NCERT Solutions For Class 8 Maths Chapter 11 Mensuration

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 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:

Perimeter of square = perimeter of rectangle

$\Rightarrow$ $4\times side = 2(length + breadth)$

$\Rightarrow$ $4\times 60 = 2(80 + breadth)$

$\Rightarrow$ $240\div 2=(80+breadth)$

$\Rightarrow$

Area of square $= side^{2}= 60^{2}=3600 m^{2}$

Area of rectangle = $= (length\times breadth)=80\times 40=3200m^{2}$

Hence, area of square is greater than area of rectangle.

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 $m^{2}$ .

Answer:

Area of plot = $side^{2}=25^{2}=625m^{2}$

Area of house = $20\times 15=300m^{2}$

Area of garden around house = $625-300=325m^{2}$

the rate = 55 per $m^{2}$ .

the total cost of developing a garden around the house= $55\times 325= Rs 17875$

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:

Length of rectangle is 20 – (3.5 + 3.5) metres=13 metres

Breadth of rectangle = 7 metres

diameter of circular side = 7metres

Area of garden = area of rectangle + 2 times area of semi circular part

Area of garden

$=\left ( 13\times 7 \right ) + \left ( \pi \times 7^{2}\div 4 \right )$

$=\left ( 91 ) + \left ( 38.5 )$

$metres^{2}$

Perimetre of garden

$=\left ( 2\times \pi \times r \right )+\left ( 2\times 13 \right )$

$=\left ( 22 \right )+\left ( 26 \right )$

metres

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 $1080 m^{2}?$ (If required you can split the tiles in whatever way you want to fill up the corners).

Answer:

Area of parallelogram

$=base \times height$

$=24\times10$

$cm^{2}$

a floor of area $1080m^{2}$ = $1080m^{2}\times 10000$ = $10800000cm^{2}$

Number of tiles = $10800000\div 240$ = tiles

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 $c = 2\pi r$ , where r is the radius of the circle.

Answer:

(a) circumference

$= 2.8+\left ( \pi \times 1.4 \right )$

(b)circumference

$=1.5+2.8+1.5+\left ( \pi \times 1.4 \right )$

$=2 +\left ( \pi \times 1.4 \right )+2$

Hence, the perimeter of (b) is highest so ant have to take a longer round in (b).

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.3 area of trapezium

Question:1 Nazma’s sister also has a trapezium-shaped plot. Divide it into three parts as shown (Fig 11.4). Show that the area of a trapezium

$WXYZ=h\frac{(a+b)}{2}$ .

Answer:

Area of trapezium WXYZ = Area of triangle with base 'c' + area of rectangle + area of triangle with base'd'

$=(\frac{1}{2}\times c\times h)+(b\times h)+\left ( \frac{1}{2}\times d\times h \right )$

Taking 'h' common, we get

$=(\frac{c}{2}+b+\frac{d}{2})h$

$=h(\frac{c+d}{2}+b)$

Replacing

$=h(\frac{a-b}{2}+b)$

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

Hence proved that the area of a trapezium

$WXYZ=h\frac{(a+b)}{2}$

Question:2 If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression $\frac{h(a+b)}{2}$ .

Answer:

Area of trapezium WXYZ = Area of traingle with base'c'+area of rectangle +area of triangle with base 'd'

$=\left ( \frac{1}{2}\times c\times h \right )+\left ( b\times h \right )+\left ( \frac{1}{2}\times d\times h \right )$

$=\left ( 30 \right )+\left ( 120 \right )+\left ( 20 \right )$

$cm^{2}$

the area WXYZ by the expression

. $\frac{h(a+b)}{2}$

$=\frac{10 \left ( 22+12 \right )}{2}$

$=\frac{10 \left ( 34 \right )}{2}$

$=10\times 17$

$=170 cm^{2}$

Hence,we can conclude area from given expression and calculated area is equal.

