Important Questions for CBSE Class 8 Maths Chapter 11 - Mensuration

Very Short Answer Question                                                                      1 Mark

1.  Find the perimeter of the given figure

(Image will be uploaded soon)

(a) ${\text{11cm}}$

(b) ${\text{8}}{\text{.5cm}}$ 

(c) ${\text{13cm}}$ 

(d) ${\text{9}}{\text{.5cm}}$

 Ans: Perimeter of the given figure = ${\text{3cm + 3cm + 2}}{\text{.5cm  =  8}}{\text{.5cm}}$

Therefore, the correct option is (b)

2. Area of a trapezium = Half of the sum of the length of parallel sides × _____? 

Ans: Perpendicular distance between them.

3. The area of a parallelogram whose base is ${\text{9cm}}$and altitude is ${\text{6cm}}$ 

(a) ${\text{45c}}{{\text{m}}^{\text{2}}}$

(b) ${\text{54c}}{{\text{m}}^2}$

(c) ${\text{48c}}{{\text{m}}^{\text{2}}}$

(d) ${\text{84c}}{{\text{m}}^2}$ 

Ans: Area of parallelogram = base × altitude 

  = ${\text{9}}\;{\text{cm } \times 6}\;{\text{cm  =  54}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

Therefore, correct option is (b)

4. The volume of a cube whose edge is ${\text{6a}}$ is

(a) ${\text{25}}{{\text{a}}^{\text{3}}}$

(b) ${\text{216}}{{\text{a}}^3}$ 

(c) ${\text{125}}{{\text{a}}^{\text{3}}}$

(d) None of these 

Ans: Volume of cube = ${\text{6}}{{\text{a}}^{\text{3}}}\;{\text{ = }}\;{\text{216}}{{\text{a}}^{\text{3}}}$

Therefore, the correct option is (c)

5. The sum of the areas of all six faces of a cuboid is the _____ of the cuboid. 

(a) Volume 

(b) Surface area 

(c) Area 

(d) Curved surface area 

Ans:  Sum of the areas of all six faces of cuboid is the surface area of cuboid which is given by  ${\text{2(l b  + b h  +  h l)}}$

Therefore, the correct option is (b)

6. The area of a. Rhombus is $240\mathrm{~cm}^{2}$ and one of the diagonals is $16 \mathrm{~cm}$ Then other diagonal is

(a) ${\text{25cm}}$

(b) ${\text{30cm}}$

(c) ${\text{18cm}}$

(d) ${\text{35cm}}$ 

Ans: Area of Rhombus= ${\text{240c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;\dfrac{{\text{1}}}{{\text{2}}}{\text{(}}{{\text{d}}_{\text{1}}}{ \times }{{\text{d}}_{\text{2}}}{\text{)}}$

         = ${\text{2} \times 240}\;{\text{c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;{\text{16}}\;{\text{cm} \times }{{\text{d}}_{\text{2}}}$

        $ \Rightarrow $${{\text{d}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{2}}\;{ \times }\;{\text{240}}\;{\text{c}}{{\text{m}}^{\text{2}}}}}{{{\text{16}}\;{\text{cm}}}}{\text{ = 30}}\;{\text{cm}}$

Therefore, correct option is (b)

7. The volume of water tank is ${\text{3}}\;{{\text{m}}^{\text{3}}}$. Its capacity in litres is 

(a) ${\text{30}}$

(b) ${\text{300}}$

(c) ${\text{3000}}$

(d) None of these 

Ans:

${\text{1}}{{\text{m}}^{\text{3}}}{\text{ = 1000}}\;{\text{litres}}$

V = ${\text{3}}{{\text{m}}^{\text{3}}}{\text{ = 3(1000) = 3000}}\;{\text{litres}}$

Therefore, correct option is (c)

