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?
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?
Solution-
It has been given that,
Side length of square = x = 60 m
Length of the rectangle l is = 80 m
Value of perimeter of the given field is = value of perimeter of given square
Note: Perimeter - sum of all sides of the given shape.
So, from the formula of perimeter of square and rectangle we get,
2× (l+b) = 4×Side (x = Side, l = length, b = breadth)
⇨ 2× (80+b) = 4×60
⇨ 160 + 2×b = 240 (transfer 160 from LHS to RHS)
⇨ 2b = 240– 160
⇨ 2b = 80 (divide LHS and RHS by 2)
⇨ b = 80/2
⇨ b = 40
The calculated breadth of rectangular field is equal to 40 m.
To calculate area of given square, formula is = (side)2
= (x)2
= (60)2
= 60×60
= 3600 m2
the area of the rectangular field can be calculated as
= length (l) ×breadth (b) = 80×40
= 3200 m2
Hence, Area of square is equal to 3600 m2 and the area of rectangle is equal to 3200 m2.
Answer- Square field contains large area.
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.
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.
Solution:
Side of square = 25 meters (Given)
We know, Area of square = (side)2
= (25)2
= 25×25
House width = 15 m
House area = length x width
= 20 × 15 (simplify)
= 300 m2
Garden area = Square area – House area
= 625 – 1×300 (simplify)
= 325 m2
∵ The per square m developing cost of the area is Rs. 55
The developing cost of a garden with 325 square m area = Rs. 325 x 55
= Rs. 17875.
So, the total cost to develop the garden is calculated to be Rs. 17,875.
Answer: 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 meters]
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 meters]
Solution-
It has been given that,
Total length of garden = 20 m
Diameter of the half-circle = 7 m
So, Radius of the half-circle = 7/2 = 3.5 m
Length of only the rectangle field
= 20 – (2×3.5) (from each side)
= 20 – 7
= 13 m
As, Breadth of rectangle field is same as the diameter of circle = 7 m
Area of the rectangle garden = (length) × (breadth)
= 13×7 (simplify)
= 91m2
Area of both the semi circles = 2×((1/2)×π×r2)
= 2×((1/2)×22/7×3.5×3.5)
= 38.5 m2
Total area of the garden will be = 91 + 38.5 = 129.5 m2
the perimeter of both the half-circles = πr + πr = 2πr
= 2×(22/7)×3.5 (simplify)
= 22 m
Hence, the calculated perimeter of the garden is = 13 + 22 + 13 = 48 m.
Answer: 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]
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]
Solution-
It has been given that,
Flooring tile base is = b = 24 cm
= 0.24 m [since, 1 m = 100 cm]
Height of one flooring tile will be = h = 10 cm = 0.10 m
So, Area of one floor tile will be = b × h
= 0.24× 0.10
= 0.024 m2
Hence, Area of one flooring tile will be 0.024 m2
Total tiles number needed to cover the whole floor = Area of total floor obtained / Area of one flooring tile = (1080)/(0.024) = 45,000
So, a total of 45000 tiles is needed to cover the whole area of the floor.
Answer: 45000.
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.
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.
Solution:
(a) Radius of the semi-circular shape will be = (Diameter)/2
= 2.8/2 (simplify)
= 1.4 cm
Circumference of the given half-circular shape will be = π×r
= (22/7)×1.4 (simplify)
= 4.4
Circumference of the given half-circle calculated is 4.4 cm
Total length to be travelled by an ant = Circumference of the semi-circle + Diameter of semi-circle
= 4.4 + 2.8
= 7.2 cm.
(b) Given semi-circular shapes diameter = x
⇨ x = 2.8 cm
So, radius of semi-circular shape = x/ 2
= 2.8/2 (simplify)
= 1.4 cm
Hence, circumference of the given half-circular shape = π×r
= (22/7) ×1.4 (simplify)
= 4.4 cm
Total length to be travelled by the ant is = 1.5+ 4.4 +1.5 + 2.8
= 10.2 cm
(c) Diameter of the given semi-circular shape = x = 2.8 cm
Hence, the radius of the given semi-circular shape is = (x)/2
= 2.8/2 (simplify)
= 1.4 cm
circumference of the given half-circular shape = πr
= (22/7)×1.4 (simplify)
= 4.4 cm
Total length to be travelled by the ant = 2 + 4.4 + 2
= 8.4 cm
Reviewing the obtained values of all the three shapes, we deduce that for diagram (b) food type larger path needs to be taken by the ant than other figures.
