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Case Study Questions for Class 8 Maths Chapter 11 Mensuration

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Here we are providing Case Study questions for Class 8 Maths Chapter 11 Mensuration.

Mensuration Case Study Questions

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NCERT Solutions for Class 8 Maths Chapter 11 - Mensuration

• NCERT Solutions
• Chapter 11 Mensuration

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration - Free PDF Download

Class 8 students have a lot on their plate in terms of academics. A subject like Maths requires them to spend enough time in solving assignments and preparing for exams. Vedantu’s team of scholars understand the needs and understanding level of Class 8 students and have designed NCERT Solution for Class 8 Maths Chapter 11 keeping all of this in mind. NCERT Solution Class 8 Maths Chapter 11 Mensuration Solution is entirely based on the latest CBSE curriculum. It will be able to clarify all the doubts that you might have in the Mensuration chapter. The problems in the Class 8 Maths Chapter 11 are well explained with diagrams and step-by-step explanations to make it simple for you to follow the solution.

You can also download Class 8 Science to help you to revise complete syllabus ans score more marks in your examinations.

Access NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration

Exercise 11.1

Q1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Ans: Let Side of a Square be ‘ \[{\text{a}}\] ’.

         Length and Breadth of Rectangle are ‘ \[{\text{l}}\] ’ and ‘ \[{\text{b}}\] ’ respectively.

         Perimeter of Square = $4$ (side of square) = $4$ ( \[{\text{a}}\] )

                                                                             = $4$ ( $60$${\text{m}}$ ) = $240{\text{m}}$          

         Perimeter of Rectangle = $2$ (Length+Breadth) = $2$ ( \[{\text{l + b}}\] )

                                                                                  = $2$ ( $80{\text{m}}$ + \[{\text{b}}\] )

         Now, It is given that the perimeter of the square and the perimeter of the rectangle are equal.

         So, $240{\text{m}}$ =  $2$ ( $80{\text{m}}$ + \[{\text{b}}\] )

                 $240{\text{m}}$ = $160{\text{m}}$ + $2$\[{\text{b}}\]

                 $80{\text{m}}$ = 2 \[{\text{b}}\]

                 $40{\text{m}}$ = \[{\text{b}}\]

        Area of square = $\mathop {{\text{(side)}}}\nolimits^{\text{2}} $ = $\mathop {{\text{(60m)}}}\nolimits^{\text{2}} $ = $\mathop {{\text{3600m}}}\nolimits^{\text{2}} $

        Area of Rectangle = Length x Breadth = $(80 x 40)$$\mathop {\text{m}}\nolimits^{\text{2}} $ = $3200$${{\text{m}}^{\text{2}}}$

        Thus, the area of the square field is larger than the area of the rectangular field.

Q2. Mrs. Kaushik has a square plot with the measurement as shown in the following 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 ${\text{Rs}}{\text{.}}\;{\text{55}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$?

Ans: Area of the Square Plot = ${{\text{(side)}}^{\text{2}}}$ = \[{{\text{(25m)}}^{\text{2}}}\] = ${\text{625}}{{\text{m}}^{\text{2}}}$

          Area of the House = Length x Breadth = ${\text{15m x 20m}}$ = ${\text{300}}{{\text{m}}^{\text{2}}}$

          Now, Area of the remaining portion = Area of the square plot – Area of the house

                                                                   =  ${\text{625}}{{\text{m}}^{\text{2}}}$ - ${\text{300}}{{\text{m}}^{\text{2}}}$

                                                                   =   ${\text{325}}{{\text{m}}^{\text{2}}}$

          The cost of developing the garden around the house is ${\text{Rs}}{\text{.}}\;{\text{55}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$ .

          Therefore, total cost of developing the area ${\text{325}}{{\text{m}}^{\text{2}}}$ is ${\text{Rs}}{\text{.}}\;{\text{(55 x 325)}}$ = ${\text{Rs}}{\text{.}}\;{\text{17875}}$ .

Q3. 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 the garden (Length of rectangle is ${\text{20 - (3}}{\text{.5 + 3}}{\text{.5)m}}$)?

Ans: As we have given that length of rectangle = ${\text{[20 - (3}}{\text{.5 + 3}}{\text{.5)]m}}$

                                                                                  =  ${\text{[20 - 7]m}}$ = ${\text{13m}}$

         Breadth = \[{\text{7m}}\] .

         Now, we have to find the circumference of both semi circles.

         As, Diameter = ${\text{7m}}$ , so, Radius(r) = $\dfrac{7}{{\text{2}}}{\text{m}}$ = ${\text{3}}{\text{.5m}}$

         Circumference of one semicircle = ${\text{$\pi$ r}}$ = $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(3}}{\text{.5)m}}$ = ${\text{11m}}$

         Circumference of both circles = ${\text{2 x 11m}}$ = ${\text{22m}}$

         Now, Perimeter of the garden = AB + CD + Length of both semi-circular regions AD & BC

                                                        = ${\text{(13 + 13 + 22)m}}$

                                                        =  ${\text{48m}}$

         Area of the garden = Area of the rectangle + $2 x $ Area of two semi-circular regions

                                       = (Length x Breadth) + ${\text{2 x }}\dfrac{{\text{1}}}{{\text{2}}}{\text{$\pi$ }}{{\text{r}}^{\text{2}}}$

                                       = ${\text{[(13 x 7) + 2 x (}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 3}}{\text{.5 x 3}}{\text{.5)]}}{{\text{m}}^{\text{2}}}$

                                       = ${\text{(91 + 38}}{\text{.5)}}{{\text{m}}^{\text{2}}}$

                                       = \[{\text{129}}{\text{.5}}{{\text{m}}^{\text{2}}}\]

Q4. A flooring tile has the shape of a parallelogram whose base is ${\text{24cm}}$and the corresponding height is ${\text{10cm}}$.How many such tiles are required to cover a floor of area ${\text{1080}}{{\text{m}}^{\text{2}}}$? (If required you can split the tiles in whatever way you want to fill up the corners).  

Ans: Given that the Base of a parallelogram is ${\text{24cm}}$ and Height of a parallelogram is ${\text{10cm}}$ .

          Therefore, Area of Parallelogram = Base x Height

                                                               = ${\text{(24 x 10)c}}{{\text{m}}^2}$

                                                               = ${\text{240c}}{{\text{m}}^2}$ .

          Therefore, Area of parallelogram = Area of one tile.

          Now, we have to find the number of tiles.

          Given that Area of floor = ${\text{1080}}{{\text{m}}^{\text{2}}}$

          So, Number of tiles = $\dfrac{{{\text{Area}}\;{\text{of}}\;{\text{floor}}}}{{{\text{Area}}\;{\text{of}}\;{\text{one}}\;{\text{tile}}}}$

                                         = $\dfrac{{{\text{1080}}{{\text{m}}^{\text{2}}}}}{{{\text{240c}}{{\text{m}}^{\text{2}}}}}$ =  $\dfrac{{{\text{1080 x 10000}}}}{{{\text{240}}}}{\text{c}}{{\text{m}}^{\text{2}}}$           \[(\because \;1{\text{m}} = 100{\text{cm}})\]

                                         = $45000$ tiles.

          Hence, $45000$ tiles are required to cover a floor of area ${\text{1080}}{{\text{m}}^{\text{2}}}$

Q5. 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 =${\text{2$\pi$ r}}$, where r is

the radius of the circle.

