Mensuration
Class 8th Mathematics CBSE Solution
- A square and a rectangular field with measurements as given in the figure have…
- Mrs. Kaushik has a square plot with the measurement as shown in the figure. She…
- The shape of a garden is rectangular in the middle and semi circular at the…
- A flooring tile has the shape of a parallelogram whose base is 24 cm and the…
- An ant is moving around a few food pieces of different shapes scattered on the…
- The shape of the top surface of a table is a trapezium. Find its area if its…
- The area of a trapezium is 34 cm^2 and the length of one of the parallel sides…
- Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD…
- The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars…
- The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
- Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If…
- The floor of a building consists of 3000 tiles which are rhombus shaped and…
- Mohan wants to buy a trapezium shaped field. Its side along the river is…
- Top surface of a raised platform is in the shape of a regular octagon as shown…
- There is a pentagonal shaped park as shown in the figure. For finding its area…
- Diagram of the adjacent picture frame has outer dimensions =24 cm 28 cm and…
- There are two cuboidal boxes as shown in the adjoining figure. Which box…
- A suitcase with measures 80 cm 48 cm 24 cm is to be covered with a tarpaulin…
- Find the side of a cube whose surface area is 600 cm^2 .
- Rukhsar painted the outside of the cabinet of measure 1 m 2 m 1.5 m. How much…
- Daniel is painting the walls and ceiling of a cuboidal hall with length,…
- Describe how the two figures at the right are alike and how they are different.…
- A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of…
- The lateral surface area of a hollow cylinder is 4224 cm^2 . It is cut along…
- A road roller takes 750 complete revolutions to move once over to level a road.…
- A company packages its milk powder in cylindrical container whose base has a…
- Given a cylindrical tank, in which situation will you find surface area and in…
- Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B…
- Find the height of a cuboid whose base area is 180 cm^2 and volume is 900 cm^3…
- A cuboid is of dimensions 60 cm 54 cm 30 cm. How many small cubes with side6 cm…
- Find the height of the cylinder whose volume is 1.54 m^3 and diameter of the…
- A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m.…
- If each edge of a cube is doubled, (i) how many times will its surface area…
- Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute.…
Exercises 11.1
Question 1.A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Answer:
Given:
Length of rectangle = l = 80 cm
side of square = a = 60 cm
Also, we know
Perimeter of rectangle = 2(l + b)
Area of rectangle = lb,
where l = length and b = breadth
Perimeter of square = 4a
area of square = a2 ,
where a is side of square.
Let breadth of the rectangle be b
∵ Perimeter of the rectangle = Perimeter of the square
∴ 2 × (80 + b) = 4 × 60
2 × (80 + b) = 240
(80 + b) = 240/2
80 + b = 120
b = 120 - 80
⇒ b = 40 m
Now, area of square = a2 = 602 = 3600 m2
And area of the rectangle = lb = 80 × 40 = 3200 m2
Clearly, area of square is greater.
Question 2.
Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs 55 per m2.
Answer:
Area of the square plot = Side2 = 252 = 625 square metre
Area of the house construction part = length x breadth
= 20 x 15 = 300 sq m
Area of garden = 252 – 15 × 20
= 625 – 300
= 325 m2
Thus, total cost of developing a garden around the house at the rate of Rs 55 per m2 = 325 × 55
= Rs 17875
Question 3.
The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].
Answer:
Here, length of rectangle = 20 – (3.5 + 3.5) = 13 m, breadth = diameter of semi circle = 7 m
Diameter of Semi Circle = 7 m, i.e. radius r = 7/2 m = 3.5 m
Area of the garden = sum of areas two semicircular shapes + area of rectangle
Area of the garden = π r2 + Length × Breadth
[putting the value of π = 22/7 ]
Area of the garden = 22 × 0.5 × 3.5 + 13 × 7
Area of the garden = 38.5 + 91
Area of the garden = 129.5 m2
And perimeter = Total Length of Outer Boundary = Circumference of Circle + 2 × Length of rectangle
Perimeter of the garden = 2 π r + 2 l
Perimeter of the garden = 22 + 26
Perimeter of the garden = 48 m
Question 4.
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).
Answer: Given,
tile is in the shape of parallelogram with base = 24 cm and height = 10 cm
We know,
area of parallelogram = Base × Height
Therefore,
area of tile = 24(10) = 240 cm2
Now, area of floor = 1080 m2
(we know that, 1m = 100cm, therefore, 1m2 = 10000cm2)
Therefore,
= 1080 × (100)2 cm2
= 10800000 cm2
No of tiles required
Question 5.
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression C = 2πr, where r is the radius of the circle.
A. 
B. 
C. 
Answer:
Food pieces with larger perimeter would make the ant to take the longer round.
Perimeter for Shape A:
Since, Shape A is Semi Circle.
