1.A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

We know that Perimeter of square=\(4\times side=4\times 60=240m\)

Also, we have perimeter of rectangle=2(l+b)

Given that these both have same parimeter.

Therefore, we can say that:

perimeter of rectangle=Perimeter of square

\(2(l+b)=240\)

\(\Rightarrow 2(80 + b) = 240\)

\(\Rightarrow 160+2b = 240\)

\(\Rightarrow 2b=240-160=80\)

\(\Rightarrow b=40m\)

Hence we have, Area of figure (a) i.e, square =\( (side)^2\) = \(60 \times 60 = 3600 m^2\)

Area of figure (b) i.e, rectangle =\( l \times b = 80 \times 40 = 3200 m^2\)

So, the area of figure (a) i.e, square is more than the area of figure (b) i.e, rectangle.

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 ₹ 55 per \(m^2\).

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

We have the area of the square plot =\( side \times side = 25 m \times 25 m = 625 m^2\)

Area of the rectangular house =\( l \times b = 20 m \times 15 m = 300 m^2\)

Therefore, Area of the garden to be developed = Area of the plot – Area of the house = \(625 m^2 – 300 m^2 = 325 m^2\)

Cost of developing the garden = \(₹ 325 \times 55 = ₹ 17875\)

3.The shape of a garden is rectangular in the middle and semicircular 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]

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

We have,Total area of the garden = Area of the rectangular portion + The sum of the areas of the pair of semi-circles

We have, Length of the rectangle = \(20 – (3.5 + 3.5) = 20 – 7 = 13 m\)

Therefore, area of the rectangle =\( l \times b = 13 \times 7 = 91 m^2\)

Area of two circular ends = \(2\times \frac { 1 }{ 2 }\pi r^2\)

\(= \pi r^2\)

\(= \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 }\)

\(=\frac { 77 }{ 2 } m^2\)

\(= 38.5 m^2\)

Total area = Area of the rectangle + Area of two ends = \(91 m^2 + 38.5 m^2 = 129.5 m^2\)

Total perimeter = Perimeter of the rectangular portion + Perimeter of two circular ends

\(= 2 (l + b) + 2 \times (\pi r) – 2(2r)\)

\(= 2 (13 + 7) + 2(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 }) – 4 \times \frac { 7 }{ 2 }\)

\(= 2 \times 20 + 22 – 14\)

\(= 40 + 22 – 14\)

\(= 48 m\)

So the total perimeter of the figure =48m

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).

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given that area of floor=\(1080 m^2=1080\times 10000cm^2=10800000cm^2\)[∵ \(1m^2=100cm^2\)]

Area of one tile= \(1 \times base \times height = 1 \times 24 \times 10 = 240 cm^2\)

So, the number of tiles required=\(\frac{Area \; of \;the \;floor}{Area\; of \;1\;tile}\)

\(=\frac{10800000}{240}\)

\(= 45000 tiles\)

Hence, the reuired number of tiles=45000 tiles

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, the circumference of a circle can be obtained by using the expression C = \(2\pi r\), where r is the radius of the circle.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

(a)Distance covered by the ant to take a round:

\(=\frac{1}2\times2\pi r+2r\)

\(=\pi r+2r\)

\(=\frac{22}7\times1.4+2\times1.4\)

\(=22\times 0.2+2.8\)

\(=4.4+2.8=7.2cm\)

(b)Distance covered by the ant to take a round:

\(=\frac{1}2 \times 2\pi r+1.5+1.5+2.8\)

\(=\pi r+5.8\)

\(=\frac{22}7\times 0.2+5.8\)

\(=4.4+5.8\)

\(=10.2cm\)

(c)Distance covered by the ant to take a round:

\(=\frac{1}2 \times 2\pi r+2+2\)

\(=\pi r+4\)

\(=\frac{22}7\times 1.4+4\)

\(=4.4+4\)

\(=8.4cm\)

So, among all the given paths, in path (b) ant has to travel a larger distance for food piece than others.

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.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Area of top trapezium surface of the table=(\frac{1}2\times (sum\; of \;parallel \;sides)\times (distances \;between\; them)\)

\(=\frac{1}2\times(1+1.2)\times0.8 m^2\)

\(=\frac{1}2\times 2.2 \times 0.8 m^2\)

\(=(1.1\times 0.8)=0.88m^2\)

So, the area of top trapezium surface of the table=\(0.88m^2\)

2.The area of a trapezium is \(34 cm^2\) 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.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, area of trapezium is \(34cm^2\).

Let the required side be x cm.

