case study questions on mensuration

• Mensuration Practice Questions

Mensuration Practice Questions section contains all the important question types that you will have to answer in the section. We will talk about volumes and areas in the following section and this will help you check where you stand in the Mensuration Practice Questions. Let us see more.

Q 1: Find the volume and surface area of a cuboid 16m long, 14 m broad and 7 m high.

A) 868 cm         B) 920 cm                C) 868 cm 2               D) 920 cm 2 

Q 2: Find the length of the longest pole that can be placed in a room 12 m long, 8m broad and 9 m high.

A) 16 m       B) 17 m           C) 18 m            D) 19 m

Q 3: The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 12.8 cu. meters. Find the breadth of the wall.

A) 0.04m        B) 4m           C) 400 cm              D) 40 cm

Example 4: Find the number of bricks, each measuring 24 cm×12 cm × 8 cm, required to construct a wall 24 m long, 8m high and 60 cm thick if 10% of the wall is filled with mortar?

A) 450        B) 4500          C) 45000              D) 450000

Q5: The area of the base of a rectangular tank is 6500 cm 2 and the volume of water contained in it is 2.6 cubic meters. The depth of water in the tank is:

A) 3.5 m         B) 4 m           C) 5 m                 D) 6 m

Q6: Given that one cubic cm of marble weighs 25 gms, the weight of a marble block 28 cm in width and 5 cm thick is 112 kg. The length of the block is:

A) 26.5 cm           B) 32 cm          C) 36 cm           D) 37.5 cm

Q7: Half cubic meter of the gold sheet is extended by hammering so as to cover an area of one hectare. The thickness of the sheet is:

A) 0.0005 cm              B) 0.005 cm                C) 0.05 cm          D) 0.5 cm

Q8: In a shower, 5 cm of rain falls. The volume of water that falls on 1.5 hectares of the ground is:

A) 75 cu m                   B) 750 cu. m               C) 7500 cu. m               D) 75000 cu. m

Find Your Answers Here

Q1: C),   Q2:  B), Q3: D), Q4:  C), Q5: B), Q6: B), Q7: B), Q8: B)

Browse more Topics under Mensuration

• Volumes and Areas
• Results on Triangles
• Results on Quadrilaterals
• Cylinder, Cone and Sphere
• Data Sufficiency

Q1: Water flows into a tank 200 m × 150 m through a rectangular pipe 1.5 m × 1.25 m @ 20 kmph. In what time (in minutes) will the water rise by 2 meters?

A) 234 minutes             B) 1.2 hours          C) 9 hours          D) 96 minutes

Q2: The dimensions of an open box are 50 cm, 40 cm, and 23 cm. Its thickness is 3 cm. If 1 cubic cm of metal used in the box weighs 0.5 gms, find the weight of the box.

A) 6. 08 kg             B) 8.04 kg          C) 8.06 kg              D) 6.04 kg

Q3: A cube of edge 15 cm is immersed completely in a rectangular vessel containing water. If the dimensions of the base of the vessel are 20 cm × 15 cm, find the rise in water level.    [RRB 2003]

A) 11 cm                B) 11.11 cm           C) 11. 22 cm            D) 11.25 cm

Q4: A conical vessel, whose internal radius is 12 cm and height 50 cm, is full of some liquid. The contents of this vessel are emptied into a cylindrical vessel with an internal radius of 10 cm. Find the height to which the liquid rises in the cylindrical vessel.

A) 22 cm           B) 23 cm                C) 24 cm           D) 25 cm

Q5: How many spherical bullets can be made out of a lead cylinder 28 cm high and with base radius 6 cm, each bullet being 1.5 cm in diameter?      [RRB 2003]

A) 1600            B) 1793           C) 1601           D) 1792

Q6: If the radius of a sphere is increased by 50%, find the increase percent in volume and the increase percent in the surface area.

A) 95%         B) 100 %        C) 115%         D) 125%

Q7: Two metallic right circular cones having their heights 4.1 cm and 4.3 cm and the radii of their bases 2.1 cm each, have been melted together and recast into a sphere. Find the diameter of the sphere.

A) 4 cm         B) 4.1 cm          C) 4.2 cm              D) 4.3 cm

Q8: A hemispherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl?

A) 23             B) 34                 C) 54                     D) 46

Q1: D), Q2: B), Q3: D), Q4: C), Q5: D), Q6: D), Q7: C), Q8: C)

Directions: Each of the questions given below consists of a statement and/or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the given question. Read both the statements and give an answer:

(a) when the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.

(b) if the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.

(c) the data either in Statement I or in Statement II alone are sufficient to answer the question.

(d) if the data even in both Statements I and II together are not efficient to answer the question.

(e) the data in both Statements I and II together are necessary to give the answer.

Q1: What is the weight of the iron beam?

I. The beam is 9 m long, 40 cm wide and 20 cm high.

II. Iron weighs 50 kg per cubic meter.

Q2: What is the volume of 32 meter high cylindrical tank?         [Bank PO 2003]

I. The area of its base is 154 sq. m.

II. The diameter of the base is 14 m.

Q3: What is the volume of a cube?                  [Bank PO 2003]

I. The area of each face of the cube is 64 sq. m.

II. Length of one side of the cube is 8 metres.

Q1: (e), Q2: (c), Q3: (c)

Customize your course in 30 seconds

Which class are you in.

Mensuration

• Results of Quadrilaterals
• Results on Triangle
• Results on Depreciation

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Download the App

Talk to our experts

1800-120-456-456

• Important Questions for CBSE Class 8 Maths Chapter 11 - Mensuration

CBSE Class 8 Maths Important Questions for Mensuration - Free PDF Download

Math is a tricky subject to study, and thus having proper notes can boost up the studying procedure and eventually your grades. Important Questions for Class 8 Maths Chapter 11 is a student lifesaver consisting of all the necessary questions that might come in the exam. Professionals at Vedantu have especially curated these Class 8 Maths Chapter 11 Important Questions, keeping in mind the question paper's easy and difficult level. Students can get a quick revision before the exam by going through these Mensuration Class 8 Important Questions. 

Vedantu is a platform that provides free (CBSE) NCERT Solution and other study materials to the students. Download Class 8 Maths and Class 8 Science NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations.

Study Important Questions for Class 8 Mathematics 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  ${\text{2(l b  + b h  +  h l)}}$

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}}$

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) ${\text{3000}}$

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

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. 

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. 

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

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.

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 ${\text{14cm}}$. 

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

$ \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.

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?

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 $\pi \;{\text{ = }}\;{\text{3}}{\text{.14}}$) 

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 

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

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

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

Download PDF of Class 8 Maths Mensuration Important Questions

Mensuration is a relatively easy chapter that can be completely studied by the student in a matter of hours, provided that he/she is determined and diligently practised. Important Questions For Class 8 Maths Mensuration can help students to prepare for their exam in no time. Below are the topics listed within the chapter and the questions are based on these topics.

Topics Included in Class 8 Maths Chapter 11

Following are the main topics in chapter mensuration as the division in the textbook. These topics help the students strengthen their base knowledge about the chapter to answer related questions.

Mensuration deals with studying kinds of geometrical shapes; for example, it's an area, length, volume, and perimeters. Below is a list of the table containing the name of shapes and the formula to calculate the questioned value.

