the lengths of equal sides of the two cubes be a and b units respectively.
So, their volumes of them are a3 and b3 cubic-units respectively.
As per question, a3 : b3 = 1:8
\(\Rightarrow \frac{a^3}{b^3}=\frac{1}{8} \Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{1}{2}\right)^3 \Rightarrow \frac{a}{b}=\frac{1}{2}\) [by taking cube roots]
\(\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{1}{2}\right)^2 \Rightarrow \frac{a^2}{b^2}=\frac{1}{4}\)⇒ a2: b2 = 1:4
Hence the ratio of total surface areas of two cubes =6a2: 6b2 = a2: b2 = 1: 4.
∴ 2. 1: 4 is correct.
The ratio of total surface areas of two cubes is 1: 4
Example 5. If total surface area of a cube is s sq.unit and the length of its diagonal is d unit, then the relation between s and d is
- s = 6d2
- 3s = 7d
- s3 = d2
- d2 = \(\frac{s}{2}\)
Solution: If the length of equal sides of the cube be q units, then s = 6a2 and d = √3a
⇒ \(\dot{a}=\frac{d}{\sqrt{3}} \Rightarrow a^2=\frac{d^2}{3}\)
⇒ \(s=6 \times \frac{d^2}{3} \Rightarrow s=2 d^2 \Rightarrow d^2=\frac{s}{2}\)
∴ 4. d2 = \(\frac{s}{2}\) is correct.
The relation between s and d is d2 = \(\frac{s}{2}\).
Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid True Or False
“Chapter 1 cuboid exercises WBBSE solutions”
Example 1. If the length of each edge of a cube is twice of that 1st cube, then the volume of this cube is 4 times more than that of the 1st cube.
Solution:
Given:
If the length of each edge of a cube is twice of that 1st cube, then the volume of this cube is 4 times more than that of the 1st cube
Let the length of each edge of the 1st cube be a cm.
Then the volume of this cube = a3 cc.
If the edge be twice, then the edge becomes 2a cm.
Then the volume of the cube = (2a)3cc. = 8a3cc.
So, the volume is \(\frac{8 a^3}{a^3}\) = 8 times more.
Hence the statement is false.
Example 2. The height of rainfall in 2 hectare land is 5 cm, the volume, of rainwater is 1000 cubic metres.
Solution:
Given:
The height of rainfall in 2 hectare land is 5 cm, the volume, of rainwater is 1000 cubic metres.
1 hectre = 100 arc = 100 x 100 sq-m. = 10000 sq-m.
2 hectre = 2 x 10000 sq-m = 20000 sq-m.
Height of rainfall = 5 cm = 0.05 m
volume of rainwater = 0 05 x 20000 cubic-metre = 1000 cubic-metres
So the required volume of rainwater = 1000 cubic-metre
Hence the statement is true.
Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid Fill In The Blanks
Example 1. Area of a rectangular cardboard = _______ x breadth.
Solution: length.
Example 2. If the length = a unit, breadth = b unit and height = c unit of a room, then the length of the diagonal of the room = _______ unit.
Solution: \(\sqrt{a^2+b^2+c^2}\)
Example 3. The length of the diagonal of a cube = _______ x length of one side.
Solution: 3.
Example 4. If the length, breadth and height of a rectangular parallelopiped are equal then the name of this solid is ________.
Solution: Cube.
Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid Short Answer Type Questions
“Class 10 Maths cuboid volume and surface area problems”
Example 1. A tank can contain 384 litres of water. The length of the tank is 3 times of its depthness and the breadth is 2 times of its depths. Find the depthness of the tank.
Solution:
Given:
A tank can contain 384 litres of water. The length of the tank is 3 times of its depthness and the breadth is 2 times of its depths.
384 litres = 384 cubic dcm.
Let the depthness of the tank = x dcm.
As per question, the length of the tank = 3x dcm. and the breadth = 2x dcm.
∴ the volume of the tank = x X 3x X 2x cubic-dcm = 6x3 cubic-dcm
As per the question, 6x3 = 384
⇒ x3 = \(\frac{384}{6}\) = 64 = 43
⇒ x = 4.
Hence the depthness of the tank = 4 dcm = 0.4 metres.
Example 2. If the number of surfaces of a cuboid is x, the number of edges is y, the number of vertices is z and the number of diagonals is p, find the value of (x – y + z + p).
Solution:
Given:
If the number of surfaces of a cuboid is x, the number of edges is y, the number of vertices is z and the number of diagonals is p
The number of surfaces of a cuboid is 6. ∴ x = 6
The number of edges of a cuboid is 12. ∴ y = 12.