Short Answer Questions                                                    2 Mark

8.  Find the area of a square, the length of diagonal is $2 \sqrt{2} m$

Ans:  Area of Square = $\dfrac{{\text{1}}}{{\text{2}}}{{\text{d}}^{\text{2}}}\;\;\;\;\;\;\;\;\;\;\;\;\;(\therefore {\text{d = 2}}\sqrt {\text{2}} {\text{m}})$

where d = diagonal length

Area of Square = $\dfrac{{\text{1}}}{{\text{2}}}{{\text{(2}}\sqrt {\text{2}} \;{\text{m)}}^{\text{2}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}\;{\times }\;{\text{4}}\;{ \times }\;{\text{2 }}{{\text{m}}^{\text{2}}}{\text{ = 4}}\;{{\text{m}}^{\text{2}}}$

9. If the parallel sides of a parallelogram are ${\text{2cm}}$apart and their sum is ${\text{12cm}}$ then find its area. 

Ans: 

Opposite sides of a parallelogram are equal 

∴ ${\text{a}}\;{\text{ + }}\;{\text{b}}\;{\text{ = }}\;{\text{12}}$

${\text{a}}\;{\text{ + }}\;{\text{a = 12}}$

${\text{2a = 12}}$ 

${\text{a = 6}}$

Area of parallelogram = ${\text{a}}\;{ \times }\;{\text{h  = 6}}\;{ \times }\;{\text{2 = 12}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

10. The length, breadth and height of a cuboid are ${\text{20cm}}$, ${\text{15cm}}$ and ${\text{10cm}}$ respectively. Find its total surface area. 

Ans: 

${\text{L = 20cm , B = 15cm , H = 10cm}}$

Surface area of cuboid = ${\text{2(lb}}\;{\text{ + }}\;{\text{bh}}\;{\text{ + }}\;{\text{hl)}}$

                                   = ${{2(20 \times 15 + 15 \times 10 + 10 \times 20)c}}{{\text{m}}^{\text{2}}}$

                                   = ${\text{2(300}}\;{\text{ + }}\;{\text{150}}\;{\text{ + }}\;{\text{200)}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

                                  = ${\text{2(650)}}\;{\text{c}}{{\text{m}}^{\text{2}}}{\text{ = 1300}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

11. Volume of Cube is ${\text{8000c}}{{\text{m}}^{\text{3}}}$. Find its surface area.

Ans:

V= ${\text{8000}}\;{\text{c}}{{\text{m}}^{\text{3}}}$

 ${{\text{l}}^{\text{3}}}{\text{ = 8000}}\;{\text{c}}{{\text{m}}^{\text{3}}}$

$ \Rightarrow \;{\text{l = }}\sqrt[{\text{3}}]{{{\text{8000 c}}{{\text{m}}^{\text{3}}}{\text{ }}}}\;$

${\text{l}}\;{\text{ = }}\;{\text{20}}\;{\text{cm}}$

Surface Area of Cube = ${\text{6}}{{\text{l}}^{\text{2}}}$

                                   = ${\text{6(20cm}}{{\text{)}}^{\text{2}}}{\text{ = 2400c}}{{\text{m}}^{\text{2}}}$

12. Find the ratio of the areas of two circles whose radii is ${\text{7cm}}$ and ${\text{14cm}}$. 

Ans:

 ${{\text{r}}_{\text{1}}}{\text{ = 7cm}}$

$ \Rightarrow $ ${{\text{A}}_{\text{1}}}{\text{ = }}\;\pi {{\text{(7)}}^{\text{2}}}{\text{ = 49}}\;\pi $

${{\text{r}}_2}{\text{ = 14cm}}$ 

$ \Rightarrow $${{\text{A}}_2}{\text{ = }}\pi \;{{\text{(14)}}^{\text{2}}}{\text{ = 196}}\;\pi $

${{\text{A}}_{\text{1}}}{\text{:}}{{\text{A}}_{\text{2}}}{\text{ = 49}}\pi \,{\text{:}}\;{\text{196}}\pi $

${{\text{A}}_{\text{1}}}{\text{:}}{{\text{A}}_{\text{2}}}{\text{ = 7:28}}$

13. Find the diameter of the circle whose circumference is ${\text{230m}}$. 

Ans: Circumference = ${\text{230m}}$

${\text{2}}\pi {\text{r  =  230}}\,{\text{m}}$

${\text{r  =  }}\dfrac{{{\text{230}}}}{{{\text{2}}\pi }}{\text{ = }}\dfrac{{{{230 \times 7}}}}{{{{2 \times 22}}}}{\text{ = 36}}{\text{.6}}\;{\text{m}}$