Ans: (a) Diameter( \[{\text{d}}\] ) = ${\text{2}}{\text{.8cm}}$ , so radius( \[{\text{r}}\] ) = $\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{cm}}$ = ${\text{1}}{\text{.4cm}}$

                Therefore, Perimeter = ${\text{d + $\pi$ r}}$

                                                  = ${\text{2}}{\text{.8cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$

                                                  = ${\text{[2}}{\text{.8 + (0}}{\text{.2 x 22)]cm}}$

                                                  = ${\text{(2}}{\text{.8 + 4}}{\text{.4)cm}}$

                                                  = ${\text{7}}{\text{.2cm}}$

          (b) Diameter( \[{\text{d}}\] ) = ${\text{2}}{\text{.8cm}}$ , so radius( \[{\text{r}}\] ) = $\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{cm}}$ = ${\text{1}}{\text{.4cm}}$

                Therefore, Perimeter = ${\text{1}}{\text{.5cm + 2}}{\text{.8cm + 1}}{\text{.5cm + $\pi$ (1}}{\text{.4cm)}}$

                                                     = ${\text{5}}{\text{.8cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$

                                                     = ${\text{5}}{\text{.8cm + 4}}{\text{.4cm}}$

                                                     = ${\text{10}}{\text{.2cm}}$ .

           (c) Radius( \[{\text{r}}\] ) = $\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{cm}}$ = ${\text{1}}{\text{.4cm}}$

                Perimeter = ${\text{2cm + $\pi$ r + 2cm}}$

                                =  ${\text{4cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$

                                = ${\text{4cm + 4}}{\text{.4cm}}$

                                = ${\text{8}}{\text{.4cm}}$ .

Thus, the ant will have to take a longer round for the food piece (b) because its perimeter is  ${\text{10}}{\text{.2cm}}$ which is the greatest among all.

Exercise 11.2

Q1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are ${\text{1m}}$and ${\text{1}}{\text{.2m}}$and perpendicular distance between them is${\text{0}}{\text{.8m}}$.

Ans: Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ ( Sum of parallel sides)  x  (Distances between parallel sides) }}$

                                           = $\dfrac{{\text{1}}}{{\text{2}}}{\text{[(1m + 1}}{\text{.2m) x 0}}{\text{.8m]}}$

                                           =  $\dfrac{{\text{1}}}{{\text{2}}}{\text{(1}}{\text{.76}}{{\text{m}}^{\text{2}}}{\text{)}}$ = ${\text{0}}{\text{.88}}{{\text{m}}^{\text{2}}}$

Q2. The area of a trapezium is ${\text{34c}}{{\text{m}}^{\text{2}}}$and the length of one of the parallel sides is ${\text{10cm}}$and its height is ${\text{4cm}}$. Find the length of the other parallel side.

Ans: It is given that Area of Trapezium =  ${\text{34c}}{{\text{m}}^{\text{2}}}$ .

          Length of one parallel side =  ${\text{10cm}}$

          Height =  ${\text{4cm}}$

          Now, let length of other parallel side = ‘ \[{\text{a}}\] ’ cm

Therefore, Area of Trapezium =  $\dfrac{{\text{1}}}{{\text{2}}}{\text{ ( Sum of parallel sides)  x  (Distances between parallel sides) }}$

                        ${\text{34c}}{{\text{m}}^{\text{2}}}$               = $\dfrac{{\text{1}}}{{\text{2}}}{\text{[(10 + a) x 4]cm}}$

                        ${\text{34c}}{{\text{m}}^{\text{2}}}$               = ${\text{(20 + 2a)cm}}$

                         $\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm}}$               = \[{\text{a}}\] cm

  Thus, Length of the other parallel side ( \[{\text{a}}\] ) = ${\text{7cm}}$ .

Q3. Length of the fence of a trapezium shaped field ABCD is ${\text{120m}}$. If BC = ${\text{48m}}$, CD =${\text{17m}}$ and AD =${\text{40m}}$, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Ans: Length of the fence of Trapezium ABCD = AB+BC+CD+DA

                                                                     ${\text{120m}}$ =${\text{(AB + 48 + 17 + 40)m}}$

                                                                     ${\text{120m}}$   = ${\text{AB + 105m}}$

          Therefore, AB = ${\text{15m}}$

          Area of the field ABCD = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(AD + BC) x AB}}$

                      =$\dfrac{{\text{1}}}{{\text{2}}}{\text{(40 + 48) x 15}}$

                      =$\dfrac{{\text{1}}}{{\text{2}}}{\text{(88) x 15}}$

                           = ${\text{660}}{{\text{m}}^{\text{2}}}$

Q4. The diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8m and 13m. Find the area of the field. 

Ans: It is given that Length of diagonal =  ${\text{24m}}$

          Length of the perpendiculars ${{\text{h}}_{\text{1}}}$ and ${{\text{h}}_{\text{2}}}$ from the opposite vertices to the diagonal are 

           ${{\text{h}}_{\text{1}}}$ = ${\text{8m}}$ and  ${{\text{h}}_{\text{2}}}$ = ${\text{13m}}$

          Area of Quadrilateral = $\dfrac{{\text{1}}}{{\text{2}}}{\text{d}}\left( {{{\text{h}}_{\text{1}}}{\text{ + }}{{\text{h}}_{\text{2}}}} \right)$

                                               = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(24m) x (13m + 8cm)}}$

                                               = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(24m)(21m)}}$

                                               = ${\text{252}}{{\text{m}}^{\text{2}}}$

          Thus, Area of field =  ${\text{252}}{{\text{m}}^{\text{2}}}$

Q5. The diagonals of a rhombus are ${\text{7}}{\text{.5cm}}$and ${\text{12cm}}$. Find its area.

Ans: Area of Rhombus = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Product of its diagonals) }}$

                                          = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 7}}{\text{.5cm x 12cm}}$

                                        = ${\text{45c}}{{\text{m}}^{\text{2}}}$

Q6. Find the area of a rhombus whose side is ${\text{6cm}}$and whose altitude is ${\text{4cm}}$. If one of its diagonals is ${\text{8cm}}$long, find the length of the other diagonal.

Ans: Let the length of other diagonal of Rhombus be ‘ \[{\text{X}}\] ’

          A Rhombus is a special case of Parallelogram

          The area of Parallelogram is given by its base and height

          Thus, Area of Rhombus = Base x Height

                                                     = ${\text{6cm x 4cm}}$ = ${\text{24c}}{{\text{m}}^{\text{2}}}$

          So, 

          Area of Rhombus = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Product of its diagonals) }}$

                            ${\text{24c}}{{\text{m}}^{\text{2}}}$   = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(8cm x X)}}$

                             ${\text{X = }}\;{\text{6cm}}$

          Therefore, Length of other diagonal of Rhombus = ${\text{6cm}}$

Q7. The floor of a building consists of $3000$tiles which are rhombus shaped and each of its diagonals are ${\text{45cm}}$and ${\text{30cm}}$in length. Find the total cost of polishing the floor, if the cost per ${{\text{m}}^{\text{2}}}$is Rs$4$.

Ans: Given that each diagonals of Rhombus are ${\text{45cm}}$ and ${\text{30cm}}$

                                          =  $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 45 x 30}}} \right){\text{c}}{{\text{m}}^{\text{2}}}$

                                        =  ${\text{675c}}{{\text{m}}^{\text{2}}}$

          So, Area of $3000$ tiles = $(675 x 3000){\text{c}}{{\text{m}}^2} = 2025000\;{\text{c}}{{\text{m}}^2} = 202.5\;{{\text{m}}^2}$

          Now, it is given that cost of polishing is ${\text{Rs}}{\text{.}}\;{\text{4}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$

          So, Cost of Polishing ${\text{202}}{\text{.5}}{{\text{m}}^{\text{2}}}$ area = ${\text{Rs(4 x 202}}{\text{.5) =  Rs 810}}$

          Hence, Cost of polishing the floor is ${\text{Rs}}{\text{.}}\;{\text{810}}$

Q8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. It the area of this field is ${\text{10500}}{{\text{m}}^{\text{2}}}$and the perpendicular distance between the two parallel sides is ${\text{100m}}$, find the length of the side along the river.