Answer: (b).
Question-6
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 the perpendicular distance between them is 0.8 m.
Question-6
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 the perpendicular distance between them is 0.8 m.
Solution-
It has been given that,
Length of 1st parallel = 1 m (of trapezium)
The length of the second parallel side (y) will be = 1.2 m
Height of the shape = h = 0.8 m
We know, Area of trapezium = (1/2)×(x+y)×h
Area of the top surface of table = (1/2)×(1+1.2)×0.8
= (1/2)×2.2×0.8
= 0.88
Hence, the top surface area of table has value equal to 0.88 m2.
Question-7
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.
Question-7
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.
Solution-
It has been given that,
Value of length of 1st parallel segment, x is = 10 cm
Value of its height = h
⇨ h = 4 cm
Trapeziums area is given to be 34 cm2
Let the length of the 2nd parallel segment be y.
trapezium’s area = (1/2)×(x + y)×h
⇨ 34 = (1/2)× (10 + y)×4
⇨ 34 = 2×(10+y)
After calculating, y = 7
Hence, the value of length of 2nd parallel segment is equal to 7 cm.
Answer: 7 cm.
Question-8
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.
Question-8
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.
Solution-
According to Question:
Value of CB = 48 m, Length of DC = 17 m,
Length of DA = 40 m
Trapezium’s perimeter is = 120 m
We know, Perimeter of trapezium ADCB = AB+DA+DC+BC
⇨ 120 = AB + 40 + 17 + 48
⇨ 120 = 105 + AB (transfer 105 from RHS to LHS and rewrite the equation)
⇨ AB = 120 –105 (simplify)
⇨ AB = 15 m
Here we get, Area of the given trapezium = (1/2)× (AD+BC) × AB
= (1/2)× (40 + 48)×15
= (1/2)×15×88
= 660 m2
So, calculated area of the given trapezium ADCB is equal to 660m2.
Answer: 660 m2.
Question-9
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.
Question-9
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.
Solution-
AC = b
b = 24 m (as shown in figure)
h1 = 13 m and h2 = 8 m.
Area of the quadrilateral ADCB will be = Area of the triangle formed by ABC + Area of the triangle formed bt ADC
= (1/2)×(h1 ×b) + (1/2)(h2×b)
= (1/2) ×b× (h1+h2)
= (1/2)×24×(8+13)
= (1/2)×21×24
= 252
So, the calculated area of the quadrilateral is equal to 252 m2.
Question-10
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Question-10
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution-
It has been given that,
Diagonal, x1 = 7.5 cm
Diagonal, x2 = 12 cm
From formula, area of the rhombus with given diagonals = (1/2)×x1×x2
= (1/2)×12×7.5
= 45
Hence, the area of the rhombus calculated is equal to 45 cm2.
Answer: 45 cm2.
Question-11
Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.
Question-11
Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.
Solution-
As we know all rhombus are called parallelogram,
The Rhombus’s area with given base and altitude = Base × Altitude
Substituting given values, we got
Rhombus’s area = 6×4 = 24
Rhombus’s area is 24 cm2
Rhombus’s area with given diagonals = (1/2)×x1×x2
Putting the values, we got
⇨ 24 = (1/2)×8×d2 (multiply LHS and RHS by 4)
⇨ d2 = 6
So, length of the second diagonal will be 6 cm.
Answer: 24 cm2, 6 cm.
Question-12
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.
Question-12
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.
Solution-
Value of length of the first diagonal x1 = 45 cm
Length of the second diagonal, x2= 30 cm
Area of only one tile = (½)×x1×x2
= (1/2)×30×45
= 675
Calculated area of only tile will be equal to 675 cm2.