We know that, area of trapezium=\(\frac{1}2\times (sum\; of \;parallel \;sides)\times (distances \;between\; them)\)

\(34=\frac{1}2\times(10+x)\times 4\)

\(34=2(10+x)\)

\(10+x=17\)

\(x=7\)

Hence, the required side is 7cm.

3.Length of the fence of a trapezium A 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.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given:

BC = 48 m, CD = 17 m, AD = 40 m

AB + BC + CD + DA = 120 m .

AB = 120 m – (48 m + 17 m + 40 m) = 120 – 105 m = 15 m

So the area of trapezium fence ABCD=\(\frac{1}2\times (sum\; of \;parallel \;sides)\times (distances \;between\; them)\)

\(=\frac{1}2\times (BC+AD)\times AB\)

\(=\frac{1}2\times (48+40)\times 15\)

\(=\frac{1}2\times 88\times 15\)

\(=44\times 15=660m^2\)

So, the required area=\(660 m^2\)

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.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

As per the figure we have area of field=area of ∆ABD+area of ∆BCD

\(=\frac{1}2\times base \times height+\frac{1}2\times base \times height\)

\(=\frac{1}2\times 24 \times 13+\frac{1}2\times 24\times 8\)

\(=12\times 13+12\times 8\)

\(=12\times(13+8)\)

\(=12\times 21\)

\(=252m^2\)

Hence the required area of the given field=\(252m^2\)

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given,d1 = 7.5 cm, d2 = 12 cm
Area of a rhombus =\( \frac { 1 }{ 2 } x (product of diagonals)\)

\(=\frac{1}2\times (d1\times d2)\)

\(=\frac{1}2\times (7.5 \times 12)\)

\(=7.5\times 6\)

\(=45 cm^2\)

So the area of rhombus=45 \(cm^2\)

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.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, Side = 5 cm, Altitude = 4.8 cm and length of one diagonal(d1) = 8 cm

We know, area of the rhombus = \(Side \times Altitude = 5 \times 4.8 = 24 cm^2\)

Area of the rhombus =\(\frac { 1 }{ 2 } \times d1 \times d2\)

\(\Rightarrow 24 = \frac { 1 }{ 2 } \times 8 \times d2\)

\(\Rightarrow 24 = 4\times d2\)

\(\Rightarrow d2 = 6 cm\)

Hence, the length of other diagonal = 6 cm.

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 \(m^2\) is ₹ 4.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, number of tiles = 3000 and the length of the two diagonals of a tile = 45 cm and 30 cm

We have area of one rhombus shaped tile =\(\frac { 1 }{ 2 }\times d1 \times d2\)

\(=\frac { 1 }{ 2 }\times 45 \times 30\)

\(= 45 \times 15\)

\(= 675 cm^2\)

So, area of one tile=675\(cm^2\)

Therefore, Area covered by 3000 tiles = \(3000 \times 675 cm^2 = 2025000 cm^2 = 202.5 m^2\)

So, the cost of polishing the floor = \(202.5 \times 4 = ₹ 810\quad\)[∵ cost per \(m^2\) is ₹ 4.
]

Hence, the required cost = ₹ 810.

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 \(m^2\) and the perpendicular distance between the two parallel sides is 100m, find the length of the side along the river.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, area of trapezium field area=10500\(m^2\), Altitude of the field(h)=100m.

Let, the side of the trapezium near road be x cm and
the opposite parallel side = 2x m

We know, area of trapezium = \(\frac { 1 }{ 2 }\times (a + b) \times h\)

\(\Rightarrow 10500 = \frac { 1 }{ 2 }\times (2x + x) \times 100\)

\(\Rightarrow 2 \times10500 = 3x \times 100\)

\(\Rightarrow 21000 = 300x\)

\(\Rightarrow x = 70 m\)

So, the other side = \(2x = 2 \times 70 = 140 m\)

Hence, the required length of the side along the river= 140 m.

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.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

We can observe, Area of the octagonal surface= area of trapezium ABCH + area of rectangle HCDG + area of trapezium GDEF

Finding area of different parts:

Area of trapezium ABCH = Area of trapezium GDEF

\(=\frac { 1 }{ 2 }\times (a + b) \times h\)

\(=\frac { 1 }{ 2 }\times (11 + 5) \times 4\)

\(=\frac { 1 }{ 2 }\times 16 \times 4\)

\(= 32 m^2\)

Area of rectangle HCDG =\( l \times b = 11 m \times 5 m = 55 m^2\)

Area of the octagonal surface = \(32 m^2 + 55 m^2 + 32 m^2 = 119 m^2\)

Hence, the required area = \(119 m^2\).