Ans: Let the length of the side along the road = ‘ \[{\text{l}}\] ’

          And Let the length of the side along the river = ‘ ${\text{2l}}$ ’

         It is given that distance between two parallel sides = ${\text{100m}}$

         and Area of Trapezium =   ${\text{10500}}{{\text{m}}^{\text{2}}}$

         Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$

                      ${\text{10500}}{{\text{m}}^{\text{2}}}$     = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(l + 2l) x (100m)}}$

                      ${\text{3l = }}\left( {\dfrac{{{\text{2 x 10500}}}}{{{\text{100}}}}} \right){\text{m = 210m}}$

                        ${\text{l = }}\;\dfrac{{{\text{210}}}}{{\text{3}}}{\text{m}}\;{\text{ = }}\;{\text{70m}}$

         Therefore, Length of the side along the river ‘ ${\text{2l}}$ ’ = ${\text{140m}}$

Q9. 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.

It is given in the figure that side of octagon = ${\text{5cm}}$

Area of Trapezium ABCH = Area of Trapezium DEFG

Area of Trapezium =  $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$

                               = $\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{(4)(11 + 5)}}} \right]{{\text{m}}^{\text{2}}}$

                               = $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 4 x 16}}} \right){{\text{m}}^{\text{2}}}{\text{ = 32}}{{\text{m}}^{\text{2}}}$

In rectangle HCDG, Length( \[{\text{l}}\] ) = ${\text{11m}}$ and Breadth( \[{\text{b}}\] ) = ${\text{5m}}$

So, Area of rectangle = ${\text{(11 x 5)}}{{\text{m}}^{\text{2}}}$ = ${\text{55}}{{\text{m}}^{\text{2}}}$

Therefore, Area of octagon = Area of Trapezium ABCH + Area of Trapezium DEFG +

                                               Area of Rectangle

                                            = ${\text{(32 + 32 + 55)}}{{\text{m}}^{\text{2}}}$

                                            = ${\text{119}}{{\text{m}}^{\text{2}}}$

Q10. 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?

Ans: From Jyoti’s Way of finding area , 

       Area of Pentagon = $2$ (Area of Trapezium ABCF)

                                      = $2$ [ $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$ ]

                                      = $\left[ {{\text{2 x }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(15 + 30)}}\left( {\dfrac{{{\text{15}}}}{{\text{2}}}} \right)} \right]{{\text{m}}^{\text{2}}}$

                                      = ${\text{337}}{\text{.5}}{{\text{m}}^{\text{2}}}$

      From Kavita’s Way of finding area , 

Area of Pentagon = Area of Triangle ABE + Area of Square BCDE

                             = [ $\dfrac{1}{2}$ (basexheight)] + (sidexside)

                             = $\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 15 x (30 - 15) + (15}}{{\text{)}}^{\text{2}}}} \right]{{\text{m}}^{\text{2}}}$

                             = $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 15 x 15 + 225}}} \right){{\text{m}}^{\text{2}}}$

                             = ${\text{(112}}{\text{.5 + 225)}}{{\text{m}}^{\text{2}}}$

                             =  ${\text{337}}{\text{.5}}{{\text{m}}^{\text{2}}}$

Q11. Diagram of the adjacent picture frame has outer dimensions = ${\text{24cm}}$× ${\text{28cm}}$and inner dimensions ${\text{16cm}}$× ${\text{20cm}}$ Find the area of each section of the frame, if the width of each section is same.

Given that, the width of each section is the same. 

IB = BJ = CK = CL = DM = DN = AO = AP

IL = IB + BC + CL

\[28\ =\ \text{IB}\ +\ 20\ +\ \text{CL}\]

$\begin{align} & 28-20\ \text{=}\ \text{IB+}\ \text{CL} \\ & 8\ \text{=}\ \text{IB+}\ \text{CL} \\ \end{align}$

IB = CL = \[{\text{4cm}}\]

Hence, IB = BJ = CK = CL = DM = DN = AO = AP = $\text{4}\,\text{cm}$

Area of section BEFC = Area of section DGHA = Area of Trapezium

  $\left[ \frac{\text{1}}{\text{2}}\text{(bas}{{\text{e}}_{1}}\text{ + bas}{{\text{e}}_{2}}\text{)(h)} \right]=\left[ \frac{1}{2}(20\text{ }+\text{ }28)(4) \right]\text{c}{{\text{m}}^{\text{2}}}\text{=}96\ \text{c}{{\text{m}}^{\text{2}}}$

Therefore, Area of section ABEH = Area of section CDGF. 

Hence area of each section of frame is $96\ \text{c}{{\text{m}}^{\text{2}}}$

Exercise 11.3

Q1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Ans: From the given figure; Length( \[{\text{l}}\] ), Breadth( \[{\text{b}}\] ) and Height( \[{\text{h}}\] ) of the Cuboid is

                                                    \[{\text{60cm}}\] , ${\text{40cm}}$ , ${\text{50cm}}$ respectively.

          And, Side of Cube is  ${\text{50cm}}$ .

          Now, Total Surface area of Cuboid(a) = $2$ ( \[{\text{l}}\] \[{\text{h}}\] + \[{\text{b}}\] \[{\text{h}}\] + \[{\text{l}}\] \[{\text{b}}\] )

                                                                       = \[{\text{[2\{ (60)(40) + (40)(50) + (50)(60)\} ]c}}{{\text{m}}^{\text{2}}}\]

                                                                       = ${\text{[2(2400 + 2000 + 3000)]c}}{{\text{m}}^{\text{2}}}$

                                                                       = ${\text{(2 x 7400)c}}{{\text{m}}^{\text{2}}}$

                                                                       = ${\text{14800c}}{{\text{m}}^{\text{2}}}$

                       Total Surface area of Cube(b) = \[{\text{6}}{{\text{l}}^{\text{2}}}\]

                                                                        = $6{(50\;{\text{cm}})^2} = 15000\;{\text{c}}{{\text{m}}^2}$

          Therefore, Cuboidal box(a) requires a lesser amount of material for making.

Q2. A suitcase with measure ${\text{80cm x 48cm x 24cm}}$is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width ${\text{96cm}}$is required to cover $100$ such suitcases? 

Ans: Given that length( \[{\text{l}}\] ),breadth( \[{\text{b}}\] ),height( \[{\text{h}}\] ) of suitcase is \[{\text{(80,48,24)cm}}\] respectively.

         Total surface area of suitcase =  $2$ ( \[{\text{l}}\] \[{\text{h}}\] + \[{\text{b}}\] \[{\text{h}}\] + \[{\text{l}}\] \[{\text{b}}\] )

                                                         = ${\text{2[(80)(48) + (48)(24) + (24)(80)]}}$

                                                         = \[{\text{2[3840 + 1152 + 1920]}}\]

                                                         = ${\text{13824}}{{\text{m}}^{\text{2}}}$

         Total surface area of  $100$ suitcases = ${\text{100 x 13824c}}{{\text{m}}^{\text{2}}}$

                                                                    = ${\text{1382400c}}{{\text{m}}^{\text{2}}}$

                                                                    = required Tarpaulin

         We have given that breadth of Tarpaulin is  ${\text{96cm}}$ and we have to find Length 

         of tarpaulin.

         Required Tarpaulin = (Length x Breadth) of Tarpaulin

                          ${\text{1382400c}}{{\text{m}}^{\text{2}}}$ = Length x   ${\text{96cm}}$

                       Length = $\left( {\dfrac{{{\text{1382400}}}}{{{\text{96}}}}} \right){\text{cm = 14400cm}}$

       Therefore, Length = \[{\text{144m}}\;\;\;\;\;(\because \;{\text{1m = 100cm)}}\]

       Thus, ${\text{144m}}$ of tarpaulin is required to cover $100$ suitcases.

