NCERT solutions class 8 maths chapter 11 mensuration is one of the most common topics from where questions are asked in the examination. The NCERT solutions for class 8 maths chapter 11 mensuration is provided below. Mensuration is one of the most important concepts of mathematics. Students must learn this chapter with seriousness to excel in their examination. The NCERT solutions for class 8 maths chapter 11 is prepared by teachers so in such a manner that students can understand the concepts with ease. Check chapter 11 for class 8 NCERT Solutions pdf given below.

### NCERT Solutions Class 8 Maths Chapter 11 Exercises

- NCERT Solutions Class 8 Maths Chapter 11 Mensuration Exercise 11.1
- NCERT Solutions Class 8 Maths Chapter 11 Mensuration Exercise 11.2
- NCERT Solutions Class 8 Maths Chapter 11 Mensuration Exercise 11.3

**Exercise 11.1**

*1. Â A rectangle and a square field with given measurements have the same perimeter are given below in the figure. Which of the following has larger area?*

**Solution:**

Perimeter of square = 4 Ã— side of square

= 4 Ã— 80

= 320 m

Perimeter of rectangle = 2 Ã— (length + breadth)

= 2 Ã— (100 + breadth)

= 200 + (2 Ã— breadth)

But perimeter of both the fields is same

Therefore,

320 = 200 + (2 Ã— breadth)

120 = 2 Ã— breadth

Breadth = 60 m

Area of square = \((side)^{2}\)

= \((80 m)^{2}\)

= 6400 \(m^{2}\)

Area of rectangle = Length Ã— breadth

= (100 Ã— 60) \(m^{2}\)

= 6000 \(m^{2}\)

Therefore, Area of square is larger compared to area of rectangle.

*2. Â Mrs. Sharma has a square plot with measurements as given in the figure. She wants to build house at the center of the plot and garden around the house. Find the cost of developing the garden at the cost of Rs 60 per \(m^{2}\).*

**Solution:**

Area of square plot = \((40 m)^{2}\)

= 1600 \(m^{2}\)

Area of house = Length Ã— breadth

= (30 Ã— 35) \(m^{2}\)

= 1050 \(m^{2}\)

Area of garden = Area of square plot â€“ area of house

= (1600 â€“ 1050) \(m^{2}\)

= 550 \(m^{2}\)

Cost of developing the garden = 60 per \(m^{2}\)

Therefore, the total cost of developing the garden area 550 \(m^{2}\)

= Rs( 550Ã— 60)

= Rs 33000

*Â *

*3. A rectangular garden is in the middle and has semicircular ends as shown in the figure. Find the perimeter and the area of the garden.*

*[Length of rectangle = 30 â€“ (7 + 7) metres]*

**Solution:**

Length of rectangle = 30 â€“ (7 + 7) metres

= 16metres

Circumference of one semi â€“ circle = Ï€r

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

= 22 metres

Circumference of two semi â€“ circle = 2Ã— Ï€r

= 2 Ã— 22

= 44 metres

Perimeter of garden = PQ + perimeter of both the semi â€“ circles + RS

= 16 + 44 + 16

= 76 metres

Area of garden = Area of rectangle + 2 Ã— area of semicircles

= [(16 Ã— 14) + (2 Ã— \(\frac{1}{2} \times \frac{22}{7} \times (7)^{2}\))] \(m^{2}\)

= (224 + 154) metres

= 378 metres

*Â *

*4. A flooring tile has parallelogram shape, its base is 30 cm, and height is 10 cm. How many tiles will be required to cover an area of 1110 \(m^{2}\)?*

*Â *

**Solution:**

Area of a tile = base Ã— height

= 30 Ã— 10

= 300 \(cm^{2}\)

Number of tiles required = \(\frac{Area of floor}{Area of each tile}\)

= \(\frac{1110 m^{2}}{300 cm^{2}}\)

= \(\frac{(1110 \times 10000) cm^{2}}{300 cm^{2}}\)

= 37000 tiles

*5. An ant is moving around food pieces of various shapes lying on the floor. For which food piece would ant have to take a longer route?*