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.3 area of trapezium

Question:1 Find the area of the following trapeziums (Fig 11.8)

Answer:

(i) Area of trapezium

$=\frac{1}{2}\ (Sum\: of\: Parallel\: sides)\times (Distance\: between \:parallel \:sides)$

$=\frac{1}{2}\times 16\times 3$

$=27cm^{2}$

(ii)

Area of trapezium

$=\frac{1}{2}\ (Sum\: of\: Parallel\: sides)\times (Distance\: between \:parallel \:sides)$

$=\frac{1}{2}\times (10+5)\times 6$

$cm^{2}$

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.4 area of a general quadrilateral

Question:1 We know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already? (Fig 11.12)

Answer:

$Area \:of \:quadrilateral \:ABCD = Area \:of( \bigtriangleup ABD \:+\:\bigtriangleup BCD)$

$=\left ( \frac{1}{2}\times b\times h \right )+\left ( \frac{1}{2}\times b\times h \right )$

This agree with the formula that we know already.

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.4.1 area of a special quadrilaterals

Question:1 Find the area of these quadrilaterals (Fig 11.14).

Answer:

(i) Area=

$=\frac{1}{2}\times d\left ( h1+h2 \right )$

$=\frac{1}{2}\times 6\times \left ( 3+5 \right )$

$=3\times \left ( 8 \right )$

$cm^{2}$

(ii) Area=

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

$=\frac{1}{2}\times 7\times \left ( 6 \right )$

$= 7\times \left (3 \right )$

$cm^{2}$

(iii)Area=

$=\frac{1}{2}\times d\left ( h1+h2 \right )$

$=\frac{1}{2}\times 8\times \left ( 2+2 \right )$

$=4\times 4$

$cm^{2}$

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.5 area of a polygon

Question:(i) Divide the following polygons (Fig 11.17) into parts (triangles and trapezium) to find out its area.

FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR

Answer:

(i)

Area of polygon EFGHI = area of $\triangle$ EFI + area of quadrilateral FGHI

Draw a diagonal FH

Area of polygon EFGHI = area of $\triangle$ EFI + area of $\triangle$ FGH + area of $\triangle$ FHI

Area of polygon MNOPQR = area of quadrilateral NMRQ+area of quadrilateral NOPQ

Draw diagonal NP and NR .

Area of polygon MNOPQR = area of $\triangle$ NOP+area of $\triangle$ NPQ +area of $\triangle$ NMR +area of $\triangle$ NRQ

Question:(ii) Polygon ABCDE is divided into parts as shown below (Fig 11.18). Find its area if , , , and perpendiculars , , . Area of Polygon ABCDE = area of $\Delta AFB+...$

Area of $\Delta AFB=\frac{1}{2}\times AF\times BF=\frac{1}{2}\times 3\times 2=....$

Area of trapezium $FBCH=FH\times \frac{(BF+CH)}{2}$

$=3\times \frac{(2+3)}{2}[FH=AH-AF]$

Area of $\Delta CHD=\frac{1}{2}\times HD\times CH=.....,$ Area of $\Delta ADE=\frac{1}{2}\times AD\times GE=...$

So, the area of polygon ABCDE = ....

Answer:

Area of Polygon ABCDE = area of $\traingle$ $\bigtriangleup AFB$ + area of trapezium BCHF + area of $\traingle$ $\bigtriangleup CDH$ +area of $\traingle$ $\bigtriangleup AED$

$=23.5 cm^{2}$

Question:(iii) Find the area of polygon MNOPQR (Fig 11.19) if , , , , NA, and are perpendiculars to diagonal MP.

Answer:

the area of polygon MNOPQR

= area of $\triangle$ MAN + area of trapezium ACON+area of $\triangle$ CPO + area of $\triangle$ MBR+area of trapezium BDQR+area of $\triangle$ DPQ

$cm^{2}$

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 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 table =

$=\frac{1}{2}\times sum \, of \, parallel\, sides\times distance\, between\, parallel\, sides$

$=\frac{1}{2}\times\left ( 1+1.2 \right )\times 0.8$

$= 0.4\times 2.2$

$m^{2}$

Question:2 The area of a trapezium is $34 cm^{2}$ and the length of one of the parallel sides is and its height is . Find the length of the other parallel side.