So, total area of all the 3000 tiles will be
= 3000×675 (simplify)
= 2025000 cm2
= 2025000/10000 (changing into metre)
= 202.5 m2 [1m2 = 104 cm2]
As, Cost to polish the given floor / square meter is = 4
Cost to polish the given floor / 202.50 square meter = 202.50×4 = 810
So, total cost of floor polishing is calculated to be Rs. 810.
Answer: Rs. 810.
Question-13
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.
Question-13
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.
Solution-
It has been given that,
Perpendicular height of the shape = h
⇨ h = 100 m
trapezium’s area = 10500 m2
Let the value of length of the side near to the road is ‘a’ m.
So, the value of length of the side which is near to the river = 2a m.
We know, trapezium’s area = (1/2)×(x+y)×h
⇨ 10500 = (1/2)×(a+2a)×100
⇨ 10500 = 3×a × 50
After solving, we get a = 70. So, the length of the side near to the road is equal to 70 m
Hence, the length of the side near to the river = 2a
= 2×70 (simplify)
= 140 m.
Answer: 140 m
Question-14
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.
Question-14
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.
Solution-
Given:
Side length of the given octagon = 5 m.
Dividing the octagon into three parts as shown in the figure below, two of them are in shape of trapeziums whose perpendicular length is 4m and parallel sides length is 11 m, one shape is a rectangle with the breadth of 5m and length of 11 m.
From formular area of both the trapeziums = 2×[(½)×(x+y)×h]
= 2×(½)×(5+11)×4
= 1×4×16
= 64
Area of both the trapezium will be 64 m2
We know, Area of the only rectangle = l × b
= 11×5 (simplify)
= 55
Rectangle’s area is calculated to be equal to 55 m2
Hence, Octagon’s total calculated area will be = area of both trapeziums + area of rectangle
= 64 + 55 (simplify)
= 119 m2
Answer: 119 m2
Question-15
Here 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?
Question-15
Here 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?
Solution-
Given: side of pentagon = 15 m and height = 30 m
First Method: From Jyoti figure,
Pentagon’s area will be = Trapezium APCB area + Trapezium APDE area
= (1/2)×(ED+AP)×DP + (1/2)× (AP+BC)×CP
= (1/2)×(15+30)×DP + (1/2)× (15+30)×CP
= (1/2)×(30+15)×(DP+CP) (from figure CD= CP+DP)
= (1/2)×45×CD
= (1/2)×45×15
= 337.5
Area of pentagon by Jyoti’s diagram is equal to 337.5 m2
Second Method: From Kavita figure,
Draw a perpendicular segment from A to the side BE. Let it be AM.
AM = Total height – Side of the square
⇨ AM = 30 – 15 (simplify)
⇨ AM = 15 m
Pentagon’s area = Triangle ABE area + Square BCDE area (as concluded through fig.)
= (1/2)×15×15 + (15×15) (simplify)
= 225 + 112.5
= 337.5
Area of pentagon by Jyoti’s diagram = 337.5 m2
Question–16
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 the same.
Question–16
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 the same.
Solution-
Dividing given shape into 4 different sub-parts like:
In this, two of the parts that is (1) and (3) are in similar shape.
Also, other parts that is (2) and (4) are in similar shape.
Area of part (1) = Area of the trapezium
= (½)×(x+y)×h
= (½)×(20+28)×4
= (½)×4×48 = 96
Area of part (1) = 96 cm2
So, Area of part (3) is also = 96 cm2
So, Area of part (2) = Area of the trapezium
= (1/2)×(x+y)×h
= (1/2)×(16+24) ×4
= (1/2)×4×40 (simplify)
= 80
Calculated area of part (2) is equal to 80 cm2
So, Area of part (4) is = 80 cm2
Question-17
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Question-17
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
Solution-
(a)
Value of length of the box l = 60 cm (Given)
Height of the box, h = 50 cm (Given)
Breadth of the box, b = 40 cm (Given)
From formula,
The surface area of a cuboid can be calculated by = 2×(b×l+h×b+h×l)
= 2×(60×40+40×50+50×60)
= 2×(2400+2000+3000)
= 14800 cm2
Surface area of the cuboidal is 14800 cm2
(b)
Value of length of the box l is = 50 cm (Given)
Value of Height of the box, h = 50 cm (Given)
Value of breadth of the box, b = 50 cm (Given)
We know,
Cube’s surface area is = 6× (side)2
= 6×(50)2
= 6×50×50 (simplify)
= 15000 cm2
Cube’s surface area is = 15000 cm2
Analyzing the results obtained in part (a) and part (b), the box taking the lesser material to make is cuboidal box.