Also diameter is the part of the figure.
Perimeter = πr + Diameter
= πr + 2r
= (22/7) × 1.4 + 2.8
= 7.2 cm
Perimeter for Shape B:
Perimeter = Sum of length of all the outer boundary
= 2.8 + 1.5 + 1.5 + Perimeter of semi circular part
= 2.8 + 1.5 + 1.5 + πr
= 2.8 + 1.5 + 1.5 + (22/7) × 1.4
= 10.2 cm
Perimeter for Shape C:
Perimeter = Sum of length of all the outer boundary
Perimeter = 2 + 2 + perimeter of semi circular part
Radius of semicircle = d/2
= 2.8/2
= 1.4 cm
= 2 + 2 + πr
Perimeter = 2 + 2 + (22/7) × 1.4
= 8.4 cm
∴ Perimeter is maximum for shape B, So Ant would have to take the longer round for Shape B.
Exercises 11.2
Question 1.The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.
Answer:
Area of trapezium = 1/2 (a + b) × h
Here, a = 1.2m, b = 1m, h = 0.8m, thus
Area = 1/2 × (1.2 + 1) × 0.8
Area = 0.5 × 2.2 × 0.8
Area = 1.1 × 0.8
Area = 0.11 × 8
= 0.88 m2
Question 2.
The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Answer:
Area of Trapezium = 1/2 × (Sum of Length of Parallel Side) × Distance Between Them
∴ Area of trapezium = 1/2 × (a + b) × h = 34 cm2
Here, a = 10cm, h = 4cm, b = length of another parallel Side
∴ 1/2 × (10 + b) × 4 = 34
(10 + b) × 2 = 34
20 + 2b = 34
2b = 34 - 20
2b = 14
b = 7 cm
Question 3.
Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC
Answer:
If a perpendicular from D to BC is drawn then it will be equal to AB,
then EC = 48-40 = 8 m.
In ΔDEC,
By Pythagoras theorem
DE2 = DC2-EC2
DE2 = 172 - 82 = 289 - 64 = 225
Or, DE = 15 m
Area of Trapezium = 1/2 × (sum of parallel sides) × height
= 1/2 × (48 + 40) × 15
= 660 m2
Question 4.
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.
Answer:
Area of upper triangle = 1/2 × height × base
= 1/2 × 13 × 24
= 156 m2
Area of lower triangle = 1/2 × height × base
= 1/2 × 8 × 24
= 96 m2
Area of the field = 156 + 96 = 252 m2
Question 5.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Answer:
Formula: Area of Rhombus = 1/2 × product of diagonals
Given: Diagonals are 7.5cm and 12cmApplying the formula, we get,
= 1/2 × 7.5 × 12
= 45 cm2
Question 6.
Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Answer:
The diagram is given below:
Now,
Area of rhombus = Base × Altitude
= CD × BF
= 5 cm × 4.8 cm
= 24.0 cm2
Also, Area of rhombus = 1/2 × AD× BC
⇒ 24 = 1/2 × 8 × BC
⇒ BC = 6 cm
Hence, the length of the other diagonal is 6 cm.
Question 7.
The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is Rs 4
Answer:
Here, d’ = 45 cm and d’’= 30 cm
∵ Area of one tile = area of rhombus= 1/2 (d’ × d’’)
= 1/2 × 45 × 30
= 675 cm2
∴ Area of 3000 tiles= 675 × 3000 = 2025000 
= 202.50 m2 ----------(∵ 1m2 = 10000cm2)
∵ Cost of polishing the floor per sq. meter = 4
∴ Cost of polishing the floor per 202.50 sq. meter = 4 × 202.50 = 810
Hence the total cost of polishing the floor is Rs810
Question 8.
Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.
Answer:
Given: Perpendicular distance(h) = 100 m
Area of the trapezium shaped field
= 10500 m2
Let side along the road be x m then, side along the river = 2 x m
Area of the trapezium field
= 1/2× (sum of parallel sides)h
⇒ 10500 = 1/2 × (x + 2x) × 100
⇒ 10500 = 1/2 × 3x × 100
⇒x =
= 70m
Hence the side along the river = 2x
= 2 × 70 = 140 m.
Question 9.
Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Answer:
Given: Octagon having eight equal sides, each 5 m
Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively.
Now Area of two trapeziums = 2 x 1/2 × (a + b) h
= (11 + 5) × 4
= 64 sq. m
And Area of rectangle = length × breadth
= 11 × 5
= 55 sq. m
∴ Total area of octagon = 64 + 55
= 119 sq. m
Question 10.
There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area?