The number of vertices of a cuboid is 8. ∴ z = 8.
The number of diagonals of a cuboid is 4. ∴ p = 4
So, x – y + z+ p = 6-12 + 8 + 4 = 6.
Example 3. The lengths of the dimensions of two cuboids are 4, 6, 4 units and 8, (2h – 1), 2 units respectively. If the volumes of two cuboids are equal, then find the value of h.
Solution:
Given:
The lengths of the dimensions of two cuboids are 4, 6, 4 units and 8, (2h – 1), 2 units respectively.
The volume of the 1st cube = 4x6x4 cubic units, and the volume of the 2nd cube = 8 x (2h – 1) x 2 cubic units.
As per question, 8 x (2h – 1) x 2 = 4 x 6 x 4
or, 2h- 1 = \(\frac{4 \times 6 \times 4}{8 \times 2}\) or, 2h – 1 = 6
or, 2h= 6+1 or, 2h = 7 or, h = \(\frac{7}{5}\) = 3.5
Hence the value of h = 3.5 units.
Example 4. If each edge of a cube is increased by 50%, then how much the total surface area of the cube will be increased in per cent?
Solution:
Given:
If each edge of a cube is increased by 50%,
Let the side of the cube be a units.
∴ the area of the total surface area = 6a2 sq-units.
The side becomes when it is increased by 50% = \(\left(a+a \times \frac{50}{100}\right)\) units = \(\frac{3a}{2}\) units.
Then the area of the total surface = 6 x \(\left(\frac{3 a}{2}\right)^2\) sq-units = \(\frac{27 a^2}{2}\) sq-units
So, the increase in total surface area = \(=\left(\frac{27 a^2}{2}-6 a^2\right)\) sq-units = \(\frac{15 a^2}{2}\) sq-units.
∴ the total surface area of the cube will be increased in per cent = \(\frac{\frac{15 a^2}{2}}{6 a^2}\) x l00% = 125%
Hence the required increment = 125%.
Example 5. If a river of depthless 2 metres and breadth 45 metres, flows at a speed of 3 km per hour, then find the quantity of water that will fall from the river to the sea in 1 minute.
Solution:
Given:
If a river of depthless 2 metres and breadth 45 metres, flows at a speed of 3 km per hour
Speed of the river = 3 km per hour = \(\frac{3 \times 1000}{60}\) metres/minute = 50 metres/minute.
So, in 1 minute the quantity of water that will fall from the river to the sea = 2 x 45 x 50 cubic metres = 4500 cubic metres.
Hence the required quantity of water = 4500 m3.
“Understanding rectangular parallelepiped in Class 10 Maths”
Example 6. Find the volume of the water deposited in a land of a surface area of 2 hectres if in a day there rains 5 cm.
Solution:
The surface area of the land = 2 hectares = 2 x 10000 sq-metres
The height of the rain-fall = 5 cm = 0.05 metres.
So, the volume of the water deposited in the land = (2 x 10000 x 0.05) m3 = 1000 m3.
Hence the required volume of water = 1000 cubic-metres.
Example 7. If the areas of three consecutive surfaces of a parallelopiped-shape box be 50 sq-cm, 40 sq-cm and 20 sq-cm respectively. Find the volume of the box.
Solution:
Given:
If the areas of three consecutive surfaces of a parallelopiped-shape box be 50 sq-cm, 40 sq-cm and 20 sq-cm respectively.
Let the length, breadth and height of the box be x cm, y cm and z cm respectively.
As per question, xy = 50, yz = 40 and zx = 20.
∴ xy X yz X zx = 50 x 40 x 20
or, (xyz)2 = 50 x 2 x 20 x 20
or, (xyz)2 = 100 x 20 x 20
or, (xyz)2 = (10 x 20)2
⇒ xyz = 10 x 20 = 200.
Hence the volume of the box = 200 cc.
Example 8. The length, breadth and height of a rectangular parallelopiped type piece of wood are 15 cm, 12 cm and 20 cm respectively. To make a cube at least how many piece of such wood will be required?
Solution:
Given:
The length, breadth and height of a rectangular parallelopiped type piece of wood are 15 cm, 12 cm and 20 cm respectively.
The length of the least side of the cube = L.C.M. of (20 cm, 15 cm and 12 cm) = 60 cm.
Then the volume of the cube = 60 x 60 x 60 cc.