Benefits of Using Important Questions of Class 8 Maths for Mensuration 

1. Expertly Curated Material: Vedantu's Class 8 Maths Mensuration Important Questions PDF is curated by top-notch math professors for a solid understanding of foundational concepts.

2. Comprehensive Coverage for Chapter 11: The important questions for Mensuration in Chapter 11 encompass all essential topics, presented in an understandable manner.

3. Historical Connection of Math: Math, a subject integral to humanity since ancient times, was utilized in systems like the Indus Valley's numbering system for trade and commerce.

4. Contemporary Importance: In today's world, math remains crucial, with daily applications such as calculating grocery bills, showcasing its significance in modern life.

Why is NCERT Class 8 Mensuration Chapter is Vital?

Class 8 Mensuration chapter enhances the core of conceptual knowledge. It gives a full view of the prescribed syllabus, as the NCERT textbook doesn't contain an intricate description of the syllabus and the solutions. Every critical theorem in the chapter is fully solved, which makes the concept significantly simpler for students.

Conclusion 

Reviewing all the crucial questions for Class 8 Maths Chapter 11 - Mensuration provides students with a solid grasp of the chapter's topics. The extra and important questions for Class 8 Maths Chapter 11 - Mensuration engage in a concept-focused discussion, encompassing all chapter themes. This question-and-answer method proves time-saving during exam prep, offering an efficient way to revise the chapter and enhance understanding. Practising these important questions streamlines preparation and boosts confidence for the upcoming exams.

FAQs on Important Questions for CBSE Class 8 Maths Chapter 11 - Mensuration

1. How can we get perfect scores on the NCERT Solutions for Class 8 Maths Chapter 11 - Mensuration in class tests?

You may get perfect scores on class tests and in board exams by following the Solutions for Class 8 Maths Chapter 11 - Mensuration offered on the Vedantu website. For quick and simple review during quizzes and exams, these solutions are really essential. For learners, this is the best study material right now.

2. What key ideas are discussed in Chapter 11 of the NCERT Solutions for Class 8 Math?

NCERT Solutions for Class 8 Chapter 11 Maths' primary subject areas are:

3. What is a trapezium?

A convex quadrilateral having exactly one set of opposite sides that are parallel to one another is called a trapezium . The trapezium is a two-dimensional shape that resembles a table when it is drawn on paper. A quadrilateral is a polygon in Euclidean geometry that has four sides and four vertices. A trapezium therefore also has four vertices, four angles, and four sides.

4. What is the difference between Volume and capacity?

The total amount of space an object occupies in three dimensions is indicated by its volume. The term "capacity" describes something's ability to contain, absorb, or receive an object (such as a solid substance, gas, or liquid). Both solid and hollow objects have volume.. The capability only applies to hollow things.

eg- Suppose, for instance, that there is a water-filled tank. Thus, the volume of a tank refers to the area that the tank and its water occupy, but the capacity relates to the amount of water required to fill the tank.

5. What is a cylinder and what is the formula for the cylinder's curved surface area?

One of the most fundamental curvilinear geometric shapes, a cylinder , has historically been a three-dimensional solid. It is regarded as a prism with a circle as its base in basic geometry. In several contemporary fields of geometry and topology, a cylinder can alternatively be characterised as an infinitely curved surface.

A cylinder's curved surface area is equal to 2πrh square units, where r is the radius of the cylinder's circular ends and h is its height.

Chapterwise Important Questions for CBSE Class 8 Maths

Cbse study materials.

• Bihar Board

SRM University

Tn sslc result 2024.

• TN Board Result 2024
• GSEB Board Result 2024
• Karnataka Board Result 2024
• CG Board Result 2024
• Kerala Board Result 2024
• Shiv Khera Special
• Education News
• Web Stories
• Current Affairs
• नए भारत का नया उत्तर प्रदेश
• School & Boards
• College Admission
• Govt Jobs Alert & Prep
• GK & Aptitude
• CBSE Class 10 Study Material

CBSE Class 10 Maths Case Study Questions for Chapter 13 - Surface Areas and Volumes (Published By CBSE)

Check case study questions for cbse class 10 maths chapter 13 - surface areas and volumes. this question bank based on case study has been published by the board for class 10 mathematics..

Case study questions for CBSE Class 10 Maths Chapter 13 - Surface Areas and Volumes are published by the CBSE board. Students can solve these questions to acquaint themselves with the new type of questions. Answers to all the questions have been provided with them. These case study questions are quite helpful to prepare for the CBSE Class 10 Maths Exam 2021-2022.

Case Study Questions for Class 10 Maths Chapter 10 - Surface Areas and Volumes

CASE STUDY 1:

Adventure camps are the perfect place for the children to practice decision making for themselves without parents and teachers guiding their every move. Some students of a school reached for adventure at Sakleshpur. At the camp, the waiters served some students with a welcome drink in a cylindrical glass and some students in a hemispherical cup whose dimensions are shown below. After that they went for a jungle trek. The jungle trek was enjoyable but tiring. As dusk fell, it was time to take shelter. Each group of four students was given a canvas of area 551m2 . Each group had to make a conical tent to accommodate all the four students. Assuming that all the stitching and wasting incurred while cutting, would amount to 1m2 , the students put the tents. The radius of the tent is 7m.

1. The volume of cylindrical cup is

a) 295.75 cm 3

b) 7415.5 cm 3

c) 384.88 cm 3

d) 404.25 cm 3

Answer: d) 404.25 cm 3

2. The volume of hemispherical cup is

a) 179.67 cm 3

b) 89.83 cm 3

c) 172.25 cm 3

d) 210.60 cm 3

Answer: b) 89.83 cm 3

3. Which container had more juice and by how much?

a) Hemispherical cup, 195 cm 3

b) Cylindrical glass, 207 cm 3

c) Hemispherical cup, 280.85 cm 3

d) Cylindrical glass, 314.42 cm 3

Answer: d) Cylindrical glass, 314.42 cm 3

4. The height of the conical tent prepared to accommodate four students is

Answer: c) 24m

5. How much space on the ground is occupied by each student in the conical tent

b) 38.5 m 2

Answer: b) 38.5 m 2

CASE STUDY 2:

The Great Stupa at Sanchi is one of the oldest stone structures in India, and an important monument of Indian Architecture. It was originally commissioned by the emperor Ashoka in the 3rd century BCE. Its nucleus was a simple hemispherical brick structure built over the relics of the Buddha. .It is a perfect example of combination of solid figures. A big hemispherical dome with a cuboidal structure mounted on it. (Take π = 22/7)

1. Calculate the volume of the hemispherical dome if the height of the dome is 21 m –

a) 19404 cu. m

b) 2000 cu .m

c) 15000 cu. m

d) 19000 cu. m

Answer: a) 19404 cu. m

2. The formula to find the Volume of Sphere is -

a) 2/3 πr 3

b) 4/3 πr 3

c) 4 πr 2

d) 2 πr 2

Answer: b) 4/3 πr 3

3. The cloth require to cover the hemispherical dome if the radius of its base is 14m is

a) 1222 sq.m

b) 1232 sq.m

c) 1200 sq.m

d) 1400 sq.m

Answer: b) 1232 sq.m

4. The total surface area of the combined figure i.e. hemispherical dome with radius 14m and cuboidal shaped top with dimensions 8m 6m 4m is

a)1200 sq. m

b) 1232 sq. m

c) 1392 sq.m

d) 1932 sq. m

Answer: c) 1392 sq.m

5. The volume of the cuboidal shaped top is with dimensions mentioned in question 4

a) 182.45 m 3

b) 282.45 m 3

Answer: d) 192 m 3

CASE STUDY 3:

On a Sunday, your Parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’s cube and strawberry ice-cream for you. Observe the figures and answer the questions:-

1. The length of the diagonal if each edge measures 6cm is

a) 3√3

b) 3√6

c) √12

d) 6√3

Answers: d) 6√3

2. Volume of the solid figure if the length of the edge is 7cm is

b) 196 cm 3

c) 343 cm 3

d) 434 cm 3

Answers: c) 343 cm 3

3. What is the curved surface area of hemisphere (ice cream) if the base radius is 7cm?

a) 309 cm 2

b) 308 cm 2

c) 803 cm 2

d) 903 cm 2

Answers: b) 308 cm 2

4. Slant height of a cone if the radius is 7cm and the height is 24 cm___

Answers: b) 25 cm

5. The total surface area of cone with hemispherical ice cream is

a) 858 cm 2

b) 885 cm 2

c) 588 cm 2

d) 855 cm 2

Answers: a) 858 cm 2

Also Check:

CBSE Class 10 Maths Case Study Questions - All Chapters

Tips to Solve Case Study Based Questions Accurately

Get here latest School , CBSE and Govt Jobs notification in English and Hindi for Sarkari Naukari and Sarkari Result . Download the Jagran Josh Sarkari Naukri App . Check  Board Result 2024  for Class 10 and Class 12 like  CBSE Board Result ,  UP Board Result ,  Bihar Board Result ,  MP Board Result ,  Rajasthan Board Result  and Other States Boards.

• SSC GD Result 2024
• RBSE 10th, 12th Result 2024
• SSLC Result 2024 Tamil Nadu
• tnresults.nic.in 10th Result 2024
• dge.tn.gov.in Result 2024
• 10th Public Exam Result 2024 Tamil Nadu
• DHSE Kerala Plus Two Result 2024
• CGBSE 10th Result 2024
• CGBSE 12th Result 2024
• NDA Result 2024
• CBSE Study Material
• CBSE Class 10

Latest Education News

GSEB SSC Result 2024 LIVE Updates: Gujarat Class 10th Results Link Online at gseb.org, Check via Seat Number

SSC GD Result 2024 Live: Constable Results Direct Link on ssc.gov.in; Check Expected Cut Off, Merit List Date

SSC GD Constable Result 2024 Live Update: जल्द जारी होने वाला है एसएससी कांस्टेबल रिजल्ट ssc.gov.in पर, यहाँ देखें एक्सपेक्टेड कटऑफ मार्क्स

Kerala SSLC Result 2024 [ഫലം] Out LIVE: Check KBPE Class 10 Results Link at Official Website: results.kite.kerala.gov.in by Roll Number, School-wise

SSLC Result 2024 Karnataka LIVE (ಫಲಿತಾಂಶ) OUT: KSEEB 10th Results Online at kseab.karnataka.gov.in, karresults.nic.in by Login and Registration Number

UGC NET Previous Year Question Paper PDF for Paper 1 & 2, Download Subject-wise PYQs

WBPSC Clerkship Salary 2024: Check In-Hand Pay, Structure, Perks and Allowances

BSEB STET Exam Date 2024: इसी माह हो सकती है बिहार एसटीईटी की परीक्षा, जल्द घोषित होगी तारीखें

[यहां देखें] Purple Cap in IPL 2024: किसके नाम सबसे ज्यादा विकेट, कौन निकलेगा सबसे आगे?

Microsoft Gears Up for Mobile Domination: Xbox Game Store Arrives on iOS and Android

Lok Sabha Elections 2024: प्रधानमंत्री पद पर रहते हुए लोकसभा चुनाव में किस नेता को मिली है हार? पढ़ें

GT IPL 2024 Match Schedule: गुजरात टाइटंस के सभी मैचों का वेन्यू, डेट और टाइम यहां देखें

IPL 2024 CSK Players: चेन्नई सुपर किंग्स के खिलाड़ियों की पूरी लिस्ट यहां देखें

ICC T20 World Cup 2024: T20 वर्ल्ड कप का शेड्यूल जारी, कब और किससे है भारत का मैच देखें यहां

India T20 World Cup 2024 Squad: ये है रोहित की ‘विराट’ टीम, किस रोल में कौन? देखें यहां

Most Sixes In IPL 2024: आईपीएल में चौकों-छक्कों की रेस में कौन सबसे आगे? देखें पूरी लिस्ट

[चेक] Most Runs In IPL 2024: दिलचस्प हो गयी है Orange Cap की रेस, Virat टॉप पर

Fastest 50s In IPL History: किसने जड़ा है आईपीएल इतिहास का सबसे तेज़ अर्द्धशतक? देखें पूरी लिस्ट

[Direct LINK] karresults.nic.in, sslc.karnataka.gov.in 2024 SSLC Results Out: Official Website to Check Karnataka 10th Result Online

IPL 2024: हिंदी हो या भोजपुरी इन चर्चित कमेंटेटरों के साथ आईपीएल का लें लुफ्त

If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

To log in and use all the features of Khan Academy, please enable JavaScript in your browser.

Unit 11: Mensuration

• Perimeter: introduction (Opens a modal)
• Finding missing side length when given perimeter (Opens a modal)
• Find perimeter when given side lengths Get 5 of 7 questions to level up!
• Find a missing side length when given perimeter Get 3 of 4 questions to level up!
• Counting unit squares to find area formula (Opens a modal)
• Finding missing side when given area (Opens a modal)
• Area of rectangles Get 5 of 7 questions to level up!
• Find a missing side length when given area Get 5 of 7 questions to level up!

Area of composite figures

• Decomposing shapes to find area: grids (Opens a modal)
• Decomposing shapes to find area: add (Opens a modal)
• Decomposing shapes to find area: subtract (Opens a modal)
• Understand decomposing figures to find area Get 3 of 4 questions to level up!
• Decompose figures to find area Get 3 of 4 questions to level up!

Mensuration word problems

• Perimeter word problem: tables (Opens a modal)
• Perimeter word problem: skating rink (Opens a modal)
• Area word problem: house size (Opens a modal)
• Perimeter word problems Get 3 of 4 questions to level up!
• Area word problems Get 3 of 4 questions to level up!

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Case Study Questions for Class 8 Maths

• Last modified on: 9 months ago
• Reading Time: 7 Minutes

Table of Contents

Here in this article, we are providing case study questions for class 8 maths.

Maths Class 8 Chapter List

Latest chapter list (2023-24).