Question-18
A suitcase with measures 80 cm x 48 cm x 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Question-18
A suitcase with measures 80 cm x 48 cm x 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution-
Given:
suitcase’s height = h = 24 cm
suitcase’s length = l = 80 cm,
suitcase’s breadth = b = 48 cm
Suitcase’s surface area = (b×l+ h×b+ l×h) × 2
= (80×48+48×24+24×80) ×2
= (3840+1152+1920) ×2
= 2×6912
= 13824
surface area of suitcase will be 13824 cm2
So, Tarpaulin cloth needed = Total Suitcase’s surface area
⇨ l×b = 13824
⇨ l×96 = 13824 (divide LHS and RHS by 96)
⇨ l = 13824/96 cm (simplify)
⇨ l = 144 cm
Length of tarpaulin needed to enclose 100 suitcases = 100×144
= 14400 cm (converting in metre)
= 144 m
Hence, the required length of tarpaulin cloth to enclose 100 suitcases is equal to 144 m.
Answer: 144 m.
Question-3:
Find the side of a cube whose surface area is 600 cm2.
Question-3:
Find the side of a cube whose surface area is 600 cm2.
Solution-
cube’s surface area = 600 cm2 (Given)
From formula, cube’s surface area = 6× (side)2
Putting all the values,
⇨ 6× (side)2 = 600 (divide LHS and RHS by 6)
⇨ (side)2 = 100 (simplify)
side = ±10
Since length is always positive, the value of all side of cube is equal to 10 cm.
Answer: 10 cm
Question-20
Rukshar 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?
Question-20
Rukshar 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?
Solution-
Given:
Cabinet’s height = h = 1.5 m
Cabinet’s length = l = 2 m
cabinet’s breadth = b = 1 m
Total area painted by Rukshar = cabinet’s surface area – cabinet’s bottom area
Cabinet’s surfaces area is = (b×l+h×b+l×h)×2
= (2×1+1×1.5+2×1.5) ×2
= 2× (6.5) (simplify)
= 13 m2
Area of the bottom of the cabinet = Length × Breadth
= 2 × 1
= 2 m2
Area painted by Rukshar = 13 – 2
= 11 m2
surface area of cabinet painted by Rukshar is 11m2.
Answer: 11 m2
Question-21
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 he need to paint the room?
Question-21
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 he need to paint the room?
Solution-
Given:
Value of length of the wall l is = 15 m
Height of the wall h = 7 m
Breadth of the wall b = 10 m
Hall’s total area = 2×(b×h+h×l )+ l×b
= 2× (10×7+7×15)+ 15×10
= 150 +2× (70+105)
= 150+350
= 500 m2
Now, number of cans needed to paint = Total area of hall / Area painted by one can
= 500/100 (simplify)
= 5
Hence, 5 cans will be required to completely paint the given cuboidal hall.
Answer: 5 cans.
Question-22
Describe how the two figures below are alike and how they are different. Which box has larger lateral surface areas?
Question-22
Describe how the two figures below are alike and how they are different. Which box has larger lateral surface areas?
Solution-
Similarities-
Both of the shapes have the same value of length.
Both of the shapes have the same value of height.
Differences-
The 1st shape contains a circular base and also circular top.
The 2nd shape contains square base and also square top.
The 1st shape is called cylinder but the 2nd shape is called cube.
Given:
Value of cylinder’s height h = 7 cm
Cylinder’s diameter = 7 cm
Cylinder’s radius = 7/2 cm
Now, Cylinder’s lateral area is = 2πrh
= (22/7)×(7/2)×7×2 (simplify)
= 22×7
= 154
So, cylinder’s lateral surface area obtained to be 154 cm2
Cube’s lateral area = 4 ×(length of sides)2
= 4×72 (simplify)
= 4×49 (simplify)
= 196
cube’s lateral surface area obtained to be 196 cm2
By comparing both values we can say that the cube has more lateral surface area than the cylinder.