Answer:

Let the length of the other parallel side be x

area of a trapezium =

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

$= \left ( 10+x \right )=17$

Hence,the length of the other parallel side is 7cm

Question:3 Length of the fence of a trapezium shaped field is . If , and , find the area of this field. Side is perpendicular to the parallel sides and .

Answer:

, and ,

Length of the fence of a trapezium shaped field = =

Area of trapezium =

$\frac{1}{2}\times sum \, of\, parallel\, sides\times height$

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

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

$= (44)\times 15$

$=660m^{2}$

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:

The diagonal of a quadrilateral shaped field is 24 m

the perpendiculars are 8 m and 13 m.

the area of the field =

$\frac{1}{2}\times d\times (h1+h2)$

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

$=12\times 21$

$=252 m^{2}$

Question:5 The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer:

Area of rhombus =

$\frac{1}{2}\times product\, of\, diagonals$

$=\frac{1}{2}\times 7.5\times 12$

$= 7.5\times 6$

$= 45 cm^{2}$

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:

Rhombus is a type of parallelogram and area of parallelogram is product of base and height.

So,Area of rhombus = base $\times$ height

$= 5\times 4.8$

Let the other diagonal be x

Area of rhombus =

$=\frac{1}{2}\times product\, of\, diagonals$

$24=\frac{1}{2}\times8\times x$

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 $m^{2}$ is 4.

Answer:

Area of 1 tile =

$\frac{1}{2}\times product\, of\, diagonals$

$=\frac{1}{2}\times45\times 30$

$=675 cm^{2}$

Area of 3000 tiles =

$=675 \times 3000$

$= 202.5 m^{2}$

Total cost of polishing =

$= 202.5\times 4$

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 $m^{2}$ and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Answer:

Let the length of the side along road be x m.

Then according to question, lenght of side along river will be 2x m.

$Area \:of \:trapezium = \frac{1}{2}h(a+b)$

So equation becomes :

<img alt="\frac{1}{2}100(x+2x) = 10500" height="37"

src="https://lh6.googleusercontent.com/0HnLqMf74XRkw5R40cffRj-y_QkSsgufsUcmW5mmaD8VCWCTwI72h5tPNtx5g5hUvN8VZUT7dgu7ZWSQ0niRe1B7k1SrC19GbilZG7B9-Ca9KM9MZ16_7Wq_lBXMO0iQ0m8cAqk" style="margin-left: 0px; margin-top: 0px;" width="172" />

So the length of the side along the river is 2x = 140m.

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:

Area of octagonal surface = Area of rectangular surface + 2(Area of trapezium surface)

Area of recatangular surface = $11\times5 = 55 m^2$

Area of trapezium surface =

$\frac{1}{2}\times4(11+5) = 2\times16 = 32 m^2$

So total area of octagonal surface = 55 + 2(32) = 55 + 64 = 119

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:

Area of pentagonal park according to Jyoti's diagram :-

= 2(Area of trapezium) = $2(\frac{1}{2}\times\frac{15}{2}\times45) = \frac{15\times45}{2} = 337.5 m^2$

Area of pentagonal park according to Kavita's diagram :-

= Area of triangle + Area of square.

$= \frac{1}{2}\times15\times15 + 15\times15 = 112.5 + 225 = 337.5 m^2$

Question:11 Diagram of the adjacent 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 same.

Answer:

Area of opposite sections will be same.