Again, the volume of each piece of wood = (20 x 15 x 12) cc
∴ the required piece of wood = \(\frac{60 \times 60 \times 60 c c}{20 \times 15 \times 12 c c}\) 60.
Hence at least 60 pieces of wood will be required to make a cube.
Example 9. By melting a large metal cube of side 4 cm each, some smaller metal cube of side 2 cm each are made. Find the total surface area of each smaller cube.
Solution:
Given:
By melting a large metal cube of side 4 cm each, some smaller metal cube of side 2 cm each are made
The volume of the large cube = 43 cc = 64 cc.
The volume of the smaller cube = 23 cc = 8 cc.
Hence the number of smaller cube = \(\frac{64 \mathrm{cc}}{8 \mathrm{cc}}\) = 8.
The total surface area of each smaller cube = 6 x 22 sq-cm = 24 sq-cm.
So, the total surface area of 8 cube = 24 x 8 sq-cm = 192 sq-cm.
Hence the required total surface area of each smaller cube =192 sq. cm.
Mensuration Chapter 1 Rectangular Parallelopiped Or Cuboid Long Answer Type Questions
“WBBSE Mensuration Chapter 1 cuboid practice questions”
Example 1. The inner length, breadth and height of a tea box are 7 dcm, 6 dcm and 5 dcm respectively. The weight of the box filled with tea is 42 kg 750 gms, but the weight of the empty box is 3.75 kg. Then what will be the weight of tea of volume 1 cu-dcm?
Solution:
Given:
The inner length, breadth and height of a tea box are 7 dcm, 6 dcm and 5 dcm respectively. The weight of the box filled with tea is 42 kg 750 gms, but the weight of the empty box is 3.75 kg.
Length of the tea-box = 7 dcm,
Breadth = 6 dcm and height = 5 dcm
∴ Volume of the tea-box = 7x6x5 cu-dcm = 210 cu-dcm.
The weight of the box full with tea = 45.750 kg and the weight of the empty box = 3.750 kg.
∴ the weight of only tea = (45.750 – 3.750) kg = 42 kg
∴ The weight of tea of volume 210 cu-dcm = 42 kg
∴ The weigth of tea of volume 1 cu-dcm = \(\frac{42}{210}\) kg = 0.2 kg = 200 gms.
Hence the required weight = 200 gms.
Example 2. The length of brass-plate of a square-shaped base is x cm, thickness 1 mm and weight 2100 gms. If the weight of 1 cu-cm brass is 8.4 gms, then what will be the value of x?
Solution:
Given:
The length of brass-plate of a square-shaped base is x cm, thickness 1 mm and weight 2100 gms. If the weight of 1 cu-cm brass is 8.4 gms
The area of the base of the brass-plate = x2 sq-cm.
Thickness = 1 mm = 0.1 cm.
∴ the volume of the plate = x2 x 0.1 cc = \(=\frac{x^2}{10}\) cc.
∴ the weight of 1 cc brass = 8.4 gms.
∴ the weight of \(=\frac{x^2}{10}\) cc brass = \(\frac{8 \cdot 4 \times x^2}{10}\) gms = \(\frac{84 \times x^2}{10 \times 10}\) gms = \(\frac{84 x^2}{100}\) gms.
As per question, \(\frac{84 \times x^2}{10 \times 10}\) = 2100 or, 84x2 = 210000
or, x2 = \(\frac{210000}{84}\) or, x2 = 2500 or, x = √2500 = 50.
Hence the value of x is 50.
Example 3. The height of a expressway is to be raised, so, 20 cuboidal holes with equal depth and of equal measure are dug out on both sides of the way and with this soil the way is elevated. If the length and breadth of each hole are 14 m and 8 m respectively and if the total quantity of soil required to make the way be 1680 cubic metre, then calculate the depth of each hole.
Solution:
Given:
The height of a expressway is to be raised, so, 20 cuboidal holes with equal depth and of equal measure are dug out on both sides of the way and with this soil the way is elevated. If the length and breadth of each hole are 14 m and 8 m respectively and if the total quantity of soil required to make the way be 1680 cubic metre
The length and breadth of each hole is 14 metres and 8 metres respectively.
Let the depth of each whole be x metres.
∴ the volume of each hole = 14 X 8 X x cubic-metres = 112 x cu-metres.
∴ the volume of 20 holes = 112x X 20 cu-metres.
As per question, 112x X 20 = 1680 or, x = \(\frac{1680}{112 \times 20}\) =0.75
Hence the depth of each-hole = 0.75 metres.