Chapter 1 Rational Numbers Chapter 2 Linear Equations in One Variable Chapter 3 Understanding Quadrilaterals Chapter 4 Data Handling Chapter 5 Squares and Square Roots Chapter 6 Cubes and Cube Roots Chapter 7 Comparing Quantities Chapter 8 Algebraic Expressions and Identities Chapter 9 Mensuration Chapter 10 Exponents and Powers Chapter 11 Direct and Indirect proportions Chapter 12 Factorisation Chapter 13 Introduction to Graphs

Old Chapter List

Chapter 1 Rational Numbers Chapter 2 Linear Equations in One Variable Chapter 3 Understanding Quadrilaterals Chapter 4 Practical Geometry Chapter 5 Data Handling Chapter 6 Squares and Square Roots Chapter 7 Cubes and Cube Roots Chapter 8 Comparing Quantities Chapter 9 Algebraic Expressions and Identities Chapter 10 Visualising Solid Shapes Chapter 11 Mensuration Chapter 12 Exponents and Powers Chapter 13 Direct and Indirect proportions Chapter 14 Factorisation Chapter 15 Introduction to Graphs Chapter 16 Playing with Numbers

Tips for Answering Case Study Questions for Class 8 Maths in Exam

1. Comprehensive Reading for Context: Prioritize a thorough understanding of the provided case study. Absorb the contextual details and data meticulously to establish a strong foundation for your solution.

2. Relevance Identification: Pinpoint pertinent mathematical concepts applicable to the case study. By doing so, you can streamline your thinking process and apply appropriate methods with precision.

3. Deconstruction of the Problem: Break down the complex problem into manageable components or steps. This approach enhances clarity and facilitates organized problem-solving.

4. Highlighting Key Data: Emphasize critical information and data supplied within the case study. This practice aids quick referencing during the problem-solving process.

5. Application of Formulas: Leverage pertinent mathematical formulas, theorems, and principles to solve the case study. Accuracy in formula selection and unit usage is paramount.

6. Transparent Workflow Display: Document your solution with transparency, showcasing intermediate calculations and steps taken. This not only helps track progress but also offers insight into your analytical process.

7. Variable Labeling and Definition: For introduced variables or unknowns, offer clear labels and definitions. This eliminates ambiguity and reinforces a structured solution approach.

8. Step Explanation: Accompany each step with an explanatory note. This reinforces your grasp of concepts and demonstrates effective application.

9. Realistic Application: When the case study pertains to real-world scenarios, infuse practical reasoning and logic into your solution. This ensures alignment with real-life implications.

10. Thorough Answer Review: Post-solving, meticulously review your answer for accuracy and coherence. Assess its compatibility with the case study’s context.

11. Solution Recap: Before submission, revisit your solution to guarantee comprehensive coverage of the problem and a well-organized response.

12. Previous Case Study Practice: Boost your confidence by practicing with past case study questions from exams or textbooks. This familiarity enhances your readiness for the question format.

13. Efficient Time Management: Strategically allocate time for each case study question based on its complexity and the overall exam duration.

14. Maintain Composure and Confidence: Approach questions with poise and self-assurance. Your preparation equips you to conquer the challenges presented.

Download CBSE Books

Exam Special Series:

• Sample Question Paper for CBSE Class 10 Science (for 2024)
• Sample Question Paper for CBSE Class 10 Maths (for 2024)
• CBSE Most Repeated Questions for Class 10 Science Board Exams
• CBSE Important Diagram Based Questions Class 10 Physics Board Exams
• CBSE Important Numericals Class 10 Physics Board Exams
• CBSE Practical Based Questions for Class 10 Science Board Exams
• CBSE Important “Differentiate Between” Based Questions Class 10 Social Science
• Sample Question Papers for CBSE Class 12 Physics (for 2024)
• Sample Question Papers for CBSE Class 12 Chemistry (for 2024)
• Sample Question Papers for CBSE Class 12 Maths (for 2024)
• Sample Question Papers for CBSE Class 12 Biology (for 2024)
• CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
• Master Organic Conversions CBSE Class 12 Chemistry Board Exams
• CBSE Important Numericals Class 12 Physics Board Exams
• CBSE Important Definitions Class 12 Physics Board Exams
• CBSE Important Laws & Principles Class 12 Physics Board Exams
• 10 Years CBSE Class 12 Chemistry Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Physics Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Maths Previous Year-Wise Solved Papers (2023-2024)
• 10 Years CBSE Class 12 Biology Previous Year-Wise Solved Papers (2023-2024)
• ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
• ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
• ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
• ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
• ICSE Important Functions and Locations Based Questions Class 10 Biology
• ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)

✨ Join our Online JEE Test Series for 499/- Only (Web + App) for 1 Year

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

1 thought on “ Case Study Questions for Class 8 Maths ”

• Pingback: Case Study Questions for CBSE Class 8 Maths - ebookpublisher.in

Leave a Reply Cancel reply

Join our Online Test Series for CBSE, ICSE, JEE, NEET and Other Exams

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading

• RD Sharma Solutions
• Chapter 20 Mensuration I

RD Sharma Solutions for Class 7 Maths Chapter 20 Mensuration - I (Perimeter and Area of Rectilinear Figures)

RD Sharma Solutions for Class 7 Maths Chapter 20 – Mensuration – I (Perimeter and Area of Rectilinear Figures) are the best study materials for students who find it difficult to solve complex problems. Students can refer to and download the PDF of RD Sharma Solutions for Class 7 Maths Chapter 20 from the given links. The solutions to all questions in the RD Sharma textbook for Class 7 are given here in a detailed and step-by-step way to help the students understand the concepts more effectively. Chapter 20 Mensuration – I (Perimeter and Area of Rectilinear Figures) is based on the Perimeter and area of rectilinear figures.

• RD Sharma Solutions Class 7 Maths Chapter 1 Integers
• RD Sharma Solutions Class 7 Maths Chapter 2 Fractions
• RD Sharma Solutions Class 7 Maths Chapter 3 Decimals
• RD Sharma Solutions Class 7 Maths Chapter 4 Rational Numbers
• RD Sharma Solutions Class 7 Maths Chapter 5 Operations on Rational Numbers
• RD Sharma Solutions Class 7 Maths Chapter 6 Exponents
• RD Sharma Solutions Class 7 Maths Chapter 7 Algebraic Expressions
• RD Sharma Solutions Class 7 Maths Chapter 8 Linear Equations in One Variable
• RD Sharma Solutions Class 7 Maths Chapter 9 Ratio and Proportion
• RD Sharma Solutions Class 7 Maths Chapter 10 Unitary Method
• RD Sharma Solutions Class 7 Maths Chapter 11 Percentage
• RD Sharma Solutions Class 7 Maths Chapter 12 Profit and Loss
• RD Sharma Solutions Class 7 Maths Chapter 13 Simple Interest
• RD Sharma Solutions Class 7 Maths Chapter 14 Lines and Angles
• RD Sharma Solutions Class 7 Maths Chapter 15 Properties of Triangles
• RD Sharma Solutions Class 7 Maths Chapter 16 Congruence
• RD Sharma Solutions Class 7 Maths Chapter 17 Constructions
• RD Sharma Solutions Class 7 Maths Chapter 18 Symmetry
• RD Sharma Solutions Class 7 Maths Chapter 19 Visualising Solid Shapes
• RD Sharma Solutions Class 7 Maths Chapter 20 Mensuration I (Perimeter and Area of Rectilinear Figures)
• RD Sharma Solutions Class 7 Maths Chapter 21 Mensuration II (Area of Circle)
• RD Sharma Solutions Class 7 Maths Chapter 22 Data Handling I (Collection and Organisation of Data)
• RD Sharma Solutions Class 7 Maths Chapter 23 Data Handling II (Central Values)
• RD Sharma Solutions Class 7 Maths Chapter 24 Data Handling III (Constructions of Bar Graphs)
• RD Sharma Solutions Class 7 Maths Chapter 25 Data Handling IV (Probability)
• Exercise 20.1 Chapter 20 Mensuration – I
• Exercise 20.2 Chapter 20 Mensuration – I
• Exercise 20.3 Chapter 20 Mensuration – I
• Exercise 20.4 Chapter 20 Mensuration – I

RD Sharma Solutions cover all the topics related to it. Some of the important topics included in this chapter are listed below.