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?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

(i) From Jyoti’s diagram:

We can observe that, Area of the pentagonal shape = Area of trapezium ABCD + Area of trapezium ADEF

\(= 2 \times Area \;of\; trapezium \;ABCD\)

\(= 2 \times \frac { 1 }{ 2 } \times (a + b) \times h\)

\(= (15 + 30) \times 7.5\)

\(= 45 \times 7.5\)

\(= 337.5 m^2\)

So, the area of the pentagonal shape=337.5\(m^2\)

(ii) From Kavita’s diagram:

We can see that, area of the pentagonal shape = Area of triangle ABE + Area of square BCDE

\(=\frac { 1 }{ 2 }\times b \times h + 15 \times 15\)

\(=\frac { 1 }{ 2 }\times 15 \times 15 + 225\)

\(= 112.5 + 225= 337.5 m^2\)

So, the area of the pentagonal shape =\(337.5 m^2\)

Yes, there is other way to find the area which is as follows:

Required area ABCDEA=Area of rect. ABPQ-2\(\times\)area of triangle CPD

\(=(15\times 30)-(2\times \frac{1}2\times \frac{15}2)\)

\(=(450-\frac{225}2)=\frac{(900-225)}2m^2\)

\(=\frac{675}2m^2=337.5m^2\)

11.Diagram of the picture frame has outer dimensions = \(24 cm \times 28 cm\) and inner dimensions \(16 cm \times 20 cm\). Find the area of each section of the frame, if the width of each section is the same.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

We have:

\(h_1=\frac{1}2(28-20)=\frac{1}2\times 8=4cm\)

\(h_2=\frac{1}2(24-16)=\frac{1}2\times 8=4cm\)

We also know, area of trapezium A=\(\frac{1}2\times(a+b)\times h_1\)

\(=\frac{1}2\times(24+16)\times4\)

\(=\frac{1}2\times 48\times 4=80cm^2\)

so, the area of trapezium A=area of trapezium C=80\(cm^2\)

Now, area of trapezium B=Area of trapezium D

\(=\frac{1}2\times(28+20)\times4\)

\(=\frac{1}2\times 48\times 4=96cm^2\)

Hence, we have areas of all the parts as:

Part A=\(80cm^2\)

Part B=\(96cm^2\)

Part C=\(80cm^2\)

Part D=\(96cm^2\)

1.There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

We have, the total surface area of first box=\( 2(lb + bh + lh)\)

\(= 2 (60 \times 40 + 40 \times 50 + 60 \times 50) cm^2\)

\(= 200 (24+20+30) cm^2\)

\(= 200 \times 74 cm^2=14800 cm^2\)

Also, the total surface area of second box= \(6 (Edge)^2\)

\( = 6 \times 50 \times 50 cm^2= 15000 cm^2\)

So, the first box i.e., (a) requires the least amount of material to make because the total surface area of first box is less than that of the second.

2.A suitcase of measures \(80 cm \times 48 cm \times 24 cm\) is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Measurement of the suitcase =\( 80 cm \times 48 cm \times 24 cm\)

We can say that l = 80 cm, b = 48 cm and h = 24 cm
So the total surface area of the suitcase = \(2[lb + bh + hl]\)

\(= 2 (80 \times 48 + 48 \times 24 + 80 \times 24) cm^2\)

\(= 2(3840 + 1152 + 1920) cm^2\)

\(= 2 \times 6912 cm^2 =13824 cm^2\)

We have, Area of tarpaulin = Area of 100 suitcase

Area of Area of 100 suitcase=\(100 \times 13824\)

Also, we have the Area of tarpaulin = \(length \times breadth = l \times 96 = 96l cm^2\)

Equating both we have:

\(96l =100 \times 13824\)

\(\Rightarrow l = 100 \times 144 = 14400 cm = 144 m\)

Hence, the required length of the cloth = 144 m.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, the surface area of cube is \(600 cm^2\)

Total surface area of a cube =\( 6l^2\)

\(\Rightarrow 6l^2 = 600\)

\(\Rightarrow l^2 = 100\)

\(\Rightarrow l = \sqrt{100} = 10 cm\)

Hence, the required length of side = 10 cm.