Example 4. If 32 water filled buckets of equal measure are taken out from a cubical water filled tank, then 1/3rd of water remains in the tank. If the length of one edge of the tank is 2.4 m, then calculate the quantity of water that can be hold in each bucket?
Solution:
Given:
If 32 water filled buckets of equal measure are taken out from a cubical water filled tank, then 1/3rd of water remains in the tank. If the length of one edge of the tank is 2.4 m
The length of one edge of the tank = 2.4 metres = 24 dcm.
∴ the volume of the whole tank = (24 x 24 x 24) cu-dcm = 13824 cu-dcm.
The tank remains full of water 1/3 rd part of it.
∴ the part from which water has been taken out is \(\left(1-\frac{1}{3}\right)\) part = \(\frac{2}{3}\)
So, the volume of the water drawn = \(\left(13824 \times \frac{2}{3}\right)\) cu-dcm = 9216 cu-dcm = 9216 litres.
So, 9216 litres water can be hold in 32 buckets.
∴ 1 bucket can be hold \(\frac{9216}{32}\) litres of water = 288 litres of water.
Hence 288 litres of water can be hold in each bucket.
Example 5. The length, breadth and height of a rectangular parallelopiped room are 5 m, 4 m and 3 m respectively. Find the length of the longest rod that can be put into that room.
Solution:
Given:
The length, breadth and height of a rectangular parallelopiped room are 5 m, 4 m and 3 m respectively.
The length breadth and height of the room are 5 m, 4 m and 3 m respectively.
So, the length of the diagonals of the room
= \(\sqrt{5^2+4^2+3^2}\) metres = \(\sqrt{25+16+9}\) metres = 50 metres = 25 √2 metres
Hence the required length of the longest rod = 5√2 metres.
Example 6. A canal of breadth 2 m and depth 8 dcm have been drugged in our village. If the volume of the soil extrated be 288 cubic-metres, then find the length of the canal.
Solution:
Given:
A canal of breadth 2 m and depth 8 dcm have been drugged in our village. If the volume of the soil extrated be 288 cubic-metres
Let the length of the canal be x metres.
The breadth of the canal = 2 metres and the depth of the canal = 8 dcm = 0.8 metre
∴ the volume of the canal = x x 2 x 0.8 cubic-metres = 1.6x cubic-metres.
Hence the length of the canal = 180 metres.
Example 7. The sum of the areas of six surfaces of a cube is 216 sq-cm, find the volume of the cube.
Solution:
Given:
The sum of the areas of six surfaces of a cube is 216 sq-cm
Let the length of each side of the cube be a cm.
∴ the total surface areas of the cube = 6a2 sq-cm
As per question, 6a2 = 216 or, a2 = \(\frac{216}{6}\) = 36
⇒ a = √36 = 6
∴ the volume of the cube = 63 cc. = 216 cc.
Hence the volume of the cube = 216 cc.
Example 8. The volume of a rectangular parallelopiped is 432 cc. If it is divided into two cubes of equal volume, then find the length of each side of the cubes.
Solution:
Given:
The volume of a rectangular parallelopiped is 432 cc. If it is divided into two cubes of equal volume
Since the volumes of the two cubes are equal, so their length of sides will also be equal.
Let the length of each side of the cubes be a cm.
∴ the volume of each cdbe = a3 cc.
As per question, 2a3 = 432 or, a3 = 216 = 63 ⇒ a = 6.
Hence the length of each side of the two equal cubes is 6 cm.
Example 9. Each side of a cube is decreased by 50%. Then what will be the ratio of volumes of the original cube and the decreased cube?
Solution:
Given:
Each side of a cube is decreased by 50%.
Let the length of each side of the cube be a units, the volume of the cube = a3 cubic-units
After decreasing the length of each side by 50%, the length of each sides becomes \(\left(a-a \times \frac{50}{100}\right)\) units = \(\frac{a}{2}\) units.
∴ the volume of the decreased cube = \(\left(\frac{a}{2}\right)^3\) cubic-units = \(\frac{a^3}{8}\) cubic-units.
∴ (volume of the original cube) : (volume of the decreased cube)
= a3 : \(\frac{a^3}{8}\) = 1 : \(\frac{1}{8}\) = 8 : 1.
Hence the required ratio = 8:1.
“Step-by-step solutions for cuboid problems Class 10”
Example 10. If the length, breadth and height of one packet of one gross match-box are 2.4 dem, 1.2 dem and 0.9 dem respectively, then calculate the volume of one match-box. [one gross = 12 dozen]. But if the length and breadth of one match box be 3 cm and 2.5 cm, then also calculate the height of it.