Answer:
By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
we know area of trapezium = 1/2 × (Sum of parallel lines) × (Distance between parallel lines)= 1/2 (AP + BC) × CP + 1/2 (ED + AP) × DP = 1/2 (30 + 15 ) × CP + 1/2 (15 + 30) × DP
= 1/2 (30 + 15) (CP + DP)
= 1/2 × 45 × CD
= 1/2 × 45 × 15
=337.5 m2
By Kavita's diagram,
Area of pentagon = Area of square ABCD + Area of triangle ABP
we know,
area of square = (side)2 = side × side
area of triangle = (1/2) × base x Height
= AB×AB + 1/2(AB×h)
= 15×15 + ½(15×15)
= 225 + 225/2
=225+112.5 = 337.5 m2
Question 11.
Diagram of the adjacent picture frame has outer dimensions =24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.
Answer:
Here two of given figures (I) and (II) are similar in dimensions.
we know area of trapezium = (1/2) × (sum of parallel sides) × height
And also figures (III) and (IV) are similar in dimensions.
For trapezium I, 2 h = 24 - 16, 2 h = 8, h = 4
Area of figure (I) = Area of trapezium
= 1/2 (a + b) h
= 1/2 (28 + 20) × 4
= 96 cm2
Also Area of figure (II) = 96 cm2
Now Area of figure (III)
= Area of trapezium = = 1/2 (24 + 16) × 4
= 80 cm2
Also Area of figure (IV) = 80 cm2
Exercises 11.3
Question 1.There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Answer:
(a) Given: Length of cuboidal box (l) = 60 cm
Breadth of cuboidal box (b) = 40 cm
Height of cuboidal box (h) = 50 cm
∴ Total surface area of cuboidal box = 2 (lb + bh + hl)
= 2 (60 × 40 + 40 × 50 + 50 × 60)
= 2 (2400 + 2000 + 3000)
= 2 × 7400 = 14800 cm2
(b) Given: Edge of cubical box (l) = 50 cm
Total surface area of cubical box= 6l2
= 6 × 502
= 6 × 2500
= 15000 cm2
Hence cuboidal box (a) requires the lesser amount of material to make since surface area of box (a) is less than that of box (b).
Question 2.
A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Answer:
Length of suitcase = 80 cm
Breadth of suitcase = 48 cm
Height of suitcase = 24 cm
Surface area of cuboid = 2(lb + bh + hl)
= 2(80 × 48 + 80 × 24 + 48 × 24)
= 13824 cm2
Area of cloth = Area of suitcase.
Hence surface area of 100 such suitcases = 1382400 cm2
Required tarpaulin = surface area of 100 such suitcases
So, required length = 
= 144 cm
Required tarpaulin for 100 suitcases = 144 x 100 = 14400 cm = 144 m
Hence, the tarpaulin cloth required to cover 100 suitcases is 144 m.
Question 3.
Find the side of a cube whose surface area is 600 cm2.
Answer:
Surface Area = 6 x side2
⇒ 600 = 6 x side2
⇒ Side2 = 100
⇒ Side = 10 cm
Question 4.
Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet.
Answer:
Given,
length, l = 1 m
Breadth, b = 2 m
Height, h = 1.5 m
Surface Area of Cabinet = 2(lb + lh + bh)
= 2(1 x 2 + 1 x 1.5 + 1.5 x 2)
= 13 m2
Area of Bottom Surface = lb = 1 x 2 = 2 m2
Area covered by painting = Area of cabinet - Area of bottom surfaceTherefore,
Hence, Area Covered by painting = 13 – 2 = 11 m2
Question 5.
Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m2 of area is painted. How many cans of paint will she need to paint the room?
Answer:
Let length, breadth and height of room be l, b and h respectively
Given,
l = 15 m
b = 10 m
h = 7 m
and we know, area of four walls = 2(l + b)h
Hence,
Area to be painted = Area of four Walls of a Room + area of ceiling
= 2(l + b)h + lb
= 2(15 + 10)7 + 15(10)
= 14(25) + 150
= 350 + 150
= 500 m2
As, one can of paint is enough for 100 sq m
So, 500/100 = 5 cans are needed to paint the hall.
Question 6.
Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?
Answer:
The figure is shown:
Similarity: Their dimensions are equal, i.e. 7 cm width, breadth and height
Difference: The figure on the left is cylindrical and that on the right is cuboidal.
So, surface area of cylinder = 2 π rh = 2 × 
× 7
= 154 sq. cm
And surface area of the cube = 4 × side2
= 4 × 72
= 196 sq. cm
Clearly, the cube has a larger surface area than the cylinder.
Question 7.
A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Answer:
Given: Radius of cylindrical tank, r = 7m
Height of cylindrical tank, h =3m
The figure of the question is:
The total surface area of cylindrical tank, S = 2 π r( h + r) ............(1)
Putting the values of r and h in equation (1), we get,
S = 2 ×(22/7)× 7(7 + 3)
S = 2 × (22/7) × 7(10)
S = 2 × 22× 10
S = 440 m2
The total surface area of the cylinder is 440 m2
∴ 440 m2 of sheet of metal is required.