• Square millimetre
• Square centimetre
• Square decimeter
• Square metre
• Square kilometre
• Perimeter and area of squares and rectangles
• Area between rectangles
• Areas of a parallelogram and rhombus
• Area of a triangle

RD Sharma Solutions for Class 7 Maths Chapter 20 Mensuration – I (Perimeter and Area of Rectilinear Figures)

carouselExampleControls111

Previous Next

Access answers to Maths RD Sharma Solutions for Class 7 Chapter 20 – Mensuration – I (Perimeter and Area of Rectilinear Figures)

Exercise 20.1 Page No: 20.8

1. Find the area, in square metres, of a rectangle whose

(i) Length = 5.5 m, breadth = 2.4 m

(ii) Length = 180 cm, breadth = 150 cm

(i) Given Length = 5.5 m, Breadth = 2.4 m

We know that area of rectangle = Length x Breadth

= 5.5 m x 2.4 m = 13.2 m 2

(ii) Given Length = 180 cm = 1.8 m, Breadth = 150 cm = 1.5 m [Since 100 cm = 1 m]

= 1.8 m x 1.5 m = 2.7 m 2

2. Find the area, in square centimetres, of a square whose side is

(ii) 1.2 dm

(i) Given side of the square = 2.6 cm

We know that area of the square = (Side) 2

= (2.6 cm) 2

= 6.76 cm 2

(ii) Given side of the square = 1.2 dm

= 1.2 x 10 cm = 12 cm [Since 1 dm = 10 cm]

= (12 cm) 2

3. Find in square meters, the area of a square of side 16.5 dam.

Given side of the square = 16.5

dam = 16.5 x 10 m = 165 m [Since 1 dam/dm (decametre) = 10 m]

Area of the square = (Side) 2

= (165 m) 2

= 27225 m 2

4. Find the area of a rectangular field in acres whose sides are:

(1) 200 m and 125 m

(ii) 75 m 5 dm and 120 m

(i) Given length of the rectangular field = 200 m

Breadth of the rectangular field = 125 m

We know that area of the rectangular field = Length x Breadth

= 200 m x 125 m

= 25000 m 2

= 250 acres [Since 100 m 2  = 1 acre]

(ii) Given length of the rectangular field =75 m 5 dm

= (75 + 0.5) m

= 75.5 m [Since 1 dm = 10 cm = 0.1 m]

Breadth of the rectangular field = 120 m

= 75.5 m x 120 m

= 90.6 acres [Since 100 m 2  = 1 acre]

5. Find the area of a rectangular field in hectares whose sides are:

(i) 125 m and 400 m

(i) Given length of the rectangular field = 125 m

Breadth of the rectangular field = 400 m

We know that the area of the rectangular field = Length x Breadth

= 125 m x 400 m

= 50000 m 2

= 5 hectares [Since 10000 m 2  = 1 hectare]

= 0.906 hectares [Since 10000 m 2  = 1 hectare]

6. A door of dimensions 3 m x 2m is on the wall of dimension 10 m x 10 m. Find the cost of painting the wall if rate of painting is Rs 2.50 per sq. m.

Given length of the door = 3 m

Also given that breadth of the door = 2 m

Side of the wall = 10 m

We know that Area of the square = (Side) 2

Area of the wall = Side x Side = 10 m x 10 m

We know that the area of the rectangle = length x breadth

Area of the door = Length x Breadth = 3 m x 2 m = 6 m

Thus, required area of the wall for painting = Area of the wall – Area of the door

= (100 – 6) m 2

Rate of painting per square metre = Rs. 2.50

Hence, the cost of painting the wall = Rs. (94 x 2.50) = Rs. 235

7. A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side. Also, find which side encloses more area?

Given length of rectangular shaped wire = 40cm

Breadth of rectangular shaped wire = 22cm

Perimeter of the rectangle = 2(Length + Breadth)

= 2(40 cm + 22 cm)

It is given that the wire which was in the shape of a rectangle is now bent into a square.

Therefore, the perimeter of the square = Perimeter of the rectangle

Perimeter of the square = 124 cm

We know that perimeter of square = 4 x side

Therefore 4 x side = 124 cm

Side = 124/4 = 31 cm

We know that area of rectangle = length x breadth

Now, Area of the rectangle

= 40 cm x 22 cm

= (31 cm) 2

= 961 cm 2 .

Therefore, the square-shaped wire encloses more area.

8. How many square meters of glass will be required for a window, which has 12 panes, each pane measuring 25 cm by 16 cm?

Given length of the glass pane = 25 cm

Breadth of the glass pane = 16 cm

Area of one glass pane = 25 cm x 16 cm

= 0.04 m 2 [Since 1 m 2  = 10000 cm 2 ]

Thus, Area of 12 panes = 12 x 0.04 = 0.48 m 2

9. A marble tile measures 10 cm x 12 cm. How many tiles will be required to cover a wall of size 3 m x 4 m? Also, find the total cost of the tiles at the rate of Rs 2 per tile.

Given area of the wall = 3 m x 4 m

Also given that area of one marble tile = 10 cm x 12 cm

= 0.012 m 2  [Since 1 m 2  = 10000 c m 2 ]

Number of tiles required to cover the wall = Area of wall/ Area of one marble tile

= 1000 tiles

Cost of one tile = Rs. 2

Total cost = Number of tiles x Cost of one tile

= Rs. (1000 x 2)

10. A table top is 9 dm 5 cm long 6 dm 5 cm broad. What will be the cost to polish it at the rate of 20 paise per square centimetre?

Given length of the table top = 9 dm 5 cm

= (9 x 10 + 5) cm

= 95 cm [Since 1 dm = 10 cm]

Also given that Breadth of the table top = 6 dm 5 cm

= (6 x 10 + 5) cm

Area of the table top = Length x Breadth

= (95 cm x 65 cm)

= 6175 c m 2

Rate of polishing per square centimetre = 20 paise = Rs. 0.20 [since 1Rs. = 100 paise]

Total cost to polish table top = (6175 x 0.20)

11. A room is 9.68 m long and 6.2 m wide. Its floor is to be covered with rectangular tiles of size 22 cm by 10 cm. Find the total cost of the tiles at the rate of Rs 2.50 per tile.

Given length of the floor of the room = 9.68 m

Breadth of the floor of the room = 6.2 m

Area of the floor = 9.68 m x 6.2 m

= 60.016 m 2

Given that length of the tile = 22 cm

Breadth of the tile = 10 cm

Area of one tile = 22 cm x 10 cm

= 0.022 m 2  [Since 1 m 2  = 10000 c m 2 ]

Thus, Number of tiles = 60.016 m 2 /0.022 m 2

We have cost of one tile = Rs. 2.50

= (2728 x 2.50)

12. One side of a square field is 179 m. Find the cost of raising a lawn on the field at the rate of Rs 1.50 per square metre.