Solution:
Given:
If the length, breadth and height of one packet of one gross match-box are 2.4 dem, 1.2 dem and 0.9 dem respectively, then calculate the volume of one match-box. [one gross = 12 dozen]. But if the length and breadth of one match box be 3 cm and 2.5 cm
1 gross =12 dozen = 144 match-box.
The length of one gross of match-box = 2.4 dcm = 24 cm.
breadth = 1-2 dem = 12 cm and height = 0.9 dcm = 9 cm.
∴ volume of one gross match-box = 24 x 12×9 cu-cm. = 2592 cc.
∴ volume of one match-box = \(\frac{2592}{144}\) cc = 18 cc.
Length of each matchbox = 3 cm, breadth = 2.5 cm.
Let the height of each matchbox be x cm.
So, the volume of each matchbox = 3 X 2.5 x x cc = 7.5x cc.
∴ volume of 144 match-boxes = 7.5x X 144 cc.
∴ 7.5x X 144 = 3780 or, x = \(\frac{2592}{7.5 \times 144}\) = 2.4
Hence the volume of one match-box is 18 cc and the height of each match-box is 2.4 cm.
Example 11. Half of a cuboidal water tank with length of 2.2 m and breadth of 1.1 m is filled with water. If 484 litres water is poured more into the tank, then calculate the depth of water that will be increased by.
Solution:
Given:
Half of a cuboidal water tank with length of 2.2 m and breadth of 1.1 m is filled with water. If 484 litres water is poured more into the tank
The length of the tank = 2.2 m = 22 dcm.
breadth =11 metres = 11 dcm.
Water poured into the tank = 484 litres = 484 cubic-dcm.
Let the depth of water of the tank after pouring water to it be increased by x dcm.
∴ 22 x 11 x x = 484 or, x = \(\frac{484}{22 \times 11}\) = 2.
Hence the depth of water of the tank will be increased by 2 dcm.
Example 12. There were 800 litres, 725 litres and 575 litres kerosine oil in three drums in a house. The oil of these three drums is poured into a cuboidal pot and for this, the depth of oil in drums becomes 7 dcm. If the ratio of the length and breadth of the cuboidal pot is 4 : 3, then calculate the length and breadth of the pot. If the depth of the cuboidal pot would be 5 dcm, calculate whether 1620 litre oil can be kept pr not in that pot.
Solution:
Given:
There were 800 litres, 725 litres and 575 litres kerosine oil in three drums in a house. The oil of these three drums is poured into a cuboidal pot and for this, the depth of oil in drums becomes 7 dcm. If the ratio of the length and breadth of the cuboidal pot is 4 : 3, then calculate the length and breadth of the pot. If the depth of the cuboidal pot would be 5 dcm,
The total quantity of oil in the three drums = (800 + 725 + 575) litres = 2100 litres = 2100 cu-dcm.
Let the length of the cuboidal pot be 4x dcm and its breadth be 3x dcm.
The depth of oil = 7 dcm.
∴ the volume of oil in the cuboidal pot = 4x X 3x X 7 cu-dcm
As per question, 4x X 3x X 7 = 2100 or, x2 = \(=\frac{2100}{4 \times 3 \times 7}\) or, x2 = 25 or, x = 5
∴ the length of the pot = 4×5 dcm = 20 dcm and breadth of the pot = 3 x 5 dcm =15 dcm.
Now, the volume of the cuboidal pot = 20 x 15 x 5 cu-dcm = 1500 cu-dcm = 1500 litres.
Hence if the depth of the cuboidal pot be 5 dcm, then 1620 litres oil can not be kept in that pot.
Example 13. The length and breadth of a rectangular field are 22 m and 18 m respectively. For construction of pillars in the 4 corners of that field 5 cubic holes having length of 4 m are dug out and the soils removed are dispersed on the remaining land. Calculate the height of the surfaces of the field that is increased by it.
Solution:
Given :
The length and breadth of a rectangular field are 22 m and 18 m respectively. For construction of pillars in the 4 corners of that field 5 cubic holes having length of 4 m are dug out and the soils removed are dispersed on the remaining land.
The volume of soil of one hole = 53 cu-m. = 125 cu-m.
∴ the volume of soil of 4 holes = 125 x 4 cu-m = 500 cu-m.
Let the height of the surface of the field be increased by h metres.