${\text{d}}\;{\text{ = }}\;{\text{2r}}\;{\text{ = }}\;{\text{2}}\;{{ \times }}\;{\text{36}}{\text{.6}}\;{\text{m}}\;{\text{ = }}\;{\text{73}}{\text{.18}}\;{\text{cm}}$

Question (14 – 18)                      3  Mark

14. Find the area of the figure if the upper portion is a semicircle

Ans: Total area = Area of semicircle + Area of Rectangle 

Area of semicircle = $\dfrac{{\text{1}}}{{\text{2}}}\pi \;{{\text{r}}^{\text{2}}}$

With ${\text{r  = }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(length}}\;{\text{of}}\;{\text{rectangle)}}\;{\text{ =  }}\dfrac{{\text{1}}}{{\text{2}}}{{ \times 14 = 7}}$ 

Area of semicircle = $\dfrac{{\text{1}}}{{\text{2}}}{{ \times }}\dfrac{{{\text{22}}}}{{\text{7}}}{{ \times 7 \times 7 = 77}}\;{\text{c}}{{\text{m}}^{\text{2}}}$ 

Area of rectangle = length × breadth 

                     = ${\text{14}}\;{{ \times }}\;{\text{8 = 112}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

Total area = ${\text{(77}}\;{\text{ + }}\;{\text{112)}}\;{\text{c}}{{\text{m}}^{\text{2}}}{\text{ = 189}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

15. A goat is tied to one corner of a square field of side ${\text{8m}}$by a rope ${\text{3m}}$ long. Find the area it can graze? Also find the area the goat cannot graze.

Ans: 

Length of side of a square = ${\text{8}}\;{\text{m}}$

Area of square = ${{\text{(8}}\;{\text{m)}}^{\text{2}}}{\text{ = 64}}\;{{\text{m}}^{\text{2}}}$

Length of rope = ${\text{3m}}$ = ${\text{r}}$ (radius of circle) 

As the goat is tied to a corner of square plot it can only graze $\dfrac{{\text{1}}}{{\text{4}}}{\text{th}}$ of circle of radius equal to length of rope inside the plot.

Area covered (or grazed) by goat = $\dfrac{{\pi \;{{\text{r}}^{\text{2}}}}}{{\text{4}}}$ 

                                       = $\dfrac{{{\text{22}}}}{{\text{7}}}{ \times }\dfrac{{{{{\text{(3)}}}^{\text{2}}}}}{{\text{4}}}{\text{ = }}\dfrac{{{22 \times 9}}}{{{\text{28}}}}$

                                                 = ${\text{7}}{\text{.07}}\;{{\text{m}}^{\text{2}}}$

Area the goat cannot graze = Area of square – Area grazed by goat 

                                            = ${\text{64}}\;{\text{ - }}\;{\text{7}}{\text{.07  =  56}}{\text{.93}}\;{{\text{m}}^{\text{2}}}$

16. If ${\text{x}}$ units are added to the length of the radius of a circle, what is the number of units by which the circumference of the circle is increased? 

Ans: Let the radius of the circle be ‘${\text{r}}$’ units 

The circumference of the circle will be ${\text{2}}\pi \;{\text{r}}$ units

If the radius is increased by ‘${\text{x}}$’ units, the new radius will be ${\text{(r + x)}}$ units. 

New circumference will be ${\text{2}}\pi \;{\text{(r}}\;{\text{ + }}\;{\text{x)  =  2}}\pi \;{\text{r}}\;{\text{ + }}\;{\text{2}}\pi {\text{x}}$

Circumference increased by ${\text{2}}\pi \;{\text{x}}$ units.