4.Rukhsar painted the outside of the cabinet of measure \(1 m \times 2 m \times 1.5 m\). How much surface area did she cover if she painted all except the bottom of the cabinet.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, l = 2 m, b = 1.5 m, h = 1 m

Area of the surface to be painted = Total surface area of box – Area of base of box

\(= 2 (lb + bh + hl) – lb\)

\(= 2(2 \times 1.5 + 1.5 \times 1 + 1 \times 2) – 2 \times 1\)

\(= 2(3 + 1.5 + 2) – 2\)

\(= 2(6.5) – 2\)

\(= 13 – 2= 11 m^2\)

Hence, the required area = 11 \(m^2\).

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 \(m^2\) of area is painted.How many cans of paint will she need to paint the room?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, l = 15 m, b = 10 m and h = 7 m

Area of the paint in one can = \(100 m^2\)

So,Surface area of a cuboidal hall without bottom = Total surface area – Area of base

\(= 2 (lb + bh + hl) – lb\)

\(= 2 (15 \times 10 + 10 \times 7 + 7 \times 15) – 15 \times 10\)

\(= 2(150 + 70 + 105) – 150\)

\(= 2 (325) – 150\)

\(= 650 – 150= 500 m^2\)

Therefore, the number of cans required =\(\frac { 500 }{ 100 }= 5 cans\).

6.Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

The two figures given differ in their shapes.One of them is cylinder and the other is cube.

Similarity: Both the given figures have the same height.

Given, Cylinder: d = 1 cm, h = 7 cm

Cube: Length of each side a = 7 cm

So, Lateral surface of cylinder = 2\(\pi \)rh

\(= 2 \times \frac { 22 }{ 7 }\times \frac { 7 }{ 2 } \times 7 = 154 cm^2\)

Also, the Lateral surface of the cube =\( 4l^2 = 4 \times (7)^2 = 4 \times 49 = 196cm^2\)

So, the cube has the larger lateral surface =\( 196 cm^2\).

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?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, r = 7 m and h = 3 m.

Sheet required to make a closed cylinder= Total surface area of the cylinder

\(= 2\pi r(r + h)\) sq. units

\(=2\times \frac { 22 }{ 7 } \times 7\times (3+7)\)

\(=2\times \frac { 22 }{ 7 } \times 7\times 10 \)

\(=440 m^2\)

So, the required metal sheet=\(440m^2\)

8.The lateral surface area of a hollow cylinder is 4224 \(cm^2\). It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, The lateral surface area of a hollow cylinder is 4224 \(cm^2\) and the width of rectangular sheet =33cm

We know that the circumference of the base of the cylinder is equal to the length of the sheet and the height of the cylinder is equal to the breadth of the sheet when a hollow cylinder is cut and a rectangular sheet is formed. Therefore, the lateral surface area of the cylinder is equal to the area of the rectangular sheet.

Let the length of the sheet be l.

∴ we have ,area of the sheet = Lateral surface area of cylinder

\(\Rightarrow l \times 33 = 4224\)

\(\Rightarrow l=\frac{4224}{33}=128cm\)

Thus, the length of the rectangular sheet=128cm\)

Now, perimeter of rectangular sheet= 2 (l + b)

\( = 2 (128 + 33) cm\)

\(= 2 \times 161 cm=322 cm\)

Hence, the perimeter of the rectangular sheet=322cm.

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.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given,diameter of the road roller=84cm and length=1m.Also revolution=750

We have radius of roller r=\(\frac{84}2=42cm=0.42m\)

So, curved surface area of the roller=\(2\pi rh\)

\(=2\times \frac{22}7 \times 0.42 \times 1\)

\(=22.64 m^2\)

∴area levelled by the roller in 750 revolutions:

\(\qquad=(750\times 2.64)=1980m^2\)

Hence, the area of the road levelled=\(1980m^2\)

10.A company packages its milk powder in cylindrical containers 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 surface area of the label.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, Diameter of the cylindrical container=14cm, height=20cm, label placed=2cm from top to bottom.

So the height covered by the label=20-(2+2)=16cm

Radius of the container=\(\frac{1}2\times d=\frac{1}2\times 14=7cmm\)

Surface area of the cylindrical shaped label =\( 2\pi rh\)

\(= 2 \times\frac { 22 }{ 7 } \times 7 \times 16\)

\(= 704 cm^2\)

Hence, the area of the label on container =\(704 cm^2.\)

1.Given a cylindrical tank, in which situation will you find the 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.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

In this situation we can find:

(a) volume.

(b)surface area.

(c)volume.

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?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Cylinder B will have a greater volume.