*(Circumference = 2Ï€r, where r is radius)*

**Solution:**

(a) Radius = \(\frac{4.2}{2}\) cm

= 2.1 cm

Perimeter of the piece = 4.2 cm + Ï€r

= (4.2 + \(\frac{27}{7} \times 2.1\)) cm

= 10.8 cm

(b) Radius = \(\frac{4.2}{2}\) cm

= 2.1 cm

Perimeter of the piece = [2.5 + 4.2 + 2.5 + Ï€(2.1)] cm

= [9.2 + (\(\frac{22}{7} \times 2.1\))] cm

= [9.2 + 6.6] cm

= 15.8 cm

(c) Radius = \(\frac{4.2}{2}\) cm

= 2.1 cm

Perimeter of the piece = 4 cm + Ï€r cm + 4 cm

= (8 + \(\frac{22}{7} \times 2.1\)) cm

= 8 + 6.6 cm

= 14.6 cm

Therefore, the ant will have to take longer route for the second food piece (b), because the perimeter of the second figure is the greatest compared to other two.

**Exercise 11.2**

*1. The upper surface of a table is trapezium. Find the area if its parallel sides are 2 m and 2.4 m. Perpendicular distance between two parallel lines is 1 m.*

**Solution:**

Area of trapezium = \(\frac{1}{2}\) Ã— (Sum of parallel sides) Ã— (distance between parallel sides)

= \([\frac{1}{2} \times (1 + 1.4) \times (1)]\)

= 0.7 \(m^{2}\)

*2. Area of trapezium is 50 \(cm^{2}\). Measurement of one of the parallel sides is 15 cm and distance between them is 5 cm. What is the measurement of other parallel side?*

**Solution:**

Area of trapezium = 50 \(cm^{2}\)

Height = 5 cm

Let the length of another side be c.

Area of trapezium = \(\frac{1}{2}\) Ã— (Sum of parallel sides) Ã— (distance between parallel sides)

50 \(cm^{2}\) = \([\frac{1}{2} \times (15 + c) \times (5)]\)

100 cm = (15 cm + c cm) Ã— 5 cm

20 cm = 15 + c

C = 5 cm

Therefore, the length of the other side is 5 cm.

* 3. Length of fence of a field which is trapezium shaped PQRS is 150 m. Length of SR = 50 m, QR = 15 m and PQ = 40 m, the area of the field. Sides PS and SR are perpendicular to each other.*

**Solution:**

Length of fence = Perimeter of trapezium

= PQ + QR + RS + SP

150 m = 40 m + 15 m + 50 m + SP m

150 m = 105 m + SP

SP = 45 m

Area of field PQRS = (\(\frac{1}{2}\) Ã— (PQ + SR ) Ã— PS) \(m^{2}\)

= (\(\frac{1}{2}\) Ã— (40 + 50 ) Ã— 45) \(m^{2}\)

= (\(\frac{1}{2}\) Ã— 90 Ã— 45) \(m^{2}\)

= 2025 \(m^{2}\)

*4. The diagonal of quadrilateral field is 26 m and the perpendiculars dropped on it from the remaining opposite vertices are 10 m and 15 m. What is the area of field?*

**Solution:**

d = 26 m

Perpendiculars,

P1 = 10 m and P2 = 15 m

Area of quadrilateral = \(\frac{1}{2}\) Ã— d Ã— (P1 + P2)

= \(\frac{1}{2}\) Ã— 26 mÃ— (10 + 15) m

= 13 Ã— 25 \(m^{2}\)

= 325 \(m^{2}\)

*5. The length of diagonals of a rhombus are 8 cm and 14 cm. What is the area of the rhombus?*

**Solution:**

Area of rhombus = \(\frac{1}{2}\) Ã— ( product of its diagonals)

= \(\frac{1}{2}\) Ã— (8 cm Ã— 14 cm)

= 56 \(cm^{2}\)

*6. Find area of rhombus whose sides are 10 m. Its altitude is 6 m. One of its diagonals is 5 m long. What is the length of the other diagonal?*

**Â **

**Solution:**

Let the length of other diagonal be d2.

Area of rhombus = Base Ã— Height

= 10 m Ã— 6 m

= 60 \(m^{2}\)

Area of rhombus = \(\frac{1}{2}\) Ã— ( product of its diagonals)

**Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â **Â Â 60 = \(\frac{1}{2}\) Ã— (5 m Ã— d2 m)

120 = 5 m Ã— d2 m

d2 = 24 m

Therefore, the length of the other diagonal d2 is 24 m.