Given that one side of the square field = 179 m

Area of the field = (Side)  2

= (179 m)  2

= 32041 m 2

Rate of raising a lawn on the field per square metre = Rs. 1.50

Therefore total cost of raising a lawn on the field = (32041 x 1.50)

= Rs. 48061.50

13. A rectangular field is measured 290 m by 210 m. How long will it take for a girl to go two times round the field, if she walks at the rate of 1.5 m/sec?

Given length of the rectangular field = 290 m

Also breadth of the rectangular field = 210 m

We know that perimeter of the rectangular field = 2(Length + Breadth)

= 2(290 + 210)

Distance covered by the girl = 2 x Perimeter of the rectangular field

The girl walks at the rate of 1.5 m/sec.

Speed of the girl = 1.5 x 60 m/min

Speed = Distance/ Time

So, time taken by the girl = Distance/Speed

14. A corridor of a school is 8 m long and 6 m wide. It is to be covered with canvas sheets. If the available canvas sheets have the size 2 m x 1 m, find the cost of canvas sheets required to cover the corridor at the rate of Rs 8 per sheet.

Given length of the corridor = 8 m

Also breadth of the corridor = 6 m

Area of the corridor of a school = Length x Breadth

= (8 m x 6 m)

We have length of the canvas sheet = 2 m

Breadth of the canvas sheet = 1 m

Area of one canvas sheet = Length x Breadth

= (2 m x 1 m)

Thus, Number of canvas sheets required to cover the corridor = 48/2 = 24

Given Cost of one canvas sheet = Rs. 8

Total cost of the canvas sheets = number of canvas sheet x cost of one canvas sheet

= (24 x 8) = Rs. 192

15. The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm respectively. Find the cost of turfing it at Rs 2.50 per square meter. How long will a man take to go three times round the field, if he walks at the rate of 2 meters per second?

Given that length of playground = 62 m 60 cm

= 62.6 m [Since 10 cm = 0.1 m]

Breadth of a playground = 25 m 40 cm

Area of a playground = Length x Breadth

= 62.6 m x 25.4 m

= 1590.04 m 2

Given rate of turfing = Rs. 2.50/ m 2

Total cost of turfing = (1590.04 x 2.50)

= Rs. 3975.10

Perimeter of a rectangular field = 2(Length + Breadth)

= 2(62.6 + 25.4) = 176 m

Distance covered by the man in 3 rounds of a field = 3 x Perimeter of a rectangular field

= 3 x 176 m = 528 m

The man walks at the rate of 2 m/sec.

Speed of man walks = 2 x 60 m/min = 120 m/min

Thus, required time to cover a distance of 528 m = 528/120

= 4 minutes 24 seconds [Since 0.1 minutes = 6 seconds]

16. A lane 180 m long and 5 m wide is to be paved with bricks of length 20 cm and breadth 15 cm. Find the cost of bricks that are required, at the rate of Rs 750 per thousand.

Given that length of lane = 180 m

Breadth of the lane = 5 m

Area of a lane = Length x Breadth

= 180 m x 5 m

Length of the brick = 20 cm

Breadth of the brick = 15 cm

Area of a brick = Length x Breadth

= 20 cm x 15 cm

= 0.03 m 2  [Since 1 m 2  = 10000 cm 2 ]

Required number of bricks = 900/0.03   = 30000

Cost of 1000 bricks = Rs. 750

Total cost of 30,000 bricks = 750×30,000/1000

= Rs. 22,500

17. How many envelopes can be made out of a sheet of paper 125 cm by 85 cm; supposing one envelope requires a piece of paper of size 17 cm by 5 cm?

Given length of the sheet of paper = 125 cm

Also given that breadth of the sheet of paper = 85 cm

Area of a sheet of paper = Length x Breadth

= 125 cm x 85 cm

= 10,625 cm 2

Length of sheet required for an envelope = 17 cm

Breadth of sheet required for an envelope = 5 cm

Area of the sheet required for one envelope = Length x Breadth

= 17 cm x 5 cm

Thus, required number of envelopes = 10,625/85

18. The width of a cloth is 170 cm. Calculate the length of the cloth required to make 25 diapers, if each diaper requires a piece of cloth of size 50 cm by 17 cm.

Given that length of the diaper = 50 cm

Breadth of the diaper = 17 cm

Area of cloth to make 1 diaper = Length x Breadth

= 50 cm x 17 cm

Thus, Area of 25 such diapers = (25 x 850) cm 2

= 21,250 cm 2

Area of total cloth = Area of 25 diapers = 21,250 cm 2

It is given that width of a cloth = 170 cm

Length of the cloth = Area of cloth

We have width of cloth = 170cm

Length of a cloth = 21,250/170 cm = 125 cm

Hence, length of the cloth will be 125 cm.

19. The carpet for a room 6.6 m by 5.6 m costs Rs 3960 and it was made from a roll 70 cm wide. Find the cost of the carpet per meter.

Given that length of a room = 6.6 m

Breadth of a room = 5.6 m

Area of a room = Length x Breadth

= 6.6 m x 5.6 m

= 36.96 m 2

Width of a carpet = 70 cm = 0.7 m [Since 1 m = 100 cm]

Length of a carpet x Width of a carpet = Area of a room

36.96 = Length of the carpet × 0.7

Length of the carpet = 36.96/0.7

Length of the carpet = 52.8 m

Cost of the carpet per metre = Total cost of the carpet/Length of the carpet

= 3960/52.8

20. A room is 9 m long, 8 m broad and 6.5 m high. It has one door of dimensions 2 m x 1.5 m and three windows each of dimensions 1.5 m x 1 m. Find the cost of white washing the walls at Rs 3.80 per square meter.

Given Length of a room = 9 m

Breadth of a room = 8 m

Height of a room = 6.5 m

Area of 4 walls = 2(1 + b) h

= 2 (9 m + 8 m) x 6.5 m

= 2 x 17 m x 6.5 m

Length of a door = 2 m

Breadth of a door = 1.5 m

Area of a door = Length x Breadth

= 2 m x 1.5 m

Length of a window = 1.5 m

Breadth of a window = 1 m

Since, area of one window = Length x Breadth

= 1.5 m x 1 m

Thus, Area of 3 such windows = 3 x 1.5 m 2

Area to be white washed = Area of 4 walls – (Area of one door + Area of 3 windows)

Area to be white washed = [221 – (3 + 4.5)] m 2

= (221 – 7.5) m 2

= 213.5 m 2

We have cost of white washing for 1 m 2  area = Rs. 3.80

Cost of white washing for 213.5 m 2  area = Rs. (213.5 x 3.80)

= Rs. 811.30

21. A hall 36 m long and 24 m broad allowing 80 m 2 for doors and windows, the cost of papering the walls at Rs 8.40 per m 2 is Rs 9408. Find the height of the hall.

Given length of the hall = 36 m

Breadth of the hall = 24 m

Let h be the height of the hall.

Now, in papering the wall, we need to paper the four walls excluding the floor and roof of the hall. So, the area of the wall which is to be papered = Area of 4 walls

= 2 h (I + b)

= 2 h (36 + 24)

Now, area left for the door and the windows = 80 m 2

So, the area which is actually papered = (120h – 80) m 2

Cost of papering the wall = Area required for papering × Cost per meter square of papering wall

9408 = (120 h – 80) × 8.40

1120 = 120 h – 80

1200 = 120 h

Hence, the height of the wall would be 10 m.