Q3. Find the side of a cube whose surface area is ${\text{600c}}{{\text{m}}^{\text{2}}}$ .

Ans: It is given that the surface area of the cube is  ${\text{600c}}{{\text{m}}^{\text{2}}}$ .

          We have to find the side of the cube ( \[{\text{a}}\] ). 

          Surface area of cube = ${\text{6}}{{\text{a}}^{\text{2}}}$

                                 ${\text{600c}}{{\text{m}}^{\text{2}}}$ =  ${\text{6}}{{\text{a}}^{\text{2}}}$

                                       $\therefore \;{{\text{a}}^{\text{2}}}{\text{ = 100c}}{{\text{m}}^{\text{2}}}$

                                           \[{\text{a}}\] = ${\text{10cm}}$

          Thus, the side of the cube is  ${\text{10cm}}$ .

Q4. Rukhsar painted the outside of the cabinet of measure${\text{1m x 2m x 1}}{\text{.5m}}$. How much surface area did she cover if she painted all except the bottom of the cabinet?

Ans: It is given that length ( \[{\text{l}}\] ), breadth( \[{\text{b}}\] ), height( \[{\text{h}}\] ) of the cabinet is 

          ${\text{2m,1m,1}}{\text{.5m}}$ respectively. 

         Area of the surface = ${\text{2h(1 + b) + lb}}$

                                        = ${\text{[2 x 1}}{\text{.5 x (2 + 1) + (2)(1)]}}{{\text{m}}^{\text{2}}}$

                                        = ${\text{[3(3) + 2]}}{{\text{m}}^{\text{2}}}$

                                        = ${\text{11}}{{\text{m}}^{\text{2}}}$

Q5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of ${\text{15m,10m and 7m}}$respectively. From each can of paint  ${\text{100}}{{\text{m}}^{\text{2}}}$ of area is painted. How many cans of paint will she need to paint the room? 

Ans: Given that length ( \[{\text{l}}\] ), breadth( \[{\text{b}}\] ), height( \[{\text{h}}\] ) of Cuboid is

           ${\text{15m,10m and 7m}}$ respectively.

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

                                                               = $2$ \[{\text{h}}\] ( \[{\text{l}}\] + \[{\text{b}}\] ) + \[{\text{l}}\]\[{\text{b}}\]

                                                            = ${\text{[2 x 7(15 + 10) + (15 x 10)]}}{{\text{m}}^{\text{2}}}$

                                                            = ${\text{[(2 x 175) + 150]}}{{\text{m}}^{\text{2}}}$

                                                            = ${\text{500}}{{\text{m}}^{\text{2}}}$

          It is given that  ${\text{100}}{{\text{m}}^{\text{2}}}$ area is to be painted from each can.

          Therefore, Number of cans required to paint an area of  ${\text{500}}{{\text{m}}^{\text{2}}}$  

                                      = $\dfrac{{500}}{{100}}{\kern 1pt} \; = \;5$

         Hence, $5$ cans of paint are required to paint the room.

Q6. Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?

Ans: The above given two figures alike for same height( \[{\text{h}}\] )= ${\text{7cm}}$

          And the difference between in these two figures is that one is cylinder and the other 

          One is a cube.

          Now, we have to find the lateral surface area for both of the given figures.

          Given that Side of cube( \[{\text{l}}\] )= ${\text{7cm}}$  

                           Height and Diameter of cylinder =  ${\text{7cm}}$ each

                           Radius of cylinder = $\dfrac{{{\text{Diameter}}}}{{\text{2}}}\;{\text{ = }}\;\dfrac{{\text{7}}}{{\text{2}}}{\text{cm = }}\;{\text{3}}{\text{.5cm}}$

          First, Lateral surface area of Cube = ${\text{4}}{{\text{l}}^{\text{2}}}$ = ${\text{4(}}{{\text{7}}^{\text{2}}}{\text{)c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;{\text{4 x 49c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;{\text{196c}}{{\text{m}}^{\text{2}}}$

          Second, Lateral surface area of cylinder = ${\text{2$\pi$ rh}}$

                                                                           = ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 3}}{\text{.5cm x 7cm}}\;{\text{ = }}\;{\text{154c}}{{\text{m}}^{\text{2}}}$

           Therefore, the Cube has a larger lateral surface area.

Q7. A closed cylindrical tank of radius ${\text{7m}}$ and height ${\text{3m}}$ is made from a sheet of metal. How much sheet of metal is required?

Ans: Given that the radius and height of Cylinder is  ${\text{7m}}$ and ${\text{3m}}$ respectively.

          Therefore, Total surface area of Cylinder = ${\text{2$\pi$ }}$\[{\text{r}}\] ( \[{\text{r}}\] + \[{\text{h}}\] )

                                                                                 = ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7m(7m + 3m)}}$

                                                                                 = ${\text{440}}{{\text{m}}^2}$

          Thus, ${\text{440}}{{\text{m}}^2}$ of metal sheet is required.

Q8. The lateral surface area of a hollow cylinder is ${\text{4224c}}{{\text{m}}^{\text{2}}}$. It is cut along its height and formed a rectangular sheet of width ${\text{33cm}}$. Find the perimeter of the rectangular?

Ans: It is given that Hollow cylinder is cut along its height and formed a 

          Rectangular sheet.

          Area of cylinder =  ${\text{4224c}}{{\text{m}}^{\text{2}}}$

          And, Breadth of rectangular sheet =  ${\text{33cm}}$ 

          So, Area of Cylinder = Area of Rectangular Sheet

                         ${\text{4224c}}{{\text{m}}^{\text{2}}}$ = Length x Breadth

                         ${\text{4224c}}{{\text{m}}^{\text{2}}}$ = Length x ${\text{33cm}}$

                        Length = \[\dfrac{{{\text{4224}}}}{{{\text{33}}}}{\text{cm}}\;{\text{ = }}\;{\text{128cm}}\]

          Now, Perimeter of Rectangle = $2$ (length + breadth)

                                                         = ${\text{2(128 + 33)cm}}\;{\text{ = }}\;{\text{2(161cm)}}\;{\text{ = }}\;{\text{322cm}}$

Q9. A road roller takes ${\text{750}}$complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is ${\text{84cm}}$ and length is ${\text{1m}}$.

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

          Here, Radius = $\dfrac{{{\text{diameter}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{84}}}}{{\text{2}}}{\text{cm = 42cm = }}\dfrac{{{\text{42}}}}{{{\text{100}}}}{\text{m}}\;\;\;\;\;\;(\because {\text{1m = 100cm}})$

                    Height = ${\text{1m}}$

          Thus, In One Revolution, 

          Area of the road covered = ${\text{2$\pi$ }}$ rh

                                                      = ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{{\text{42}}}}{{{\text{100}}}}{\text{ x 1}}{{\text{m}}^{\text{2}}}$

                                                      = $\dfrac{{{\text{264}}}}{{{\text{100}}}}{{\text{m}}^{\text{2}}}$

           In ${\text{750}}$ revolutions, area of road covered = ${\text{750 x }}\dfrac{{{\text{264}}}}{{{\text{100}}}}{{\text{m}}^{\text{2}}}$

                                                                            = \[{\text{1980}}{{\text{m}}^{\text{2}}}\]

Q10. A company packages its milk powder in a cylindrical container whose base has a diameter of ${\text{14cm}}$and height ${\text{20cm}}$. Company places a label around the surface of the container (as shown in the figure). If the label is placed ${\text{2cm}}$from top and bottom, what is the area of the label?