* 7. Floor of a house consists of 2000 tiles. These tiles are rhombus in shape and measure of its diagonals are 25 cm and 20 cm. What is the cost of polishing the floor if it costs Rs 5 perÂ \(m^{2}\)?*

**Solution:**

Area of rhombus = \(\frac{1}{2}\) Ã— ( product of its diagonals)

Area of a tile = (\(\frac{1}{2}\) Ã— (25 cm Ã— 20 cm)

= 250 \(cm^{2}\)

Area of 2000 tiles = (2000 Ã— 250) \(cm^{2}\)

= 500000 \(cm^{2}\)

= 50 \(m^{2}\)

Cost of polishing per \(m^{2}\) = Rs 5

Cost of polishing 50 \(m^{2}\) = Rs (5 Ã— 50)

= Rs 250

Therefore, the cost of polishing the floor is Rs 250.

*Â *

*8. Rohan wants a field in the shape of trapezium. Its side along the pond is parallel to and double the side along the road. Area of the field is 10200 \(m^{2}\). The perpendicular distance between two parallel sides is 200 m. What is the length of the side along the river?*

**Solution:**

Length of the field along the road = a

Length of the field along the pond = 2a

Area of trapezium = \(\frac{1}{2}\) Ã— (Sum of parallel sides) Ã— (distance between parallel sides)

10200 \(m^{2}\) = \(\frac{1}{2}\) Ã— (a + 2a) Ã— (200 m)

3a = (\(\frac{10200 \times 2}{200}\)) m

3a = 102

a = 34 m

*Â *

*9. Top surface of a table is regular octagon in shape as shown in the figure. What is the area of octagonal surface table?*

*Â Â Â Â Â Â Â Â Â Â Â *

**Solution:**

Side of the regular octagon = 10 cm

Area of PQRW = Area of STUV

Area of PQRW = \([\frac{1}{2} \times (6) \times (15 + 10)] m^{2}\)

= \([\frac{1}{2} \times (6) \times (25)] m^{2}\)

= 75 \(m^{2}\)

Area of rectangle WRSV = 15 Ã— 10

= 150 \(m^{2}\)

Area of octagon = Area of PQRW + Area of STUV + Rectangle WRSV

= (75 + 75 + 150) \(m^{2}\)

= 300 \(m^{2}\)

*10. There is a cafÃ© in the shape of pentagon as shown in the figure. To find the area of the cafÃ©, Sita and Gita divided it into two parts.Â **Find the area of cafÃ© using both the ways. Can you think about any other way to find the area of cafÃ©?*

**Â ****Â **

**Solution:**

Sita

**Â **Area of cafÃ© = 2 (Area of trapezium PQRO)

= [2 Ã— \(\frac{1}{2} \times (20 + 40) \times (\frac{20}{2}) ]m^{2}\)

= (60 m) Ã— (10 m)

= 600 \(m^{2}\)

Gita

Area of cafÃ© = Area of \(\Delta PQT\) + Area of square QRST

= \([\frac{1}{2} \times 20 \times (40-20)]\) + \([20^{2}]\) \(m^{2}\)

= [200 + 400] \(m^{2}\)

= 600 \(m^{2}\)

*11. Diagram of the adjacent picture has inner dimensions 20 cmÂ Ã—Â 24 cm and outer dimensions 26 cmÂ Ã—Â 30 cm. What will be the area of each section of frame if the width is same in every section?*

**Solution:**

Width is same for every section as it is given.

NE = EA = FB = FH = GI = GJ = KD = DM

NH = NE + EF + FH

30 = NE + 24 + FH

NE + FH = 30 cm â€“ 24 cm = 6 cm

NE = FH = 3 cm

Therefore, NE = EA = FB = FH = GI = GJ = KD = DM = 3 cm

Area of EQRF = Area of DGSP

= \([\frac{1}{2} (24 + 30)(3)]\) \(m^{2}\)

= 81 \(m^{2}\)

Area of PDEQ = Area of GFRS

**EXERCISE â€“ 11.3**

**Â **

*1. Two cuboidal boxes as shown in the given figure. Find out, which box would use up less material?*

**Solution:**

As we know that,

Total surface area of the cuboid = 2 (lhÂ +Â bhÂ +Â lb)

Total surface area of the cube = 6 (l) \(^{2}\)

(a) From the figure-:

Total surface area of the cuboid = [2{(70) (30) + (30) (45) + (70) (45)}] cm\(^{2}\)

= [2(2100 + 1350 + 3150)]Â cm\(^{2}\)

= (2 x 6600) cm\(^{2}\)

= 13200 cm\(^{2}\)

(b) From the figure-:

Total surface area of the cube = 6 \(\left ( 70 cm\right )^{2}\) = 29400Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â cm\(^{2}\)

Thus, the cuboidal box (a) would use less material.