Answer: Cube.
Question-23
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.
Question-23
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.
Solution-
Height of cylinder, h = 3 m
Radius of the cylinder, r = 7 m
Cylinder’s total area = 2πr2 + 2πrh
= 2×(22/7)×72+ 2×(22/7)×7×3
= 308+132
= 440
Hence, are of metal sheet required 440 m2.
Answer: 440 m2.
Question-24
The lateral surface area of a hollow cylinder is 4224cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet?
Question-24
The lateral surface area of a hollow cylinder is 4224cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet?
Solution-
Width of the rectangle sheet = 33 cm
Lateral surface area of the hollow cylinder is = 4224 cm2 (Given)
Let, the length of the rectangular sheet needed is l.
Since, sheet is formed from the cylinder so
Sheet’s area = Cylinder’s lateral area
⇨ l × b = 4224
⇨ l × 33 = 4224 (divide LHS and RHS by 33)
⇨ l = 4224/33
⇨ l = 128 cm
So, size of rectangular sheet formed is 128 cm.
Now, rectangular sheet’s perimeter formed = 2× (l+b)
= 2× (33+128) (simplify)
= 322 cm
The rectangular sheet’s perimeter will be equal to 322 cm.
Answer: 322 cm.
Question-25
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 1 m.
Question-25
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 1 m.
Solution-
Given:
Roller’s length h = 1 m = 100 cm
Roller’s diameter x = 84 cm
Roller’s radius r = x/2
⇨ r = 84/2
⇨ r = 42 cm
Considering the road roller to be a cylinder,
We know, cylinder’s curved area = 2πrh
= 2×(22/7)×100×42 (simplify)
= 26400
So, roller’s curved area obtained is equal to 26400 cm2
Road area covered by the roller in 750 complete revolutions = 750×26400 cm2
= 19800000cm2
= 1980 m2 [Since, 1 m2= 10000 cm2]
So, Road’s area is 1980 m2.
Answer: 1980 m2
Question-26
A company packages its milk powder in a 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 figure). If the label is placed 2 cm from top and bottom, what is the area of the label?
Question-26
A company packages its milk powder in a 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 figure). If the label is placed 2 cm from top and bottom, what is the area of the label?
Solution-
Given:
Container’s height h = 20 cm
Container’s diameter x = 14 cm
So, Radius of the container, r = x/2
⇨ r = 7 cm
Let, label’s height to be h (from the above figure)
⇨ h = 20 – 2×2
⇨ h = 20 – 4
⇨ h = 16 cm
We know, label’s curved area will be = 2πrh
= 2×(22/7)×16×7
= 704
Hence, label’s area on the cylindrical container is 704 cm2.
Question-27
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.
Question-27
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.
Solution-
An area of an object is found when we want to measure the size of the surface of that object. The area can be calculated for various shapes such as a square, rectangle, triangle, circle, or any irregular shape.
The volume of an object is found when we want to measure the amount of space that object occupies. The volume can be calculated for various shapes such as a cube, pyramid, cone, cylinder, or any irregular shape.
Answers:
(a) Volume needs to be calculated.
(b) Surface area needs to be calculated.
(c) Volume needs to be calculated.
Question-28
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.
Question-28
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.
Solution-
The answer is Yes, we can suggest whose volume is more. The volume of cylinder B is more because cylinder B’s radius is larger than the cylinder-A’s radius.
Finding the volume of cylinder-A and Cylinder-B
Cylinder-A’s height= 14 cm (Given)
Cylinder-A’s diameter = 7 cm (Given)
So, Cylinder-A’s radius = 7/2 cm
We know, Volume of the cylindrical shapes = πr2h
= (22/7)×(7/2)2×14
= 11×7×7
= 539
So, cylinder-A’s volume is equal to 539 cm3
Now,
Cylinder-B’s height = 7 cm
Cylinder-B’s diameter = 14 cm (Given)
So, Cylinder-B’s radius = 14/2 = 7 cm
We know, Volume of cylindrical shapes = πr2h
= (22/7)×72×7 = 1078
So, cylinder-B’s volume is equal to 1078 cm3.