17. Find the area of the shaded portion if diameter of circle is ${\text{16cm}}$ and ABCD is a square.

Ans: Area of shaded portion = (Area of circle with radius =${\text{8cm}}$) – (Area of square with diagonal length = ${\text{16cm}}$)

= $\pi \;{{\text{r}}^{\text{2}}}{\text{ - }}\dfrac{{\text{1}}}{{\text{2}}}{{\text{d}}^{\text{2}}}$

= $\dfrac{{22}}{7} \times {(8)^2} - \dfrac{1}{2}{(16)^2}$

= $\dfrac{{22 \times 64}}{7} - 128$

= ${\text{201}}{\text{.1}}\;{\text{ - }}\;{\text{128}}\;{\text{ = }}\;{\text{73}}{\text{.1}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

18. How many ${\text{c}}{{\text{m}}^{\text{3}}}$ of juice can be poured in a cuboidal can whose dimensions are $\text{15cm} \times \text{10cm} \times \text{25cm}$. How many cubical packs of ${\text{25c}}{{\text{m}}^{\text{3}}}$ volume can be made? 

Ans: Volume of cuboid = Length $\times$ Breadth $\times$ Height

                              =  ${\text{15}}\;{\text{cm}}\;{ \times }\;{\text{10}}\;{\text{cm}}\;{\times}\;{\text{25}}\;{\text{cm}}$

Volume of juice in cuboidal can =  ${\text{3750}}\;{\text{c}}{{\text{m}}^{\text{3}}}$

Each volume of cubical packet =   ${\text{25}}\;{\text{c}}{{\text{m}}^{\text{3}}}$ 

Number of such cubical packets made from the volume of juice in cuboidal can 

${\text{n  =  }}\dfrac{{{\text{volume}}\;{\text{of}}\;{\text{juice}}\;{\text{in}}\;{\text{cuboidal}}\;{\text{can}}}}{{{\text{each}}\;{\text{cubical}}\;{\text{pack}}\;{\text{volume}}}}$

${\text{n  =  }}\dfrac{{{\text{3750}}\;{\text{c}}{{\text{m}}^{\text{3}}}}}{{{\text{25}}\;{\text{c}}{{\text{m}}^{\text{3}}}}}$

${\text{n  =  150}}$ packets

Question (19 – 25)                      5  Mark

19. A rectangular piece of paper${\text{66}}\;{\text{cm}}$long and${\text{10}}\;{\text{cm}}$broad is rolled along the length to form a cylinder. What is the radius of the base and calculate volume of cylinder?

Ans:

When the rectangular piece is rolled in the form of a cylinder then the length became the circumference of the base of cylinder

C = ${\text{66}}$,

${\text{2}}\pi \;{\text{r  =  66}}$

$\pi \;{\text{r  =  }}\dfrac{{{\text{66}}}}{{\text{2}}}{\text{ = 33}}$

${\text{r  =  }}\dfrac{{{\text{33}}\;{{ \times }}\;{\text{7}}}}{{{\text{22}}}}{\text{ =  10}}{\text{.5}}\;{\text{cm}}$

Volume of Cylinder with radius = ${\text{10}}{\text{.5cm}}$ ang height = ${\text{10cm}}$

${\text{V = }}\pi \;{{\text{r}}^{\text{2}}}{\text{h = }}\;\dfrac{{{\text{22}}}}{{\text{7}}}\;{{ \times }}\;{{\text{(10}}{\text{.5)}}^{\text{2}}}{{ \times }}\;{\text{10}}$

   = ${\text{3465}}\;{\text{c}}{{\text{m}}^{\text{3}}}$

20. ABCD has area equal to ${\text{28}}$. BC is parallel to AD. BA is perpendicular to AD. If BC is ${\text{6}}$ and AD is ${\text{8}}$, then what is CD? 

Ans: The shape of the given figure is a trapezium

Area of Trapezium = $\dfrac{1}{2}\text{(sum of parallel sides)} \times \text{height}$

Given area of ABCD = ${\text{28}}$, BC = ${\text{6}}$, AD = $8$, CD = ?