*2. Â A suitcase with measures 70 cm x 36 cm x 14 cm need to be covered with a blanket cloth. How many meters of cloth having width of 100 cm is required to cover 150 such suitcases?*

**Â **

**Solution:**

Here,Â Total surface area of suitcase = 2 (lhÂ +Â bhÂ +Â lb)

= 2[(70) (36) + (36) (14) + (14) (70)]

= 2[2520 + 504 + 980]

= 8008 cm\(^{2}\)

Total surface area of 150 suitcases = (8008 x 150) cm\(^{2}\) = 1201200 cm\(^{2}\)

\(\ therefore\) Required blanket cloth = Length Ã— Breadth

1201200 cm\(^{2}\) = Length Ã— 100 cm

Length = \(\left ( \frac{1201200}{100} \right )cm=12012cm=120.12m\)

Hence, 120.12m of blanket cloth is required to cover 150 suitcases.

*3. A cube having surface area 1200 cm\(^{2}\). Calculate the side of the cube.*

**Solution:**

Given, surface area of cube = 1200 cm\(^{2}\)

Let us assume the length of each side of the cube be â€˜mâ€™.

Now, Surface area of cube = 6 \(\left (side \right )^{2}\)

1200 cm\(^{2}\) = \(6m^{2}\)

\(m^{2}\) = 200 cm\(^{2}\)

m = 14.142 cm

Hence, the side of the cube is 14.142 cm.

*4. Tejus painted the exterior of the house of measure 3 m x 6 m x 2.5 m. How much surface area did he cover if he painted all except the bottom of the house?*

**Â **

**Solution:**

Length (l) of the house = 6 m

Breadth (b) of the house = 3 m

Height (h) of the house = 2.5 m

Area of the house that was painted = 2hÂ (lÂ +Â b) +Â lb

= [2 X 2.5 X (6 + 3) + (6) (3)] \(m^{2}\)

= [5 X 9 + 18] \(m^{2}\)

= 135 \(m^{2}\)

*5. Raghav is painting the interior walls and the ceiling of a cuboidal hall having length (l), breadth (b) and height (h) of 30 m, 20 m and 10 m respectively. From each tin of paint 200 m ^{2}of area is covered. How many tins of paint will he need to paint the total interior of the room?*

**Â **

**Solution:**

Here,Â Length (l) = 30 m, breadth (b) = 20 m, height (h) = 10 m

Area of the hall to be painted = Area of the wall + Area of the ceiling

= 2hÂ (lÂ +Â b) +Â lb

= Â [2(10) (30 + 20) + 30 Ã—20] \(m^{2}\)

= [20 x 50 + 600] \(m^{2}\)

= 1600 \(m^{2}\)

Given that, 200 \(m^{2}\) area can be painted from each tin.

Number of tins required to paint an area of 1600Â \(m^{2}\)

= \(\frac{1600}{200}=8\)

Hence, 8 tins are required to paint the interior walls and the ceiling of the cuboidal hall.

**6. Describe what are the similarities and the difference between these two figures. Which box has larger lateral surface area?**

**Â **

**Solution:**

From the given two figures, the similarity is that both of them have same heights.

And the difference between both the figures is that one is cylinder and the other one is a cube.

Lateral surface area of the cube = \(4l^{2}\) = \(4\left ( 10cm \right )^{2}=400cm^{2}\)

Lateral surface area of the cylinder = \(2\pi rh\) = \(\left (2 \times \frac{22}{7} \times \frac{10}{2} \times 10 \right ) cm^{2}\)

= 314.285 cm^{2}

Therefore, the cube has larger lateral surface area.

**Â **

*7. A closed cylindrical container of radius and height 14m and 4m is made of steel sheet, Calculate the quantity of sheet required to make the container.*

**Solution:**

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

= \(2\times \frac{22}{7}\times 14 \left ( 14 + 4 \rightÂ )\)

= 1584 m^{2}

Therefore, the required metal sheet would be 1584 m^{2}.