So area of horizontal sections,

$=\frac{1}{2}\times4(16+24) = 2(40) = 80\ cm^2$

And area of vertical sections,

$= \frac{1}{2}\times8(20+28) = 4(48) = 96\ cm^2$

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.7.1 cuboid

Question:1 Find the total surface area of the following cuboids (Fig 11.31):

Answer:

(i) Total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl)

$=2(6\times4 + 4\times2+6\times2) = 2(24+8+12)$

$= 2(24+8+12) = 88\ cm^2$

(ii) Total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl)

$= 2(4\times4+4\times10+10\times4) = 2(16 + 40+40)$

$= 2(96) = 192\ cm^2$

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.7.2 cube

Question: Find the surface area of cube A and lateral surface area of cube B (Fig 11.36).

Answer:

Surface area of cube A =

= $6(10)^2 = 600\ cm^2$

Lateral surface area of cube B =

$=4(8)^2 = 4\times64 = 256\ cm^2$

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.7.3 cylinders

Question:1 Find the total surface area of the following cylinders following figure

Answer:

Total surface area of cylinder = 2πr (r + h)

(i) Area =

$2\Pi \times14(14+8) = 2\Pi\times 14(22) = 1935.22\ cm^2$

(ii) Area =

$2\Pi \times1(1+2) = 2\Pi\times 1(3) = 18.84\ cm^2$

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 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:

Surface area of cuboid (a) $= 2(60\times40 + 40\times50 + 50\times60) =14800\ cm^2$

Surface area of cube (b) $= 6(50)^2 = 15000\ cm^2$

So box (a) requires the lesser amount of material to make.

Question:2 A suitcase with measures $80 cm \times 48 cm \times 24 cm$ is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width is required over such suitcases?

Answer:

Surface area of suitcase $= 2(80\times48+48\times24+24\times80) = 13824\ cm^2$ .

Area of such 100 suitcase will be $1382400\ cm^2$

So lenght of tarpaulin cloth $= \frac{1382400\ cm^2 }{96\ cm} = 14400\ cm\ or\ 144m$

Question:3 Find the side of a cube whose surface area is $600 cm^{2}$ .

Answer:

Surface area of cube $=\ 6l^2$

So,

or $l = 10\ cm$ .

Thus side of cube is 10 cm.

Question: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.

Answer:

Required area = Total area - Area of bottom surface

Total area $= 2(1\times2+ 2\times1.5 + 1.5\times1 ) = 13\ m^2$

Area of bottom surface $= 1\times2 = 2\ m^2$

So required area = $13-2\ m^2 = 11\ m^2$

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 $100 m^{2}$ of area is painted. How many cans of paint will she need to paint the room?

Answer:

Total area painted by Daniel :

$= 2(15\times10+10\times7+7\times15) - 15\times10$ ( $\because$ Bottom surface is excluded.)

So, Area

$= 650 - 150\ m^2 = 500\ m^2$

No. of cans of paint required

$=\frac{ 500\ m^2}{100\ m^2} = 5$

Thus 5 cans of paint are required.

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

Answer:

The two figures have same height.The diference between them is one is cylinder and another is cube.

lateral surface area of cylinder = $2\times \pi \times r\times h$

$=2\times \pi \times 3.5\times 7$

$=154 cm^{2}$

lateral surface area of cube = $4\times side^{2}$

$=4\times 7^{2}$

$=4\times 49$

$=196 cm^{2}$

Cube has a larger lateral surface area.

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:

Total surface area of cylinder = 2πr (h + r)

$= 2\Pi \times7(7+3) = 14\Pi \times10 = 440\ m^2$

Question: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 . Find the perimeter of rectangular sheet?

Answer:

Lenght of rectangular sheet

$= \frac{4224}{33} = 128\ cm$

So perimeter of rectangular sheet = 2(l + b)

$= 2(128 + 33) = 322\ cm$ .

Thus 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:

Area in one complete revolution of roller = 2πrh.

$= 2\Pi \times0.42\times1 = 2.638\ m^2$

So area of road $= 2.638\ m^2\times 750 = 1979.20\ m^2$

Question:10 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 the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Answer:

Area of label = 2πrh

$= 2\Pi \times(7)\times(16) = 703.71\ cm^2$

Here height is found out as = 20-2-2 = 16 cm.