${\text{28  =  A  =  }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(6}}\;{\text{ + }}\;{\text{8)}}\;{\text{h}}$

${\text{h  =  }}\dfrac{{{\text{28}}\;{{ \times }}\;{\text{2}}}}{{{\text{14}}}}{\text{ = 4}}\;{\text{units}}$

To find CD: let DE perpendicular to AD (construction done) 

In triangle CED, 

ED = AD – AE 

ED = ${\text{8 - 6}}$

ED = ${\text{2}}$

${\text{C}}{{\text{D}}^{\text{2}}}{\text{ = C}}{{\text{E}}^{\text{2}}}{\text{ + E}}{{\text{D}}^{\text{2}}}$

${\text{C}}{{\text{D}}^{\text{2}}}{\text{ = }}\;{{\text{(4)}}^{\text{2}}}{\text{ + }}\;{{\text{(2)}}^{\text{2}}}{\text{ = 16}}\;{\text{ + }}\;{\text{4}}\;{\text{ = }}\;{\text{20}}$

${\text{CD  =  }}\sqrt {{\text{20}}} \;{\text{ = }}\;{\text{2}}\sqrt {\text{5}} $

21. From the adjoining figure find the area of shaded portion

Ans: From the figure, 

Area of shaded portion = [Area of rectangle with ${\text{l = 24}}$, ${\text{b = 10}}$] – [Area of rectangle with ${\text{b = 6}}$, ${\text{l = 10}}$+ Area of square with side = ${\text{4}}$]

Area of big rectangle = ${\text{l}}\;{{ \times }}\;{\text{b}}$

                                   = ${\text{24}}\;{{ \times }}\;{\text{10}}\;{\text{  =  240}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

Area of small rectangle =  ${\text{l}}\;{{ \times }}\;{\text{b}}$

                                      =   ${\text{6}}\;{{ \times }}\;{\text{10 }}\;{\text{ =  }}\;{\text{60}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

Area of squares = ${\text{4}}\;{{ \times }}\;{\text{4}}\;{\text{ = }}\;{\text{16}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

Therefore, Area of shaded portion = ${\text{240}}\;{\text{ - }}\;{\text{(60}}\;{\text{ + }}\;{\text{16)}}$

                                                    = ${\text{240}}\;{\text{ - }}\;{\text{76  =  164}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

22. A flooring tile has a shape of a parallelogram whose base is ${\text{28cm}}$ and the corresponding height is ${\text{20cm}}$. How many such tiles are required to cover a floor of area ${\text{2800}}{{\text{m}}^{\text{2}}}$. 

Ans: Given, Base = ${\text{28cm}}$ , height = ${\text{20cm}}$

Area of floor =  ${\text{2800}}\;{{\text{m}}^{\text{2}}}$

                    = ${\text{2800}}\;\;{{ \times }}\;{\text{1}}{{\text{0}}^{\text{4}}}\;{\text{c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;\;{\text{28}}\;{{ \times }}\;{\text{1}}{{\text{0}}^{\text{6}}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

Area of each parallelogram tile = base × height 

                                                   = ${\text{28}}\;{{ \times }}\;{\text{20  =  560}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

Number of tiles required =  $\dfrac{{{\text{Area}}\;{\text{of}}\;{\text{floor}}}}{{{\text{Area}}\;{\text{of}}\;{\text{tiles}}}}{\text{ = }}\dfrac{{{\text{28} \times 1}{{\text{0}}^{\text{6}}}}}{{{\text{560}}}}$  = $\dfrac{{{\text{1}}{{\text{0}}^{\text{5}}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{100000}}}}{{\text{2}}}{\text{ = 50000}}$

23. Rain water which falls on a flat rectangular surface of length ${\text{6m}}$ and breadth ${\text{4m}}$ is transferred into a cylindrical vessel of internal radius ${\text{20cm}}$. What will be the height of water in the cylindrical vessel if the rain fall is ${\text{1cm}}$ (Take $\pi \;{\text{ = }}\;{\text{3}}{\text{.14}}$) 

Ans:

Since the water in the rectangular surface is transferred to the cylindrical vessel. 