Ans: It is given that Radius = $\dfrac{{{\text{diameter}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm = 7cm}}$

                                   Height of label = ${\text{(20 - 2 - 2)cm}}\;\;\;\;\;\;(\because {\text{2cm}}\;{\text{deducted}}\;{\text{from}}\,{\text{top}}\;{\text{bottom}}\;{\text{each)}}$

                                                           = ${\text{16cm}}$

          As shown in the figure, the label is in the shape of a cylinder.

          So, Area of label(cylinder)= ${\text{2$\pi$ }}$ rh

                                                   = ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7cm x 16cm}}\;{\text{ = }}\;{\text{44 x 16c}}{{\text{m}}^{\text{2}}}$

                                                   = ${\text{704c}}{{\text{m}}^{\text{2}}}$

Exercise 11.4

Q1. 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.

Ans: (a) In this situation, we will find the volume.

         (b) Number of cement bags required to plaster cylindrical bags so for that 

              situation, we will find the surface area.

         (c) Number of smaller tanks that can be filled with so for that situation, we will

              find the volume.

Q2. Diameter of cylinder A is ${\text{7cm}}$, and the height is ${\text{14cm}}$. Diameter of cylinder B is  ${\text{14cm}}$ and height is ${\text{7cm}}$. 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?

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

         We know that,

         Volume of cylinder = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

         If measures of radius(r) and height(h) are same, then the cylinder with greater

         radius will have greater area.

        Here, Radius of cylinder A = $\dfrac{{\text{7}}}{{\text{2}}}{\text{cm}}$

                   Radius of cylinder B = $\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm = 7cm}}$

       As the radius of cylinder B is greater, therefore, the volume of cylinder B will 

        be greater.

        Let us verify it by calculating the volume of both the cylinders.

         Volume of Cylinder A =  ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

                                                = $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x 14c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{11 x 49c}}{{\text{m}}^{\text{3}}}\;{\text{ = 539c}}{{\text{m}}^{\text{3}}}$

         Volume of Cylinder B = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

                                                =  $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7 x 7 x 7c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{22 x 49c}}{{\text{m}}^{\text{3}}}\;{\text{ = 1078c}}{{\text{m}}^{\text{3}}}$

         Therefore, the volume of cylinder B is greater.

          Now, Surface area of cylinder A = ${\text{2$\pi$ r(r + h)}}$

                                                              = ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x (}}\dfrac{{\text{7}}}{{\text{2}}}{\text{ + 14)c}}{{\text{m}}^{\text{2}}}$

                                                              = \[{\text{22 x }}\dfrac{{{\text{35}}}}{{\text{2}}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{ = }}\;{\text{385c}}{{\text{m}}^{\text{2}}}\]

                   Surface area of cylinder B = ${\text{2$\pi$ r(r + h)}}$

                                                              = ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7 x (7 + 7)c}}{{\text{m}}^{\text{2}}}$

                                                              = \[{\text{44 x 14c}}{{\text{m}}^{\text{2}}}{\text{ = }}\;{\text{616c}}{{\text{m}}^{\text{2}}}\]

Q3. Find the height of a cuboid whose base area is ${\text{180c}}{{\text{m}}^{\text{2}}}$ and volume is ${\text{900c}}{{\text{m}}^{\text{3}}}$?

Ans. Here, we have given that Base Area of Cuboid = Length x Breadth

                                                                                              = ${\text{180c}}{{\text{m}}^{\text{2}}}$

          Volume of Cuboid = Length x Breadth x Height

                          ${\text{900c}}{{\text{m}}^{\text{3}}}$ =             ${\text{180c}}{{\text{m}}^{\text{2}}}$ x Height

                        Height = $\dfrac{{{\text{900}}}}{{{\text{180}}}}{\text{cm = 5cm}}$

Q4. A cuboid is of dimensions ${\text{60cm x 54cm x 30cm}}$. How many small cubes with side${\text{6cm}}$can be placed in the given cuboid?

Ans. From given condition, 

          Volume of Cuboid =  ${\text{60cm x 54cm x 30cm}}$

                                      = ${\text{97200c}}{{\text{m}}^{\text{3}}}$

         Given that side of cube = ${\text{6cm}}$

         So, Volume of cube = ${\text{(6 x 6 x 6)c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{216c}}{{\text{m}}^{\text{3}}}$

         Required number of cubes = \[\dfrac{{{\text{volume}}\;{\text{of}}\;{\text{cuboid}}}}{{{\text{volume}}\;{\text{of}}\;{\text{cube}}}}{\text{ = }}\dfrac{{{\text{97200}}}}{{{\text{216}}}}{\text{ = 450}}\]

         Therefore, ${\text{450}}$ cubes can be placed in the given Cuboid.

Q5. Find the height of the cylinder whose volume is ${\text{1}}{\text{.54}}{{\text{m}}^{\text{3}}}$and diameter of the base is ${\text{140cm}}$?

Ans. It is given that Radius of Cylinder = $\dfrac{{{\text{140}}}}{{\text{2}}}{\text{cm = 70cm}}$

                                   Volume of Cylinder = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

                                        ${\text{1}}{\text{.54}}{{\text{m}}^{\text{3}}}$                 = $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{{\text{70}}}}{{{\text{100}}}}{\text{m x }}\dfrac{{{\text{70}}}}{{{\text{100}}}}{\text{m x h}}$

                                                        Height = $\dfrac{{{\text{1}}{\text{.54 x 100}}}}{{{\text{22 x 7}}}}{\text{m = 1m}}$

          Hence, the height of cylinder = ${\text{1m}}$

Q6. A milk tank is in the form of cylinder whose radius is ${\text{1}}{\text{.5m}}$and length is ${\text{7m}}$. Find the quantity of milk in litres that can be stored in the tank?

Ans. It is given that Radius and height of the cylinder is ${\text{1}}{\text{.5m}}$ and ${\text{7m}}$ respectively .

          Therefore, Volume of Cylinder =   ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

                                                                            = $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 1}}{\text{.5m x 1}}{\text{.5m x 7m}}$

                                                                         = ${\text{22 x 2}}{\text{.25}}{{\text{m}}^{\text{3}}}$

                                                                         = ${\text{49}}{\text{.5}}{{\text{m}}^{\text{3}}}$

           As, ${\text{1}}{{\text{m}}^{\text{3}}}{\text{ = 1000}}\;{\text{litre}}$

           So, Required Quantity = ${\text{49}}{\text{.5 x 1000}}\;{\text{litre}}\;{\text{ = }}\;{\text{49500}}\;{\text{litre}}$

          Therefore, ${\text{49500}}\;{\text{litre}}$ can be stored in the tank.

Q7. 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?

Ans. (i) Let the edge of the cube be ‘ \[{\text{a}}\] ’.

             Surface area of cube = ${\text{6}}{{\text{a}}^{\text{2}}}$

            If each edge of the cube is doubled, then it becomes ${\text{2a}}$

            Therefore, New surface area = ${\text{6(2a}}{{\text{)}}^{\text{2}}}{\text{ = 24}}{{\text{a}}^{\text{2}}}{\text{ = 4(6}}{{\text{a}}^{\text{2}}}{\text{)}}$

           Clearly, the surface area will be increased by ${\text{4}}$ times.

        (ii) Let Volume of the cube = ${{\text{a}}^{\text{3}}}$

            When each edge of the cube is doubled, it becomes ${\text{2a}}$ .

           New volume = \[{{\text{(2a)}}^{\text{3}}}{\text{ = 8}}{{\text{a}}^{\text{3}}}{\text{ = 8 x }}{{\text{a}}^{\text{3}}}\]

           Clearly, the volume of the cube will be increased by ${\text{8}}$ times.

Q8. Water is pouring into a cubiodal reservoir at the rate of $60$litres per minute. If the volume of reservoir is ${\text{108}}{{\text{m}}^{\text{3}}}$, find the number of hours it will take to fill the reservoir.