**Â **

**8. Given a hollow cylinder of lateral surface area as 5000 cm ^{2}. When we will cut it along the height then it will form a rectangular sheet of width 50 cm. Calculate the perimeter of the new formed sheet.**

**Â **

**Solution:**

When a rectangle sheet will form by cutting a hollow cylinder, then the area for both will remain the same.

=> Area of Cylinder = Area of Rectangular sheet

=> 5000 cm^{2}Â = 50 cm X Length

=> \(Length = \frac{5000}{50}\) = 100 cm

Therefore, the length of the rectangle sheet will be: 100 cm

Now, the perimeter will be = 2 ( length + width)

= [2 ( 100 + 50 )]cm

= ( 2 X 150 ) cm

= 300 cm

**9. It takes a road roller 700 complete revolution to move once over a labeled road. Calculate the total area of road while the diameter of the road roller is 90 cm and its length is 15 m.**

**Â **

**Solution:**

In one revolution, the roller will cover an area equal to its lateral surface area.

Thus, in 1 revolution, area of the road covered = =Â \(2\pi rh\)

= \(2 \times \frac{22}{7} \times 45 cm \times 15m\)

= \(2 \times \frac{22}{7} \times \frac{45}{100} m \times 15m\)

= \(\frac{29700}{700}\) m^{2}

In 700 revolutions, area of the road covered

= \(\left (700 \times \frac{29700}{700} \right )\) m2

= 29700 m^{2}

**10. A manufacturer packages its chocolate powder in a cylindrical box, which is has a diameter of 22 cm and height 30 cm. The manufacturer places its sticker around the body of the box (Refer to the figure). If the sticker placed is 3 cm from top and bottom, calculate the area of the sticker.**

**Solution:**

Height of the sticker = 30cm – 3cm – 3cm = 24cm

Radius of the Sticker = \(\left ( \frac{22}{2} \right )\) = 11cm

Sticker is in the form of a cylinder with radius 11cm and height 24cm.

Area of label = \(2\pi \left ( Radius \right ) \left ( Height \right )\)

= \(\left (2 \times \frac{22}{7} \times 11 \times 24 \right ) cm^{2}\)

= 1659.428 cm^{2}

**Exercise 11.4**

*1. You are given two cylinders 1 and 2. The diameter and height of cylinder 2 is 14cm and 21cm respectively. For Cylinder 1, diameter and height is 21 cm and 14 cm respectively. Without calculating, can you suggest which one of the cylinder will have greater volume? Also, verify the volumes by doing the calculations. Check whether the cylinder with greater volume have greater surface area?*

**Solution:**

The heights and diameters of these cylinders A and B are interchanged.

We know that,

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

If measures ofÂ radiusÂ and heightÂ are same, then the cylinder with greater radius will have greater area.

Radius of Cylinder 1 = \(\frac{14}{2}\) = 7cm

Radius of Cylinder 2 = \(\frac{21}{2}\) cm

Here the radius of cylinder 2 is larger, then, the cylinder 2 will have larger volume.

Now, verifying by calculating.

Volume of cylinder 1 = \(\pi r^{2}h\)

= \(\left (\frac{22}{7} \times \frac{14}{2} \times \frac{14}{2} \times 21 \right )cm^{3}\)

= 3234cm^{3}

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

= \(\left (\frac{22}{7}\times \frac{21}{2} \times \frac{21}{2}\times 14 \right ) cm^{3}\)

= 4851 cm^{3}

Volume of Cylinder 2 is greater.

Surface area of cylinder 1 = \(2\pi r \left ( r + h \right )\)

= \(\left [ 2 \times \frac{22}{7} \times \frac{14}{2}\times \left ( \frac{14}{2} + 21 \right ) \right ]cm^{2}\)

= (44 X 28) cm^{2}

= 1232 cm^{2}

Surface area for cylinder 2Â = \(2\pi r \left ( r + h \right )\)

= \(\left [ 2 \times \frac{22}{7} \times \frac{21}{2}\times \left ( \frac{21}{2} + 14 \right ) \right ]cm^{2}\)

= (33 X 49) cm^{2}

= 1617 cm^{2}

Thus the surface area of cylinder 2 is also greater.