Exercise 20.2 Page No: 20.13

1. A rectangular grassy lawn measuring 40 m by 25 m is to be surrounded externally by a path which is 2 m wide. Calculate the cost of leveling the path at the rate of Rs 8.25 per square meter.

Let PQRS be the rectangular grassy lawn and let length = AB and breadth = BC

Given Length PQ = 40 m and breadth QR = 25 m

Area of lawn PQRS = 40 m x 25 m = 1000 m 2

Length AB = (40 + 2 + 2) m = 44 m

Breadth BC = (25 + 2 + 2) m = 29 m

Area of ABCD = 44 m x 29 m = 1276 m 2

Now, Area of the path = Area of ABCD – Area of the lawn PQRS

= 1276 m 2  – 1000 m 2

Rate of levelling the path = Rs. 8.25 per m 2

Cost of levelling the path = (8.25 x 276)

2. One meter wide path is built inside a square park of side 30 m along its sides. The remaining part of the park is covered by grass. If the total cost of covering by grass is Rs 1176, find the rate per square metre at which the park is covered by the grass.

Given that side of a square garden = 30m = a

We know that area of square = a 2

Area of the square garden including the path = a 2

= (30) 2  = 900 m 2

From the figure, it can be observed that the side of the square garden, when the path is not included, is 28 m.

Area of the square garden not including the path = (28) 2

Total cost of covering the park with grass = Area of the park covering with green grass x Rate per square metre

1176 = 784 x Rate per square metre

Rate per square metre at which the park is covered with grass = (1176/784)

= Rs. 1.50 per m 2

3. Through a rectangular field of sides 90 m x 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the center of the field. If the width of the roads is 3 m, find the total area covered by the two roads.

Given length of rectangular field = 90m

Breadth of rectangular field = 60m

Area of the rectangular field = 90 m x 60 m = 5400 m 2

Area of the road PQRS = 90 m x 3 m

Area of the road ABCD = 60 m x 3 m

Clearly, area of KLMN is common to the two roads.

Thus, area of KLMN = 3 m x 3 m

Hence, area of the roads = Area (PQRS) + Area (ABCD) – Area (KLMN)

= (270 + 180) m 2  – 9 m 2

4. From a rectangular sheet of tin, of size 100 cm by 80 cm, are cut four squares of side 10 cm from each corner. Find the area of the remaining sheet.

Given that length of rectangular sheet = 100 cm

Breadth of rectangular sheet = 80 cm

Area of the rectangular sheet of tin = 100 cm x 80 cm

= 8000 c m 2

Side of the square at the corner of the sheet = 10 cm

Area of one square at the corner of the sheet = (10 cm) 2

Area of 4 squares at the corner of the sheet = 4 x 100 cm 2

Hence, Area of the remaining sheet of tin = Area of the rectangular sheet – Area of the 4 squares

Area of the remaining sheet of tin = (8000 – 400) cm 2

= 7600 cm 2

5. A painting 8 cm long and 5 cm wide is painted on a cardboard such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.

Given length of cardboard including margin = 8 cm

Breadth of the cardboard including margin = 5 cm

Area of the cardboard including the margin = 8 cm x 5 cm = 40 c m 2

From the figure, we can observed that,

New length of the painting without margin = 8 cm – (1.5 cm + 1.5 cm)

= (8 – 3) cm

New breadth of the painting without margin = 5 cm – (1.5 cm + 1.5 cm)

= (5 – 3) cm

Area of the painting not including the margin = 5 cm x 2 cm = 10 cm 2

Hence, Area of the margin = Area of the cardboard including the margin – Area of the painting

= (40 – 10) cm 2

6. Rakesh has a rectangular field of length 80 m and breadth 60 m. In it, he wants to make a garden 10 m long and 4 m broad at one of the corners and at another corner, he wants to grow flowers in two floor-beds each of size 4 m by 1.5 m. In the remaining part of the field, he wants to apply manures. Find the cost of applying the manures at the rate of Rs 300 per area.

Given that length of a rectangular field = 80 m

Breadth of a rectangular field = 60 m

Area of rectangular field = length x breadth

= (80 x 60) m

Again, Area of the garden = 10 m x 4 m = 40 m 2

Area of one flower bed = 4 m x 1.5 m = 6 m 2

Thus, Area of two flower beds = 2 x 6 m 2  = 12 m 2

Remaining area of the field for applying manure = Area of the rectangular field – (Area of the garden + Area of the two flower beds)

Remaining area of the field for applying manure = 4800 m 2  – (40 + 12) m 2

= (4800 – 52) m 2

Since 100 m 2  = 1 acre

Therefore by using the above

4748 m 2  = 47.48 acres

So, cost of applying manure at the rate of Rs. 300 per area will be = Rs. (300 x 47.48)

= Rs. 14244

7. Each side of a square flower bed is 2 m 80 cm long. It is extended by digging a strip 30 cm wide all around it. Find the area of the enlarged flower bed and also the increase in the area of the flower bed.

Given side of the flower bed = 2 m 80 cm = 2.80 m [since 100cm = 1m]

Area of the square flower bed = (Side) 2

= (2.80 m) 2

Side of the flower bed with strip = 2.80 m + 30 cm + 30 cm

= (2.80 + 0.3 + 0.3) m

Area of the enlarged flower bed with the digging strip = (Side) 2

= 11.56 m 2

Thus, Increase in the area of the flower bed = Area of the enlarged flower bed with the digging strip – Area of the square flower bed

= 11.56 m 2  – 7.84 m 2

8. A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies an area of 22 m 2 , find the width of the verandah.

Let the width of the verandah be x m.

Given Length of the room AB = 5 m and breadth of the room, BC = 4 m

Area of the room = 5 m x 4 m

From the figure, it is clear that

Length of the veranda PQ = (5 + x + x) = (5 + 2x) m

Breadth of the veranda QR = (4 + x + x) = (4 + 2x) m

Area of veranda PQRS = (5 + 2x) x (4 + 2x)

= (4× 2 + 18x + 20) m 2

Area of veranda = Area of PQRS – Area of ABCD

22 = 4x 2  + 18x + 20 – 20

22 = 4x 2  + 18x

On dividing above equation by 2 we get,

11 = 2x 2  + 9x

2x 2  + 9x – 11 = 0

2x 2  + 11x – 2x – 11 = 0

x (2x+11) – 1 (2x+11) = 0

(x- 1) (2x+11)= 0

When x – 1 = 0, x = 1

When 2x + 11 = 0, x = (-11/2)

The width cannot be a negative value. So, width of the veranda = x = 1 m

9. A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 m 2 , find the area of the lawn.

Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let side of the lawn AB be x m.

Area of the square lawn = x 2

Given that Length PQ = (x + 2 + 2) = (x + 4) m

Area of PQRS = (x + 4) 2  = (x 2  + 8x + 16) m 2

Now, Area of the path = Area of PQRS – Area of the square lawn

136 = x 2  + 8x + 16 – x 2

136 = 8x + 16

136 – 16 = 8x

x = 120/ 8 = 15

Side of the lawn = 15 m

Hence, Area of the lawn = (Side) 2

10. A poster of size 10 cm by 8 cm is pasted on a sheet of cardboard such that there is a margin of width 1.75 cm along each side of the poster. Find (i) the total area of the margin (ii) the cost of the cardboard used at the rate of Rs 0.60 per c m 2 .