Now, Finding the surface area of the cylinders A and B
For cylinder A, Surface area = 2πr2 + 2πhr
= 2 × (22/7) × (7/2)2 + 2 × (22/7) × (7/2) × 14 = 385
Surface area of the cylinder A is equal to 385 cm2
For cylinder B, Surface area = 2πr2 + 2πhr
= 2 × (22/7) × (7/2)2 + 2 × (22/7) × (7/2) × 7 = 616
Surface area of the cylinder B is equal to 616 cm2
Yes, the cylinder containing more volume will have more surface area too.
Question-29
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?
Question-29
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?
Solution-
Volume of the given cuboid = 900 cm3
Given cuboid’s base area is = 180 cm2
We know, cuboid’s volume is = h×l×b
Also, cuboid’s base area = l×b
⇨ 900 = 180×h (putting all values)
⇨ h = 900/180 (simplify)
⇨ h = 5
So, the height obtained is 5 cm.
Answer: 5 cm
Question-30
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?
Question-30
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?
Solution-
Height of the shape, h = 30 cm (Given)
Length of the shape, l = 60 cm (Given)
Breadth of the shape, b = 54 cm (Given)
From formula, cuboid’s volume is = b×h×l
= 54×30×60
= 97200 cm3
Also, cube’s volume is = (side)3
= 63
= 6×6×6
= 216 cm3
Now, The Number of small cubes required = (cuboid’s volume) / (each cube’s volume)
= 97200/216
= 450
Hence, the number of cubes that can be put inside this cuboid is 450.
Answer: 450.
Question-31
Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm.
Question-31
Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm.
Solution-
Cylinder’s volume = 1.54 m3 (Given)
Cylinder’s diameter, x = 140 cm (Given)
So, Radius of the cylinder, r = x/2
⇨ r = 70 cm
⇨ r = 0.7 m
We know that, cylinder’s volume is = πr2h
⇨ 1.54 = (22/7)×0.72×h
⇨ h = (1.54×7)/(22×0.72) (simplify)
⇨ h = 1
After solving, we get the value of h to be, h = 1 m
So, cylinder’s calculated height is 1 m.
Answer: 1 m.
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.
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.
Solution-
Height of the milk container, h = 7 m (Given)
Radius of the milk container, r = 1.5 m (Given)
We know that, cylindrical container’s volume is = πr2h
= (22/7)×1.52 ×7
= 49.5 m3
= 49.5×1000 litres [as, 1 m3 = 1000 litres]
= 49500 litres
Hence, the volume of milk that can be put in the container is 49500 litres.
Answer: 49500 liters.
Question-33
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?
Question-33
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?
Solution-
(i) Assuming the cube's edge is equal to “a”.
We know, Surface area of cube = 6 × a2
When its edge is doubled, edge length becomes (2a)
New cube’s surface area = 6× (2×a)2
= 6×4×a2
= 24×a2
So, the newer surface area is 4 times more than the older surface area.
Hence, the cube's area increases four times.
(ii) We know, cube’s volume = a3
When its edge is doubled
New cube’s volume = (2×a)3 = 8× (a3)
So, new cube’s volume is 8 times the old cube’s volume.
Hence, the volume of cube increases by 8 times.
Question-34
Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of the reservoir is 108 m3, find the number of hours it will take to fill the reservoir.
Question-34
Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of the reservoir is 108 m3, find the number of hours it will take to fill the reservoir.
Solution-
Reservoir’s volume = 108 m3 (Given)
Rate at which water is pouring inside the reservoir = 60 litres/minute (given)
= 60/1000 m3/minute [since, 1 litre = (1/1000) m3 and 60 minute = 1 hr]
= (60×60)/1000 m3/hr
= 3.6 m3/hr
So, filling of 3.6 m3 water inside the reservoir takes = 1 hour
So, filling of 1 m3 water inside the reservoir takes = 1/(3.6) hr
So, filling of 108 m3 water inside the reservoir should take = (108)×(1/3.6) hr
= 30 hr
Hence 30 hours is required to fill the reservoir of volume 108 m3.
Answer: 108 m3
Also Read : Class 8 Mensuration Extra Question