Length of surface = ${\text{6}}\;{\text{m  =  600}}\;{\text{cm}}$ 

Breadth of surface = ${\text{4}}\;{\text{m  =  400}}\;{\text{cm}}$

Height of water level = ${\text{1}}\;{\text{cm}}$

Volume of water on the surface = ${\text{l}}\;{{ \times }}\;{\text{b}}\;{{ \times }}\;{\text{h}}$

$\begin{align} & =\text{600}\ \times \ \text{400}\ \times \ \text{1} \\ & =\text{240000}\ \text{c}{{\text{m}}^{\text{3}}} \\ \end{align}$

Let ‘${\text{h}}$’ be the height of the cylindrical vessel, ${\text{r  =  20}}\;{\text{cm}}$ (radius of cylindrical vessel) 

Volume of cylindrical vessel = $\pi \;{{\text{r}}^{\text{2}}}{\text{h}}$

                                            = $\;\pi {{\text{(20)}}^{\text{2}}}\;{{ \times }}\;{\text{h}}$

Volume of water on surface = Volume of water in cylindrical vessel

${\text{24000  =  }}\pi {{\text{(20)}}^{\text{2}}}{{ \times h}}$

${\text{h  =  }}\dfrac{{{\text{24000}}}}{{\pi \;{{ \times }}\;{\text{20}}\;{{ \times }}\;{\text{20}}}}{\text{ = 191}}{\text{.08}}\;{\text{cm}}$

24. If each edge of a cube is doubled 

(a) how many times will its surface area increases 

(b) how many times will its volume increases 

Ans:

For side of ‘x’ units, surface area ${{\text{s}}_{\text{1}}}\;{\text{ = }}\;{\text{6}}{{\text{x}}^{\text{2}}}$

When side of cube is doubled (${\text{2x}}$ units) 

Surface area ${{\text{s}}_{\text{2}}}{\text{ = }}\;{\text{6(2x}}{{\text{)}}^{\text{2}}}$

${{\text{s}}_{\text{2}}}{\text{ = }}\;{\text{6}}\;{{ \times }}\;{\text{4(x}}{{\text{)}}^{\text{2}}}\;{\text{ = }}\;\;{\text{4(6}}{{\text{x}}^{\text{2}}}{\text{)}}\;{\text{ = }}\;{\text{4}}{{\text{s}}_{\text{1}}}$

Surface area increases by ${\text{4}}$ times. 

Volume for edge of ‘${\text{x}}$’ units ${{\text{v}}_{\text{1}}}\;{\text{ = }}\;{{\text{x}}^{\text{3}}}$

Volume of cube when edge is doubled ${\text{(2x)}}$, ${{\text{v}}_{\text{2}}}\;{\text{ = }}\;{{\text{(2x)}}^{\text{3}}}$

                                                                     ${{\text{v}}_{\text{2}}}{\text{ = }}\;\;{\text{8(x}}{{\text{)}}^{\text{3}}}\;{\text{ = }}\;{\text{8}}{{\text{v}}_{\text{1}}}$

Therefore, volume increases by 8 times.

25. A box with measures ${\text{80cm} \times \text{48cm} \times \text{24cm}}$ is to be covered with a tarpaulin cloth how many metres of tarpaulin cloth of width ${\text{96cm}}$ is required to cover ${\text{50}}$ such boxes?  

Ans: The box with ${\text{l}}\;{\text{ = }}\;{\text{80}}\;{\text{cm,}}\;{\text{b}}\;{\text{ = }}\;{\text{48}}\;{\text{cm,}}\;{\text{h}}\;{\text{ = }}\;{\text{24}}\;{\text{cm}}$

Total surface area = ${\text{2(lb}}\;{\text{ + }}\;{\text{bh}}\;{\text{ + }}\;{\text{hl)}}$

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

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

                             = ${\text{2[6912]  =  13824}}\;{\text{c}}{{\text{m}}^{\text{2}}}$

Length of cloth required = ${\text{(}}\dfrac{{{\text{Area}}\;{\text{of}}\;{\text{box}}}}{{{\text{breadth}}}}{\text{)}}\;{{ \times }}\;{\text{50}}$

                                       = ${\text{(}}\dfrac{{13824}}{{96}}{\text{)}}\;{{ \times }}\;{\text{50  =  144}}\; \times \;{\text{50}}$

                                       = ${\text{7200}}\;{\text{cm = 72}}\;{\text{m}}$

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