∴ the volume- of the soil thus increased = 22 X 18 X h cu-metres
As per condition, 22 X 18 X h = 500 or, h = \(\frac{500}{22 \times 18}\) = 1.26 (approx.)
Hence the height of the surface of the field is increased by 1.26 metre (approx.).
Example 14. The daily requirements of water of three families in our three-storyed flat are 1200 litres, 1050 litres and 950 litres respectively. After fulfilling these requirements in order to put up a tank again and to deposit to store 25% of the required water, only a land having the length of 2.5 m and breadth of 1.6 m has been procured. Calculate the depth of the tank in metre that should be made. If the breadth of the land would be more by 4 dcm, then calculate the depth of the tank to be made.
Solution:
Given:
The daily requirements of water of three families in our three-storyed flat are 1200 litres, 1050 litres and 950 litres respectively. After fulfilling these requirements in order to put up a tank again and to deposit to store 25% of the required water, only a land having the length of 2.5 m and breadth of 1.6 m has been procured.
The total requirements of water of three families = (1200 + 1050 + 950) litres = 3200 litres.
As per question, the quantity of water to be stored
= \(\left(3200+3200 \times \frac{25}{100}\right)\)litres = (3200 + 800) litres = 4000 litres = 4000 cu-dcm.
The length of the tank = 2.5 m = 25 dcm
Breadth = 1.6 m = 16 dcm
Let the depth of the tank = x m = 10 x dcm.
the volume of the tank = 10x X 25 x 16 cu-dcm = 4000x cu-dcm
As per condition, 4000 x = 4000
or, x = \(\frac{4000}{4000}\) = 1.
∴ the depth of the tank = 1 m.
If the breadth be 4 dcm more then let the depth of it be y dcm.
∴ 25 X (16 + 4) X v = 4000 or, 25 X 20 X y = 4000 or, y = \(=\frac{4000}{25 \times 20}\) = 8
Hence in the first case, the depth of the tank will be 1 metre and in the second case the depth of the tank will be 8 dcm
Example 15. The weight of a wooden box made of wooden planks with the thickness of 5 cm along with its covering is 115.5 kg. But the weight of the box filled with rice is 880.5 kg. The length and breadth of inner side of the box are 12 dcm and 8.5 dcm respectively and the weight of 1 cubic dcm rice is 1.5 kg. Determine the inner height of the box. Also determine the total expenditure to colour the outside of the box, if the rate is ₹ 1.50 per sq-dcm.
Solution:
Given:
The weight of a wooden box made of wooden planks with the thickness of 5 cm along with its covering is 115.5 kg. But the weight of the box filled with rice is 880.5 kg. The length and breadth of inner side of the box are 12 dcm and 8.5 dcm respectively and the weight of 1 cubic dcm rice is 1.5 kg.
The weight of the box filled with rice = 880.5 kg
The weight of the empty box = 115.5 kg.
∴ the weight of rice = (880.5 – 115.5) kg = 765 kg
The volume of rice of 1.5 kg = 1 cu-dcm
∴ The volume of rice of 1 kg = \(\frac{1}{1 \cdot 5}\) cu-dcm
The volume of rice of 765 kg = \(\frac{1}{1 \cdot 5}\) x 765 cu-dcm = 510 cu-dcm
Let the inner height of the box be x dcm.
∴ the inner volume of the box = 12 X 8.5 X x cu-dcm
∴ 12 X 8.5 X x = 510 ⇒ \(x=\frac{510}{12 \times 8 \cdot 5}\) = 5
∴ the inner height of the box = 5 dcm.
The thickness of the wooden plank = 5 cm = 0.5 dcm
⇔ The outer length of the box = (12 + 2 x 0.5) dcm = 13 dcm
Breadth = (8.5 + 2 x 0.5) dcm = 9.5 dcm
Height = (5 + 2 x 0.5) dcm = 6 dcm
So, the total surface area of the outside of the box = 2 (13 x 9.5 + 9.5 x 6 + 13 x 6) sq-dcm = 2 (123.5 + 57 + 78) sq-dcm.
= 2 x 258.5 sq-dcm = 517 sq-dcm.