Verification:

Volume of cylinder A=\(\pi r^2h\)

\(=\frac{22}7\times \frac{7}2\times\frac{7}2\times 14\)

\(=539cm^3\)

Volume of cylinder B=\(\pi r^2h\)

\(=\frac{22}7\times7\times7\times 7\)

\(=22\times49=1078cm^3\)

So, we can see that cylinder B has more volume.Hence the assumption was correct.

Now comparing the surface areas of both the cylinders:

Total surface area of cylinder A=\(2\pi r(r+h)\)

\(=2\times\frac{22}7\times\frac{7}2(\frac{7}2+14)\)

\(=2\times\frac{22}7\times\frac{7}2\times \frac{35}2\)

\(=385cm^2\)

Total surface area of cylinder B=\(2\pi r(r+h)\)

\(=2\times\frac{22}7\times7(7+7)\)

\(=2\times\frac{22}7\times7\times 14\)

\(=616cm^2\)

Hence we can observe that cylinder B has more surface area.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given,Volume of the cuboid =900 \(cm^3\) and Base area=180 \(cm^2\)
As we know, \( (Area of the base) \times Height = 900 cm^3\)

\(\Rightarrow 180 \times Height = 900\)

\(\Rightarrow Height = \frac { 900 }{ 180 } = 5\)

Hence, the required height of the cuboid= 5 cm.

4. A cuboid is of dimensions \(60 cm \times 54 cm \times 30 cm\). How many small cubes with side 6 cm can be placed in the given cuboid?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, Volume of the cuboid \((l \times b \times h )= 60 cm \times 54 cm \times 30 cm = 97200 cm^3\) and sie of cube=6cm

So, we have Volume of the cube= \((Side)^3= (6)^3 = 216 cm^3\)

We know, Number of the cubes from the cuboid=\(\frac{Volume\;of\;the\;cuboid}{volume\;of\;cube}\)

\(=\frac{97200}{216}=450\)

Hence, the required number of cubes=450

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, V = 1.54 \(m^3\), d = 140 cm = 1.40 m[∵1m=100cm]

We have,Volume of the cylinder = \(\pi r^2h\)

\(\Rightarrow 1.54=\frac{22}7\times{1.4}2\times\frac{1.4}2\times h\quad\)[(∵\(r=\frac{d}2=\frac{1.4}2\)]

\(\Rightarrow h=\frac{1.5\times2\times2\times7}{1.4\times22\times1.4}=1m\)

Hence, the height of the cylinder =1m.

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 liters that can be stored in the tank?

,img src="https://trustudies-app-public.s3.ap-south-1.amazonaws.com/11.4(q6).png">

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given, r = 1.5 m, h = 7 m

We have, Volume of the milk tank =\(\pi r^2h\)

\(=\frac { 22 }{ 7 }\times 1.5 \times 1.5 \times 7\)

\(= 22 \times 2.25= 49.50 m^3\)

∴ Volume of milk in litres = \(49.50 \times 1000 L\quad\) [∵ 1 \(m^3\) = 1000 litres]

\(= 49500 L\)

Hence, the required volume of milk = 49500 L.

7.If each edge of a cube is doubled,

(i) how many times will it be surface area increase?

(ii) how many times will its volume increase?

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Let the edge of the cube = x cm

As per the question, edge is doubled so the new edge = 2x cm

(i)We have original surface area =\( 6x^2 cm^2\)

And, the new surface area = \(6(2x)^2 = 6 \times 4\times2 = 24\times2\)

∴ Ratio = \(6x^2 : 24x^2 = 1 : 4\)

Hence, the new surface area will be four times the original surface area.

(ii)We have, original volume of the cube = \(x^3 cm^3\)

And, the new volume of the cube = \((2x)^3 = 8x^3 cm^3\)

∴Ratio =\( x^3 : 8x^3 = 1 : 8\)

Hence, the new volume will be eight times the original volume.

8.Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108 \(m^3\), find the number of hours it will take to fill the reservoir.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

Answer :

Given,Volume of the reservoir = 108 \(m^3\) = 108000 L [∵1 \(m^3\) = 1000 L]

Water flowing into the reservoir per minute = 60 L

Time taken to fill the reservoir=\(\frac{Volume\;of\;reservoir}{Rate of flowing water}\)

\(=\frac{10800}{60}minutes=1800\;minutes\;
or \;\frac{11800}{60}=30\; hours\)

Hence, the required hour to fill the reservoir = 30 hours.s

There are total 34 questions present in ncert solutions for class 8 maths chapter 11 mensuration

There are total 1 long question/answers in ncert solutions for class 8 maths chapter 11 mensuration

There are total 4 exercise present in ncert solutions for class 8 maths chapter 11 mensuration