Ans. Volume of cuboidal reservoir =  ${\text{108}}{{\text{m}}^{\text{3}}}$   = $(108 x 1000)$ \[{\text{L}}\] = $108000$ \[{\text{L}}\]

      It is given that water is being poured at the rate of $60$ L per minute.

      That is, $(60 x 60)$ \[{\text{L}}\] = $3600$ \[{\text{L}}\] per hour

      Required number of hours = $30$ hours

      Thus, it will take $30$ hours to fill the reservoir.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration - PDF Download

It is not compulsory to be connected to the internet to access our solutions as Vedantu has made NCERT Solutions for Class 8 Maths Mensuration available in the PDF format on its official website. Now download the NCERT Solutions Class 8 Maths Chapter 11 PDF on your devices or get a printout and access it from anywhere, anytime. This mode of revision is quick and easy and can be done at your pace during a crucial exam period.

All Topics of NCERT Class 8 Maths Chapter 11 - Mensuration

The topics covered under chapter 11 Mensuration are given below.

Table of Important Formulas

Below given are the list of important formulas that you must remember to solve the exercise of chapter 11 NCERT Maths Class 8 th .

You should also remember some of the basic conversion parameters. For your ease, the most important parameters are given below.

1 m 3 = 1000000 cm 3 = 1000 L

1 cm 3 = 1 mL

1 L = 1000 cm 3

List of Exercises in class 8 Maths Chapter 11:

Chapter 11 – Mensuration

Introduction

Mensuration is the process applied to different 2-D and 3-D solids of various shapes and figures to measure their lengths, volumes, area, heights, perimeters and several other dimensions. In this section of Ch 11 Maths Class 8, you will recall the areas of plane figures like triangles, circles, rectangles, etc., that you learned in the previous chapter. In NCERT Solutions for Class 8 Maths Ch 11, you would learn how to calculate perimeter and areas of other closed figures like Quadrilaterals.

Let us Recall

In this section of Mensuration Class 8 NCERT, students would recall the formula for calculating the perimeter and area of a park, flower bed, and amount of cement required to cover a given area. All the problems are based on areas of the following shape:

Rectangle - The area of a rectangle is x*y where x is the length and y is the breadth of the rectangle.

Square - The area of a square is x 2 where x is the length of one side of a square.

Triangle - The area of a triangle is ½ *b*h where b is the length of the base and h is the height of the triable

Circle - The area of a circle is πr 2 where r is the radius of the circle.

Area of a Trapezium

A trapezium is a quadrilateral where two of the sides are parallel to each other. In this portion of Mensuration Class 8 NCERT Solutions, students will learn how to derive the area of a trapezium which is given by:

Area of a trapezium = h * (x + y)/2, where h is the height of the trapezium, x and y are the lengths of its two sides.

Area of a general Quadrilateral

A simple definition of a general quadrilateral is a closed 2-D shape having 4 straight sides. If we break the word Quadrilateral, Quad means 4, and lateral means sides. The area of a quadrilateral in Class 8th Maths Chapter 11 is calculated by splitting it into two triangles and then calculating and adding the areas of the two triangles. 

(Image to be added soon)

So if PQRS is a quadrilateral, then its area = (area of triangle PQR) + (area of triangle PRS) = ½ * d * (h 1 + h 2 ), where d is the length of the diagonal from P to R and h 1 and h 2 are heights of perpendiculars dropped from Q and Pr and S or PR respectively.

11.8 Volume of Cube, Cuboid, and Cylinder

The amount of space that a 3-D object occupies gives the volume of that object. To take examples from real life, the volume of a cupboard in a room is less than the volume of the room where it is placed. 

11.8.1 - Volume of a Cuboid - A 3-D structure with 6 rectangular faces is a cuboid. Its volume is given by “l * h* b”, here l = length, b = breadth, and h= height.

11.8.2 - Volume of a Cube - A cube is a special type of cuboid where its length, breadth, and height are the same. Hence the volume of a cube = length 3 .

11.8.3 - Volume of a Cylinder - A cylinder has two circular bases that are parallel to each other and separated by a distance. To measure the volume of a cylinder we use the formula πr 2 * h. Here r is the radius of the circular base and h is the distance between the bases. 

Key Features of NCERT Solutions for Class 8 Maths Chapter 11

You will find the NCERT Solutions of Class 8 Maths Chapter 11 by Vedantu extremely beneficial for your exams. The key features are:

Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.

Clear and structured presentation for easy comprehension.

Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

Visual aids like diagrams and illustrations to simplify complex concepts.

Additional tips and insights to enhance students' performance.

Chapter summaries for quick revision.

Online accessibility and downloadable resources for flexible study and revision.

NCERT Solutions play a crucial role in Class 8 exam prep. Start by thoroughly reading the textbook chapter. After that, solve the NCERT questions for Class 8 Chapter 11. You can find detailed solutions on Vedantu, aligning with CBSE guidelines. Download the free NCERT Solutions for Class 8 Chapter 11 to guide your exam preparation with expert-reviewed answers.

FAQs on NCERT Solutions for Class 8 Maths Chapter 11 - Mensuration

1. What are some of the applications of mensuration in our everyday lives?

In real life, mensuration is applied in many fields like:

Measuring floor and site areas required for purchasing or selling land.

Measuring agricultural fields.

Measuring surface areas of a house required for estimating painting cost.

To know the volume of the level of water in rivers and tanks.

To find out the amount of carpet required for covering a specific room.

The volume of soil that is needed to fill a ditch.

2. What are solid shapes and what is the method of measuring the surface area of a solid shape?

Any 3-D shapes occupying some space are called solid shapes like cube, sphere, cylinder, etc. To measure the surface area of a solid shape, we need to draw the net of that solid shape. From the net, we can see all the faces of the solid clearly. Then, we calculate the areas of each of the faces and add them up to get the total surface area of the solid shape. Surface area is measured in a square unit.

3. What are the concepts covered under Chapter 11 of Class 8 Maths?

The ideas or topics that are included in Chapter 11 “Mensuration” of Class 8 Maths are given below:

Let Us Recall

Area of Trapezium

Area of a General Quadrilateral

Area of Special Quadrilaterals

Area of a Polygon

Solid Shapes

Surface Area of Cylinder, Cube and Cuboid

Cuboid Cube Cylinders

Volume of Cuboid, Cube and Cylinder

Cuboid Cube Cylinder

Volume and Capacity

What Have We Discussed?

4. The base area of the cuboid is 25cm sq. Its volume is 275 cubic cm. What will be the height of the cuboid?

In the question, we are given the base area of the cuboid which is equal to 25cm sq.

The volume of the cuboid is 275 cubic cm.

We know that according to the formula,

The volume of a cuboid = Height × Base Area

Therefore, the height of the cuboid will be = Volume of cuboid/ Base Area

Height = 275/25 = 11cm

Thus, 11cm is the height of the cuboid.

5. The distance between two parallel sides is 15m and the length of one parallel side is 20m. 480m sq. Is the area of the trapezium-shaped field. What is the length of the other parallel side?

Let one parallel side be a = 20m and the other parallel side is ‘b’.

The height of the field is 15m.

Given, the area of the trapezium is 480m sq.

The formula of trapezium is,

Area of trapezium = ½ (a + b) * h

480 = ½ (20 + b) * 15

20 + b = 480 × 2/ 15

64 = 20 + b

Thus, the length of the other parallel side is 44m.

6. What are the perks of NCERT Solutions of Chapter 11 of Class 8 Maths?

The perks of the NCERT Solutions of Chapter 11 of Class 8 Maths are given below:

The NCERT Solutions of Chapter 11 of Class 8 Maths offers comprehensive learning.