So, the required expenditure = ₹ 517 x 1.50 = ₹ 775.50
Hence the inner height of the box is 5 dcm and the required expenditure = ₹ 775.50
Example 16. The depth of a cuboidal pond with length of 20 m and breadth of 18.5 m is 3.2 m, determine the time required to irrigate whole water of the pond with a pump having the capacity to irrigate 160 kilolitres water per hour. If that quantity of water is poured on a paddy field with a ridge having the length of 59.2 m and breadth of 40 m, then what is the depth of water in that land? [Let 1 cubic metre = 1 kilolitre]
Solution:
Given:
The depth of a cuboidal pond with length of 20 m and breadth of 18.5 m is 3.2 m, determine the time required to irrigate whole water of the pond with a pump having the capacity to irrigate 160 kilolitres water per hour. If that quantity of water is poured on a paddy field with a ridge having the length of 59.2 m and breadth of 40 m
1 kilolitre = 1 cubic metre
∴ 160 kilolitre = 160 cubic metre
The length of the pond = 20 metres, breadth = 18.5 metres and depth of water in the pond = 3.2 metres.
∴ the volume of water = (20 x 18.5 x 3.2) cu-metres – 1184 cu-metres.
The pump irrigate 160 cu-metres in 60 minutes
∴ The pump irrigate 1 cu-metres in \(\frac{60}{160}\) minutes
The pump irrigate 1184 cu-metres in \(\frac{60 \times 1184}{160}\) minutes
= 444 minutes = 7 hours 24 minutes.
Again, let the depth of water after pouring water in the paddy field be x metre.
Then, the volume of water in the field = 59.2 X 40 X x cubic-metres.
As per question 59.2 X 40 X x = 1184
or, x = \(x=\frac{1184}{59 \cdot 2 \times 40}=0 \cdot 5\) = 0.5
Hence the required time to irrigate the water of the pond is 7 hours and 24 minutes and the depth of water in the paddy field after pouring the water of the pond in it will be 0.5 metre = 5 dcm.
Example 17. 8 wooden rectangular parallelopiped of same shape when arranged one over another, the volume becomes 128 cc. If the base of each rectangular parallelopiped be square and of height 1 cm, then find the length, breadth and height of each wooden rectangular parallelopiped.
Solution:
Given:
8 wooden rectangular parallelopiped of same shape when arranged one over another, the volume becomes 128 cc. If the base of each rectangular parallelopiped be square and of height 1 cm
Let the length and breadth of each wooden rectangular parallelopiped be x cm.
Since there are 8 such parallelopipeds of height 1 cm each have been arranged one over another, the total height of this soild object is 8 cm.
So, the volume of whole solid object = x X x X 8 cc = 8x2 cc.
As per the question, 8x2 = 128
or, x2 = \(\frac{128}{8}\) = 16 or, x = √16 or, x = 4
Hence the length, breadth and height of each of the cuboidal solid are 4 cm, 4 cm and 1 cm respectively.
Example 18. The length and breadth of a rectangular field are 154 m and 121 m respectively. If by digging a hole of length 14 m and breadth 11 m in the middle of the field, the soil thus obtained is dispersed in the remaining land, then the height of the field is increased by 25 cm. Determine the depth of the hole.
Solution:
Given:
The length and breadth of a rectangular field are 154 m and 121 m respectively. If by digging a hole of length 14 m and breadth 11 m in the middle of the field, the soil thus obtained is dispersed in the remaining land, then the height of the field is increased by 25 cm.
The total surface area of the field = 154 x 121 sq-metres = 18634 sq-metres.
Again, the surface area of the hole = (14 x 11) sq-cm =154 sq-cm.
So, the surface area of the remaining field = (18634 – 154) sq-cm = 18480 sq-cm.
The height of the field is raised by 25 cm. = \(\frac{25}{100}\) m = \(\frac{1}{4}\) m.
So, the volume of the dispersed soil = \(\left(18480 \times \frac{1}{4}\right)\) cu-m = 4620 cu-m
Let the depth of the hole be x m.
∴ 14 X 11 X x = 4620 or, x = \(\frac{4620}{14 \times 11}\) = 30
Hence the depth of the hole was 30 m.
Example 19. If two cubes of sides 10 cm each be joined side by side, then what will be the total surface area of the produced cuboid?
Solution:
Given:
If two cubes of sides 10 cm each be joined side by side,
The length of the produced cuboid = (10 + 10) cm = 20 cm and the breadth of it remains 10 cm.
Obviously, height of the cuboid = 10 cm.
So, the total surface area of the cuboid = 2 (20 X 10 + 10 X 10 + 10 X 20) sq-cm
= 2 X (200 + 100 + 200) sq-cm = 2 X 500 sq-cm = 1000 sq-cm
Hence the required total surface area = 1000 sq-cm
Example 20. If the sum of length, breadth and height of a cuboid be 19 cm and the length of its diagonal be 11 cm, then what will be the total surface area of the cuboid?