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.8.1 cuboid

Question:1 Find the volume of the following cuboids

Answer:

(i) Volume of cuboid is given as:

$Volume\ of\ cuboid = length\times breadth\times height$ ,

So, Given that Length = 8cm, Breadth = 3cm, and height = 2 cm so,

its volume will be = $8cm\times 3cm\times 2cm = 48cm^3$ .

Aslo for Given Surface area of cuboid and height = 3 cm we can easily calculate the volume:

$Volume = Surface\ area\times height$ ;

So, Volume = $Volume = 24m^2\times \frac{3m}{100} = 0.72m^3$

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.8.2 cube

Question: Find the volume of the following cubes

(a) with a side 4 cm (b) with a side 1.5 m

Answer:

(a) Volume of cube having side equal to 4cm will be

$Volume = Side\times side\times side$ or $Volume = 4cm\times 4cm\times 4cm= 64cm^3$ .

(b) When having side length equal to 1.5m then ,

$Volume = 1.5m\times 1.5m\times 1.5m= 3.375m^3$ .

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.8.3 cylinder

Question:1 Find the volume of the following cylinders.

Answer:

(i) The volume of a cylinder given as = $\Pi \times r^2\times length$ .

or given radius of cylinder = 7cm and length of cylinder = 10cm.

So, we can calculate the volume of the cylinder = $\Pi \times (7cm)^2\times 10cm$ (Take the value of $\Pi = \frac{22}{7}$ )

The volume of cylinder = .

(ii) Given for the Surface area = and height = 2m .

we have .

$Volume\ of\ cylinder = 250m^2\times 2m = 500m^3.$

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 11.4

Question:1(a) Given a cylindrical tank, in which situation will you find surface area and in
which situation volume.

To find how much it can hold.

Answer:

(a) To find out how much the cylindrical tank can hold we will basically find out the volume of the cylinder.

Question:1(b) Given a cylindrical tank, in which situation will you find surface area and in
which situation volume.

Number of cement bags required to plaster it.

Answer:

(b) if we want to find out the cement bags required to plaster it means the area to be applied, we then calculate the surface area of the bags .

Question:1(c) Given a cylindrical tank, in which situation will you find surface area and in
which situation volume.

To find the number of smaller tanks that can be filled with water from it.

Answer:

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:

Given the diameter of cylinder A = 7cm and the height = 14cm.

Also, the diameter of cylinder B = 14cm and height = 7cm.

We can easily suggest whose volume is greater without doing any calculations:

As volume is directly proportional to the square of the radius of cylinder and directly proportional to the height of the cylinder hence

B has more Volume as compared to A because B has a larger diameter.

Verifying:

Volume of A : $\pi\times r^2\times height = \pi\times (\frac{7cm}{2})^2\times 14cm =539cm^3$

and Volume of B : $\pi\times r^2\times height = \pi\times (\frac{14cm}{2})^2\times 7cm =1,078cm^3$ .

Hence clearly we can see that the volume of cylinder B is greater than the volume of cylinder A.

The cylinder B has surface area of = $2\times \pi\times r(r+h) = 2\times \pi\times \frac{14cm}{2}(\frac{14}{2}+7) = 616cm^2$ .

and the surface area of cylinder A = $2\times \pi\times r(r+h) = 2\times \pi\times \frac{7cm}{2}\times (\frac{7}{2}+14) = 385cm^2$ .

The cylinder with greater volume also has greater surface area.

Question:3 Find the height of a cuboid whose base area is 180 cm 2 and volume is 900 cm 3?

Answer:

Given that the height of a cuboid whose base area is and volume is ;

As $Volume\ of\ cuboid = Base\ area\times height$

So, we have relation: $900cm^3= 180cm^2\times height$

or, Height = 5cm.

Question: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?