Question 8.
The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Answer:
Area of Rectangular Sheet = Surface Area of the Cylinder
Area of Rectangle = l x b
⇒ l x b = 4224
⇒ l x 33 = 4224
⇒ l = 128 cm
Perimeter of rectangular sheet = 2( l + b)
= 2 (128 + 33)
= 2 x 161 = 322 cm
Hence perimeter of rectangular sheet is 322 cm.
Question 9.
A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.
Answer:
Given: Diameter of road roller = 84 cm
∴ Radius of road roller = 42 cm
Length of road roller, h = 1 m = 100 cm
Curved surface area of road roller = 2 π r h
= 2 × 22/7 × 42 × 100
= 26400 cm2
Area covered by road roller in 750 revolutions = 26400 × 750
= 1,98,00,000 cm2 or 1980 m2
∴ the area of the road is 1980 sq. m.
Question 10.
A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label.
Answer:
Given: Diameter of cylindrical container = 14 cm
∴ Radius of cylindrical container, r = 7 cm
Height of cylindrical container = 20 cm
Height of the label, h = 20 – 2 – 2
= 16 cm
Curved surface area of label = 2 π r h
= 2 × (22/7) × 7 × 16
= 704 cm2
Hence the area of the label of 704 cm2.
Exercises 11.4
Question 1.Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
A. To find how much it can hold.
B. Number of cement bags required to plaster it.
C. To find the number of smaller tanks that can be filled with water from it.
Answer:
(a) We need to calculate the volume to find the capacity
(b) As plastering will cover the surface so we need surface area to know this
(c) Volume will give the capacity and that can be compared with capacity of smaller tanks
Question 2.
Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?
Answer:
As cylinder A’s radius is half of radius of cylinder B so its volume will be lesser than that of cylinder B.
Although Cylinder B’s height is half of height of cylinder A but as you know
while calculating the volume we need to square the radius so halving the radius has a greater impact than halving the height.
While calculating surface area, the curved surface area in both will be same and
the total surface area will be greater in the cylinder with the greater radius.
Now, volume of the cylinder = π r2 h
Thus, volume of cylinder A = 
= 539 cub. cm
Volume of cylinder B = = 1078 cub. cm
Also, curved surface area of a cylinder = 2πrh
Curved surface area of cylinder A = 
= 308 sq. cm
Curved surface area of cylinder B = 
= 308 sq. cm
Total surface area of Cylinder = 2 π r h + 2 π r2
Total surface area of Cylinder A = 2 × 
= 385 sq. cm
Total surface area of Cylinder B = 2 × 
= 616 sq. cm
Hence the surface area of cylinder B is greater.
Question 3.
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?
Answer:
Given: Area of base = 180cm2
Height of cuboid = ?
We know that,
Volume of cuboid = Area of base × Height
900cm3 = 180cm2 × Height of cuboid
Hence,
Height = 
Question 4.
A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Answer:
Given: Dimensions of cuboid = 60 cm × 54 cm × 30 cm
Side of cube = 6 cm
We know that, Volume of cuboid = l × b × h
Volume of cube = side × side × side
Therefore
Number of cubes = 
= 450 cubes
Question 5.
Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm ?
Answer:
Volume of cylinder = π r2 h
Given,
Diameter = 140 cm
Now, 
Hence, radius = r = 70 cm
Also, 100 cm = 1 m
70 cm = 0.7 m
volume of cylinder given = 1.54 m3
Hence, we have
Question 6.
A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in litres that can be stored in the tank?
Answer:
Volume of milk tank = π r2 h
= 49.5 m3
As we know, 1m3 = 1000 litres
So, 49.5 m3 = 49.5(1000)
49500 litres
Question 7.
If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Answer:
(i) ∵ Surface Area of a Cube SA= 6(Side)2
If Side is doubled then New Surface Area,
SA' = 6(2×Side)2
= 4× 6(Side)2
= 4 × SA
∴ SA of new Cube will be 4 times of SA of Original Cube.
(ii) ∵ Volume of a Cube V= (Side)3
If Side is doubled then New Volume of Cube,V'= (2× Side)3
= 8× (Side)3
= 8× V
∴ Volume of new Cube will be 8 times the Volume of Original Cube.
Question 8.
Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir.
Answer:
The volume o the reservoir is 108 m3 = 108000 L
(∵ 1 m3 = 1000 Litre)
And, water if poured into the reservoir at the rate = 60 L per minute.
⇒ In 1 min, the water poured in tank = 60 L
⇒ In 60 min (1 hour), the water poured in tank = 60 × 60 = 3600 L
∴Time to fill the reservoir,
= 30 hours