Solution:
Given:
If the sum of length, breadth and height of a cuboid be 19 cm and the length of its diagonal be 11 cm
Let the length, breadth and height of the cuboid be a cm, b cm and c cm respectively.
As per question, a + b + c = 19 and \(\sqrt{a^2+b^2+c^2}\) = 11 or, a2 + b2 + c2 = 121
Now, (a + b + c)2 = 192 = 361
⇒ a2 + b2 + c2 + 2 (ab + bc + ca) = 361
⇒ 121 +2 (ab + bc + ca) = 361 [a2 + b2 + c2 = 121]
⇒ 2 (ab + bc + ca) = 381 – 121 = 240.
Hence the required total surface area of the cuboid is 240 sq-cm.
Example 21. To make a wall of length 6 metres, of height 5 metres and of thickness 0-5 metre, how many bricks of size 25 cm x 12.5 x 7.5 cm will be required when 1/20 part of the wall will be made of a mixture of sand and cement?
Solution:
Given:
To make a wall of length 6 metres, of height 5 metres and of thickness 0-5 metre,
The volume of the wall = 6 X 5 X 0.5 cubic-metres = 15 cu-metres.
Also, volume of each brick = 25 X 12.5 X 7.5 cc. = 0.25 X 0.125 X 0.075 cu-metres
= \(\frac{25}{100} \times \frac{125}{1000} \times \frac{75}{1000}\) cu-metres = \(\frac{3}{1280}\) cu-metres
Let number of bricks to be required is x.
As per question, x X \(\frac{3}{1280}\) = (1- \(\frac{1}{20}\)) X 15
or, x X \(\frac{3}{1280}\) = \(\frac{19}{20}\) X 15
or, x = \(\frac{19}{20}\) X 15 x \(\frac{3}{1280}\) or, x = 6080
Hence 6080 bricks will be required.
Example 22. The length and breadth of a cuboid-shaped piece of wood are 2.3 metres and 0.75 metres respectively. If the volume of the wooden piece be 1.104 cubic-metres, then how many piece of wood of size 2.3 m x 0.75 m x 0.04 m can be cut off from this piece of wood?
Solution:
Given:
The length and breadth of a cuboid-shaped piece of wood are 2.3 metres and 0.75 metres respectively. If the volume of the wooden piece be 1.104 cubic-metres,
Let the thickness of the wooden piece by x metres.
Its volume = 1.104 cu-metres.
As per question, 2.3 x 0.75 X x = 1.104
⇒ x = \(\frac{1.104}{2.3 \times 0.75}=\frac{10 \times 1104 \times 100}{23 \times 75 \times 1000}\) = 0.64
∴ the thickness of the wooden piece = 0.64 m.
Let the number of such piece of wood be y.
∴ 2.3 x 0.75 x 0.04 x y = 2.3 x 0.75 x 0.64
⇒ y = \(\frac{2.3 \times 0.75 \times 0.64}{2.3 \times 0.75 \times 0.04}\) = 16
Hence the required number of piece of wood = 16.
Example 23. The length of a cuboidal tank is 20 metres. The height of the water-level in the tank decreases by 15 cm when 18 kilolitres water is taken out from the tank. Then find the breadth of the tank.
Solution:
Given:
The length of a cuboidal tank is 20 metres. The height of the water-level in the tank decreases by 15 cm when 18 kilolitres water is taken out from the tank.
18 kilolitres = 18 cubic-metres.
The height of water level in the tank decreases by 15 cm = 0.15 metre.
Let the breadth of the tank be x metres,
∴ 20 X x X 0.15 =18 or, x = \(\frac{18}{20 \times 0 \cdot 15}\) = 6
Hence the breadth of the tank = 6 metres.
Example 24. A solid cube is divided into two cuboidal objects of equal volumes, then what will be the ratio of the total surface area of the cube and the surface area of each cuboidal object?
Solution:
Given:
A solid cube is divided into two cuboidal objects of equal volumes
Let the length of each side of the cube be a unit.
Since the cube is divided into two cuboidal object, so the length of the cuboidal object will be a unit, breadth a unit and height \(\frac{a}{2}\) unit.
Now, the total surface area of the cube = 6a2 sq-units and the total surface area of each cuboidal object
= \(\) sq-units = \(\) sq-units
= \(\) sq-units = \(\) sq-units = 4a2 sq-units
So, (the total surface area of the cube) : (the total surface area of each cuboidal object) = 6a2: 4a2 = 3:2
Hence the required ratio = 3:2.