Answer:

So given the dimensions of cuboid $60 cm \times 54 cm \times 30 cm$ hence it's the volume is equal to =

We have to make small cubes with side 6cm which occupies the volume = $6cm\times 6cm\times 6cm = 216cm^3$

Hence we have now one cube having side length = 6cm volume = .

So, total numbers of small cubes that can be placed in the given cuboid = $\frac{97,200}{216} =450$

Hence 450 small cubes can be placed in that cuboid.

Question:5 Find the height of the cylinder whose volume is 1.54 m 3 and diameter of the base is 140 cm ?

Answer:

Given that the volume of the cylinder is and having its diameter of base = 140cm.

So, as $Volume\ of\ cylinder = Base\ area\times height$ ;

hence putting in the relation we get;

$1.54m^3= (\pi\times (\frac{1.4m}{2})^2)\times height$

The height of the cylinder would be = .

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 the cylinder = V = $\Pi r^2h$

$\Pi (1.5)^2\times7 = 49.48\ m^3$

So the quantity of milk in litres that can be stored in the tank is 49500 litres.

$\left ( \because 1\ m^3 = 1000\ litres \right )$

Question:7(i) If each edge of a cube is doubled,

how many times will its surface area increase?

Answer:

The surface area of cube =

So if we double the edge l becomes 2l.

New surface area =

Thus surface area becomes 4 times.

Question:7(ii) If each edge of a cube is doubled,

how many times will its volume increase?

Answer:

Volume of cube =

Since l becomes 2l, so new volume is : .

Hence volume becomes 8 times.

Question:8 Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is $108\ m^3$ , find the number of hours it will take to fill the reservoir.

Answer:

Given that the water is pouring into a cuboidal reservoir at the rate of 60 litres per minute.

Volume of the reservoir is $108\ m^3$ , then

The number of hours it will take to fill the reservoir will be:

As we know .

Then $108\ m^3$ = 108,000 Litres ;

Time taken to fill the tank will be:

$\frac{1,08,000litres}{60litres\ per\ minute} = 1,800 minutes$

or $\frac{1,800hours}{60} =30hours$ .

NCERT Class 8 Mathematics Solutions

Chapter 01 - Rational Numbers

Chapter 02 - Linear Equations in One Variable

Chapter 03 -Understanding Quadrilaterals

Chapter 04 - Practical Geometry

Chapter 05 - Data Handling

Chapter 06 - Squares and Square Roots

Chapter 07 - Cubes and Cube Roots

Chapter 08 - Comparing Quantities

Chapter 09 - Algebraic Expressions and Identities

Chapter 10 - Visualising Solid Shapes

Chapter 11 - Mensuration

Chapter 12 - Exponents and Powers

Chapter 13 - Direct and Indirect proportions

Chapter 14 - Factorisation

Chapter 15 - Introduction to Graphs

Chapter 16 - Playing with Numbers

BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET & IIT JEE COACHING

[LATEST]$type=sticky$show=home$rm=0$va=0$count=4$va=0

NCERT Solutions For Class 8 Maths Chapter 11 Mensuration

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 11.1

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.3 area of trapezium

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.3 area of trapezium

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.4 area of a general quadrilateral

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.4.1 area of a special quadrilaterals

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.5 area of a polygon

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 11.2

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.7.1 cuboid

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.7.2 cube

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.7.3 cylinders

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 11.3

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.8.1 cuboid

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.8.2 cube

NCERT solutions for class 8 maths chapter 11 mensuration topic 11.8.3 cylinder

NCERT solutions for class 8 maths chapter 11 mensuration-Exercise: 11.4

NCERT Class 8 Mathematics Solutions

Want to know more

INNER POST ADS

Latest IITJEE Articles$type=three$c=3$author=hide$comment=hide$rm=hide$date=hide$snippet=hide

Latest NEET Articles$type=three$c=3$author=hide$comment=hide$rm=hide$date=hide$snippet=hide

Request a Call Back

Testimonials

Nothing beats the satisfaction of a job well done! Here are some testimonials from Clear Exam.