*2. A cuboid having dimensions 80 cm x 40 cm x 25 cm. Calculate the how many of small cubes with sides 8 cm can be placed inside the given cuboid.*

**Solution:**

Volume of cuboid = 80 cm Ã— 40 cm Ã— 25 cm = 80000 cm^{3}

Side of the cube = 8 cm

Volume of the cube = (8)^{3}Â cm^{3}Â = 512 cm^{3}

Required number of cubes = \(\frac{Volume\:of\:the\:cuboid}{Volume\:of\:the\:cube}\)

= \(\frac{80000}{512}=156.25\)

Hence, 156 (approx.) cubes can be placed in the given cuboid.

**Â **

**3. Obtain the height of a cylinder whose volume is 2m ^{3}and diameter of the base is 200 cm?**

**Solution:**

Diameter of the base = 200 cm

Radius (r) of the baseÂ = \(\left ( \frac{200}{2} \right )cm=100cm=\frac{100}{100}m=1m\)

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

2m^{3}Â = \(\frac{22}{7}\times 1m\times 1m\times h\)

h = \(\frac{7\times 2}{22}=\frac{14}{22}m=0.636m\)

Hence, the height of the cylinder isÂ 0.636m

**Â ****Â **

*4. A milk tank with radius and length of 2m and 8m respectively, is in the form of a cylinder. Calculate the quantity of milk (in liters) which can be stored in the tank.*

**Solution:**

Given, Radius of cylinder(r) = 2m

Length of cylinder(h) = 8m

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

= \(\left ( \frac{22}{7} \times 2\times 2\times 8\right )m^{3}\)

= 100.57\(m^{3}\)

\(1m^{3}=1000L\)

Quantity required = (100.57 x 1000)L = 100570 L

Hence therefore, 100570 L of milk can be stored in the tank.

*Â *

*5. If each edge of a cube is doubled,*

*(a) Find out the number of time its surface area will increase.*

*(b) Find out the number of times its volume will increase.*

*Â *

**Solution:**

(a) Let us assume the edge of the cube as l.

Initial surface area = \(6l^{2}\)

If each edge of the cube will be doubled, then it becomes 2l.

So, New surface area = \(6\left ( 2l \right )^{2}=24l^{2}=4\times 6l^{2}\)

Hence, the surface area will be increased by 4 times.

(b) Let us assume the initial volume as \(l^{3}\)

If each edge of the cube will be doubled, then it becomes 2l.

So, New volume = \(\left ( 2l \right )^{3}\) = 8\(l^{3}\) = \(8\times l^{3}\)

Hence, the volume will increase 8 times.

*6. Find the height of a cuboid whose base area is \(200cm^{2}\) and volume is \(1200cm^{3}\)?*

*Â *

**Solution:**

Given: Base area of cuboid = 200 \(cm^{2}\)

Volume of cuboid = 1200 \(cm^{3}\)

We know that,

Volume of cuboid = l b hÂ Â Â Â Â Â Â Â Â Â Â Â Â [\(\ because\) Base area = l x b = 200(given)]

=> 1200 = 200 Â´h

=> h = \(\frac{1200}{200}=6m\).

*7. Water is pouring into a cuboidal tank at the rate of 90 liters per minute. If the volume of tank is 270 \(m^{3}\), calculate the number of hours it will take to fill the tank.*

*Â *

*Solution:*

Volume of tank = 270 \(m^{3}\) =(270 x 1000)L = 270000 L

Rate of pouring water into cuboidal tank = 90 liters/minute

=> (90 x 90)L = 8100 L per hour

âˆ´Â Required number of hours = \(\frac{270000}{8100} = 33.33\) hours = 33 hours (approx..)

It will take 33 hours to fill the tank.

Class 8 is an important phase of a students life, in class 8 mathematics many new concepts are introduced like ,Area of trapezium, area of general quadrilaterals, area of special quadrilaterals and area of polygon. The topics provided here are very much important from from examination point of view. The advanced level of these topics will be taught in further classes, so students are asked to solve the NCERT questions provided in the textbooks. While solving the NCERT questions if students face any kind of difficulties they can refer to the NCERT solutions provided here. The NCERT solutions provided here are prepared by the subject matters experts of BYJUâ€™S.

Mathematics is a subject which requires lots and lots of practice. So apart from solving NCERT solutions students are also advised to solve the questions which were asked in the last 2-3 years, solving the previous year questions will help the students to know about the types of questions asked in the examinations. While solving the previous year questions and sample papers students should keep an eye on the clock so that they can know in how much time they are solving a particular type of question.

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