It enables students to develop their reasoning and logical skills.

These solutions assist students in understanding the difficult concepts.

By practising these, students will have a strong grip over the chapter.

You will get a hint of how to answer the questions in the proper format.

It helps in scoring good marks in Chapter 11 of Class 8 Maths.

The NCERT Solutions for  Chapter 11 of Class 8 Maths are available free of cost on the Vedantu website and on the Vedantu app.

7. How can I make the best study plan for Chapter 11 of Class 8 Maths?

Keep the following points in mind while making an effective study plan for Chapter 11 of Class 8 Maths:

Have a timetable or schedule to manage your time.

Centralize the NCERT book to read Chapter 11 of Class 8 Maths.

Practice the NCERT Solutions to comprehend the Chapter. 

Give yourself a break.

Do meditation and exercise to keep your body and mind fit.

Attend all your school lectures.

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Chapter 9 Class 8 Mensuration

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Get solutions of all NCERT Questions and Examples of Chapter 9 Class 8 Mensuration free at teachoo. Answers to all questions have been solved in an easy way with detailed explanation of each and every solution.

In this chapter, we will

• First revise our formulas for Perimeter and Area of Rectangle, Square, Triangle, Parallelogram, Circle
• and do some questions using those formulas
• Then, we will derive the formula for Area of Trapezium
• and formula for Area of Quadrilateral
• We also learn the formula for Area of Rhombus
• and find Area of any general polygon by dividing it into different shapes and adding the area
• Then, we see some common 3 Dimensional shapes
• And see what area of 3 Dimensional Shapes is
• and Curved and Total Surface Area of Cylinder
• We also learn what Volume of 3 Dimensional shape is
• And study formula of Volume of Cube, Cuboid and Cylinder
• We also learn how to convert cm 3  into mL, L and m 3

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Important Questions Class 8 Maths Chapter 11 Mensuration

Some of the most important mensuration class 8 questions, i.e. chapter 11 are given here. These chapter 11 class 8 maths questions cover several short answer type questions, long answer type questions and HOTS questions that are crucial for CBSE class 8 exams. Here, some of the important mensuration questions from NCERT class 8 are also included.

Also Check:

• Important 2 Marks Questions for CBSE 8th Maths
• Important 3 Marks Questions for CBSE 8th Maths
• Important 4 Marks Questions for CBSE 8th Maths

Mensuration Important Questions For Class 8 (Chapter 11)

These class 8 mensuration questions are categorized into short answer type questions and long answer type questions. These questions cover various concepts which will help class 8 students to develop problem-solving skills for the exam.

Short Answer Type Questions:

1. The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm.

Area of trapezium = ½ × perpendicular distance between parallel sides × sum of parallel sides

= ½ × 15 × (12 + 20)

= 1/2 × 15 × 32

= 240  cm 2

2. Calculate the height of a cuboid which has a base area of 180 cm 2 and volume is 900 cm 3 .

Volume of cuboid = base area × height

900 = 180 × height

So, height = 900/180 = 5 cm

3. A square and a rectangle have the same perimeter. Calculate the area of the rectangle if the side of the square is 60 cm and the length of the rectangle is 80 cm.

Perimeter of square formula = 4 × side of the square

Hence, P (square) = 4 × 60 = 240 cm

Perimeter of rectangle formula = 2 × (Length + Breadth)

Hence, P (rectangle) = 2 (80 + Breadth)

= 160 + 2 × Breadth

According to the given question, 160 + 2 × Breadth = 240 cm 2 × Breadth = 240 – 160 Breadth = 80/2 The breadth of the rectangle = 40 cm

Now, the area of rectangle = Length × Breadth = 80 × 40 = 3200 cm 2

4. A lawnmower takes 750 complete revolutions to cut grass on a field. Calculate the area of the field if the diameter of the lawnmower is 84 cm and length is 1 m.

Given, length of lawnmower = 1m = 100cm

Its circumference = π × D = 22/7 × 84 = 264 cm

Length of field will be = 264 × 750 = 198000 cm

Here, the width of field = length of the lawnmower i.e. 100 cm

So, area of field = 198000 × 100 = 19,800,000 cm²

Or, 1980 m²

5. The area of a rhombus is 16 cm 2 and the length of one of its diagonal is 4 cm. Calculate the length of other the diagonal.

Area of rhombus = ½ × d 1 × d 2

⇒ 16 = ½ × 4 × d 2

So, d 2 = 32/4 = 8 cm

Long Answer Type Questions:

6. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed.

The area of the remaining sheet after the smaller circle is removed will be = Area of the entire circle with radius 4 cm – Area of the circle with radius 3 cm

Area of circle = πr²

Area of the entire circle = π(4)² = 16π cm 2

Area of the circle with radius 3 cm which is cut out = π(3)² = 9π cm 2

Thus, the remaining area = 16π – 9π = 7π cm 2

7. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted.

Length of the box, l = 2 m,

Breadth of box, b = 1 m

Height of box, h = 1.5 m

We know that the surface area of a cuboid = 2(lb + lh + bh)

But here the bottom part is not to be painted.

Surface area of box to be painted = lb + 2(bh + hl)

= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)

= 2 + 2 (1.5 + 3.0)

Hence, the required surface area of the cuboidal box = 11 m 2

8. In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.

From the question statement draw the diagram.

Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.

Now, CM will be the distance between the two parallel sides or the height of the trapezium.

Area of trapezium = ½ × sum of parallel sides × height.

So, height has to be found.

In the diagram, draw CL || AD

Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm

As AD = CB,

CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height.

Here, BL = AB – AL = (40 – 20) = 20 cm. So,

LM = MB = ½ BL = ½ × 20 = 10 cm

Now, in ΔCLM,

CL 2 = CM 2 + LM 2 (Pythagoras Theorem)

26 2 = CM 2 + 10 2

CM 2 = 26 2 – 10 2

Using algebraic identities, we get; 26 2 – 10 2  = (26 – 10) (26 + 10)

CM 2 = (26 – 10) (26 + 10) = 16 × 36 = 576

CM = √576 = 24 cm

Now, the area of trapezium can be calculated.

Area of trapezium, ABCD = ½ × (AB + CD) × CM

= ½ × (20 + 40) × 24

Or, Area of trapezium ABCD = 720 cm 2

Class 8 Maths Chapter 11 Extra Question

• A flooring tile is in the shape of a parallelogram with 24 cm base and the corresponding 10 cm height. Calculate the number of tiles 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).
• Two cubes are joined end to end. Now, calculate the volume of the resulting cuboid, if each side of the cubes is 6 cm.
• How many bricks each 25 cm by 15 cm by 8 cm, are required for a wall 32 m long, 3 m high and 40 cm thick?
• Find the area of a rhombus whose one side measures 5 cm and one diagonal as 8 cm.

More Topics Related to Class 8 Mensuration:

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 11 are in accordance with the latest CBSE guidelines and marking schemes.

Class 8 Maths Chapter 11 Exercise 11.1 Solutions

Class 8 Maths Chapter 11 Exercise 11.2 Solutions

Class 8 Maths Chapter 11 Exercise 11.3 Solutions

Class 8 Maths Chapter 11 Exercise 11.4 Solutions

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CBSE Class 8th Maths Value Based Questions Chapter 11 Mensuration PDF Download

CBSE Class 8th Maths Value Based Questions Chapter 11 Mensuration are the easiest questions which you see in your question paper and the scoring one all student who attempt it surely get they are just little bit difficult and examine your basic knowledge regarding the particular chapter. Maths Value Based Questions for Class 8th are available here at Free of cost. These questions are expected to be asked in the Class 8th board examination. These Maths Value Based Questions are from complete CBSE Syllabus.

CBSE Class 8th Maths Value Based Questions Chapter 11 Mensuration

Most of these Maths Value Based Questions are quite easy and students need only a basic knowledge of the chapter to answer these questions. Download CBSE Maths Value Based Questions for board examinations. These Maths Value Based Questions are prepared by Directorate of Education, Delhi.

CBSE Maths Value Based Questions Class 8th Chapter 11 Mensuration PDF

The purpose of the Maths Value Based Questions is to make students aware of how basic values are needed in the analysis of different situations and how students require to recognize those values in their daily lives. Some questions are subject related. But even if they are not, that one-minute awareness of what we write about value without any specific preparation is a good step indeed.

CBSE Maths Value Based Questions for Class 8th Chapter 11 Mensuration download here in PDF format. The most CBSE Maths Value Based Questions for annual examination are given here for free of cost. The additional questions for practice the Class 8th exam are collected from various sources. It covers questions asked in previous year examinations.

CBSE Maths Value Based Questions for Class 8th Chapter 11 Mensuration Free PDF

Class 8th books have many questions. These questions are regularly asked in exams in one or other way. Practising such most CBSE Maths Value Based Questions Chapter 11 Mensuration certainly help students to obtain good marks in the examinations. 

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NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

• Class 8 Maths Mensuration Exercise 11.1
• Class 8 Maths Mensuration Exercise 11.2
• Class 8 Maths Mensuration Exercise 11.3
• Class 8 Maths Mensuration Exercise 11.4
• Mensuration Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.1

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NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

October 4, 2019 by Sastry CBSE

• Class 8 Maths Mensuration Exercise 11.1
• Class 8 Maths Mensuration Exercise 11.2
• Class 8 Maths Mensuration Exercise 11.3
• Class 8 Maths Mensuration Exercise 11.4
• Mensuration Class 8 Extra Questions

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.1

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Unit 9: Mensuration

Mensuration 9.1.

• Area of trapezoids (Opens a modal)
• Area of composite shapes (Opens a modal)
• Mensuration 9.1 Get 7 of 10 questions to level up!

Mensuration 9.2

• Surface area of a box (cuboid) (Opens a modal)
• Cylinder volume & surface area (Opens a modal)
• Mensuration 9.2 Get 7 of 10 questions to level up!

Mensuration 9.3

• Volume of a rectangular prism (Opens a modal)
• Mensuration 9.3 Get 7 of 10 questions to level up!

• Math Article

Mensuration Class 8

Mensuration Class 8 Chapter 11 notes,  important questions and formulas are mentioned here. Class 8 Chapter 11 Mensuration materials are provided here as per CBSE syllabus (2022-2023) and latest exam pattern. Get through the formulas and practise the concept of mensuration with the help of given examples. Mensuration chapter has been further introduced in Class 10, hence it is necessary that each student learns its basic concept first. Mensuration basically means the measurement of the area and perimeter of different geometrical shapes such as triangles, trapeziums, rectangles, etc. Preparing this chapter from this article will help students in scoring good marks in the final exams.

Mensuration Class 8 Notes

Mensuration deals with the measurement of area, perimeter, surface area and volume of different types of shapes.

Let us recall the area of all two-dimensional shapes.

Mensuration Class 8 – Area of Trapezium

By constructing EC || AB, we can split the given figure (AEDCBA) into two parts (Triangle ECD right-angled at C and Rectangle AECB), Here, b = a + c = 30 m

Now, Area of Triangle DCE:

1/2 × CD × EC= 1/2 × c × s

h = 1/2 ×10× 12 = 60 m 2

Also, Area of rectangle AECB = AB × BC = h × a=12 × 20=240 m 2

Therefore, Area of trapezium AEDB = Area of Triangle DCE + Area of rectangle AECB = 60 + 240 = 300 =300m 2

Area of Trapezium = Height/2 (Sum of parallel sides) = 1/2 h(a+b)

Mensuration Class 8 – Area of General Quadrilateral

Diagonal AC divides the given quadrilateral into two triangles i.e. Triangle ABC and Triangle ADC.

Now, Area of Quadrilateral ABCD = Area of Triangle ABC + Area of Triangle ADC.

=1/2 × AC × h 1 + 1/2 × AC × h 2 =1/2 × d × (h 1 +h 2 )

Where, d = The length of diagonal of a quadrilateral.

The area of other polygons (pentagon, hexagon, etc.) can be determined, by splitting them into a number of triangles and finding the respective areas.

Mensuration Class 8 Formulas

The important formulas covered in this chapter are as follows:

• d1 and d2 are the diagonals of the rhombus
• d is the diagonal of a special quadrilateral which is divided into two triangles
• h 1  and h 2  are the two perpendiculars from the vertices of a quadrilateral to its diagonal
• l, b and h denotes length, breadth and height of the cuboid
• r represents the radius of the cylindrical base
• a is the side of the cube

Class 8 Mensuration Important Points

• The trapezium and rhombus are the figures represented in the two-dimensional plane. Whereas Cuboid, Cube and Cylinder are three-dimensional solid shape.
• The surface area of a solid shape is the sum of the areas of its faces.
• Amount of region occupied by a solid shape is called its volume.

Volume and Capacity

Volume and capacity are almost similar to each other.

• A volume is the amount of space occupied by a three-dimensional object
• Capacity is the quantity of liquid contained in an object
• 1 mL = 1 cm 3
• 1 L = 1000 cm 3
• 1 m 3 = 1000000 cm 3 = 1000 L

More Articles for Class 8

Solved Questions for Mensuration Class 8

Example 1: Find the area of a square whose side is 30 m.

Solution: The area of square = (Side) 2

Given, side = 30 m.

Then, area of square = (30) 2 = 900 m 2 .

Example 2: Find the perimeter of a triangle with two equal sides of 5 cm and one side of 10 cm.

Solution: We know that,

perimeter of triangle = sum of all the three sides

= (5 + 5 + 10)

Example 3: Calculate the area of a trapezium whose height is 4 cm and length of parallel sides are 5 cm and 3 cm, respectively. Solution: Given, Height of the trapezium = 4cm and length of parallel sides are 5cm and 3cm respectively.

Area of trapezium =  height x (sum of parallel sides)/2

A = 4 x (5 + 3)/2

A = 4 x 8/2

A = 16 sq.cm.

Example 4: Find the area of special quadrilateral whose diagonal length is 6 cm. And the lengths of two perpendiculars on given diagonal from the vertices are 3 cm and 4 cm.

Solution: Given, diagonal of quadrilateral, d = 6 cm.

Let the length of two perpendiculars be h 1 = 3 cm and h 2 = 4 cm.

As per the formula of quadrilateral, we can write,

Area of quadrilateral ABCD =   ½ x d x (h 1  + h 2 )

A = 1/2 x 6 x (3 + 4)

= 1/2 x 6 x 7

Example 5: Find the area of rhombus in which the length of its two diagonals is 8.5cm and 10 cm.

Solution: We know;

Area of rhombus =  ½ x d 1 x d 2

A = 1/2 x 8.5 x 10

A = 42.5 sq.cm.

Frequently Asked Questions on Mensuration Class 8

How many cubic centimetres make 1 litre.

1 litre = 1000 cm 3 and 1 millilitre = 1 cm 3 .

What is the curved surface area of a cylinder?

The curved surface area of a cylinder is given by 2𝜋rh square units, where r is the radius of circular ends and h is the height of the cylinder.

What is the area of a special quadrilateral?

The area of a special quadrilateral is given by ½ x d x (h 1 + h 2 ) square units, where d is the diagonal which divided the quadrilateral into two triangles and h 1 , h 2 are the two perpendiculars from the vertices of a quadrilateral to its diagonal.

What is the area of a parallelogram?

The area of a parallelogram is given by b × h, where h is the altitude of the parallelogram on the base b.

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Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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