Given length of poster = 10 cm and breadth of poster = 8 cm

Area of the poster = Length x Breadth

= 10 cm x 8 cm

Length of the cardboard when the margin is included = 10 cm + 1.75 cm + 1.75 cm

Breadth of the cardboard when the margin is included = 8 cm + 1.75 cm + 1.75 cm

Area of the cardboard = Length x Breadth

= 13.5 cm x 11.5 cm

= 155.25 c m 2

(i) Area of the margin = Area of cardboard including the margin – Area of the poster

= 155.25 c m 2  – 80 c m 2

= 75.25 c m 2

(ii) Cost of the cardboard = Area of cardboard x Rate of the cardboard Rs 0.60 per cm 2

= (155.25 x 0.60)

= Rs. 93.15

11. A rectangular field is 50 m by 40 m. It has two roads through its center, running parallel to its sides. The widths of the longer and shorter roads are 1.8 m and 2.5 m respectively. Find the area of the roads and the area of the remaining portion of the field.

Let ABCD be the rectangular field and KLMN and PQRS the two rectangular roads with width 1.8 m and 2.5 m, respectively.

Given that length of rectangular field = 50 m

Breadth of rectangular field = 40m

Area of rectangular field ABCD = 50 m x 40 m

Area of the road KLMN = 40 m x 2.5 m

Area of the road PQRS = 50 m x 1.8 m

Clearly area of EFGH is common to the two roads.

Thus, Area of EFGH = 2.5 m x 1.8 m = 4.5 m 2

Hence, Area of the roads = Area of KLMN + Area of PQRS – Area of EFGH

= (100 m 2  + 90 m 2 ) – 4.5 m 2

= 185.5 m 2

Area of the remaining portion of the field = Area of the rectangular field ABCD – Area of the roads

= (2000 – 185.5) m 2

= 1814.5 m 2

12. There is a rectangular field of size 94 m x 32 m. Three roads each of 2 m width pass through the field such that two roads are parallel to the breadth of the field and the third is parallel to the length. Calculate: (i) area of the field covered by the three roads (ii) area of the field not covered by the roads.

Let ABCD be the rectangular field.

From the figure it is clear that two roads which are parallel to the breadth of the field KLMN and EFGH with width 2 m each.

One road which is parallel to the length of the field PQRS with width 2 m.

We have from the question, length of the rectangular field AB = 94 m and breadth of the rectangular field BC = 32 m

Area of the rectangular field = Length x Breadth

= 94 m x 32 m

Area of the road KLMN = 32 m x 2 m

Area of the road EFGH = 32 m x 2 m

Area of the road PQRS = 94 m x 2 m

Clearly area of VTUI and WXYZ is common to these three roads.

Thus, Area of VTUI = 2 m x 2 m = 4 m 2

Area of WXYZ = 2 m x 2 m = 4 m 2

(i) Area of the field covered by the three roads = Area of KLMN + Area of EFGH + Area (PQRS) – (Area of VTUI + Area of WXYZ)

= [64+ 64 + 188 – (4 + 4)] m 2

= 316 m 2  – 8 m 2

(ii) Area of the field not covered by the roads = Area of the rectangular field ABCD – Area of the field covered by the three roads

= 3008 m 2  – 308 m 2

13. A school has a hall which is 22 m long and 15.5 m broad. A carpet is laid inside the hall leaving all around a margin of 75 cm from the walls. Find the area of the carpet and the area of the strip left uncovered. If the width of the carpet is 82 cm, find the cost at the rate of Rs 18 per meter.

Given that length of hall PQ = 22 m and breadth of hall QR = 15.5 m

Area of the school hall PQRS = 22 m x 15.5 m

Length of the carpet AB = 22 m – (0.75 m + 0.75 m)

= 20.5 m [Since 100 cm = 1 m]

Breadth of the carpet BC = 15.5 m – (0.75 m + 0.75 m)

Area of the carpet ABCD = 20.5 m x 14 m

Area of the strip = Area of the school hall PQRS – Area of the carpet ABCD

= 341 m 2  – 287 m 2

Now, area of the 1 m length of carpet = 1 m x 0.82 m

Thus, Length of the carpet whose area is 287 m 2  = 287 m 2  + 0.82 m 2

Cost of the 287.82 m long carpet = Rs. 18 x 287.82

= Rs. 5180.76

14. Two cross roads, each of width 5 m, run at right angles through the center of a rectangular park of length 70 m and breadth 45 m parallel to its sides. Find the area of the roads. Also, find the cost of constructing the roads at the rate of Rs 105 per m 2 .

Let ABCD be the rectangular park then EFGH and IJKL are the two rectangular roads with width 5m.

Given that length of rectangular park = 70m

Breadth of rectangular park = 45m

Area of the rectangular park = Length x Breadth

= 70 m x 45 m

Area of the road EFGH = 70 m x 5 m

Now, Area of the road JKLI = 45 m x 5 m

From the figure, it is clear that area of MNOP is common to the two roads.

Thus, Area of MNOP = 5 m x 5 m = 25 m 2

Area of the roads = Area of EFGH + Area of JKLI – Area of MNOP

= (350 + 225) m 2 – 25 m 2

Again, it is given that the cost of constructing the roads = Rs. 105 per m 2

Cost of constructing 550 m 2  area of the roads

= Rs. (105 x 550)

= Rs. 57750.

15. The length and breadth of a rectangular park are in the ratio 5: 2. A 2.5 m wide path running all around the outside the park has an area 305 m 2 . Find the dimensions of the park.

Given that area of the park be 305m 2 .

Let the length of the park be 5x m and breadth be 2x m.

Area of the rectangular park = (5x) x (2x)

= 10x 2  m 2

Width of the path = 2.5 m

From the figure,

Outer length PQ = 5x m + 2.5 m + 2.5 m

= (5x + 5) m

Outer breadth QR = 2x + 2.5 m + 2.5 m

= (2x + 5) m

Area of PQRS = (5x + 5) m x (2x + 5) m

= (10x 2  + 25x + 10x + 25) m 2

= (10x 2  + 35x + 25) m 2

Area of the path = [(10x 2  + 35x + 25) – 10x 2 ] m 2

By solving above equation we get

305 = 35x + 25

305 – 25 = 35x

x = 280/35 = 8

Length of the park = 5x

= 5 x 8 = 40 m

Breadth of the park = 2x

= 2 x 8 = 16 m

16. A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m 2 , find the area of the lawn.

Let the side of lawn be x m.

Given that width of the path = 2.5 m

Side of the lawn including the path = (x + 2.5 + 2.5) m

= (x + 5) m

Therefore, area of lawn = (Area of the lawn including the path) – (Area of the path)

We know that the area of a square = (Side) 2

Area of lawn (x 2 ) = (x + 5) 2  – 165

x 2  = (x 2  + 10x + 25) – 165

165 = 10x + 25

165 – 25 =10x

Therefore x = 140/10 = 14

Thus the side of the lawn = 14 m

The area of the lawn = (14 m)  2

Exercise 20.3 Page No: 20.20

1. Find the area of a parallelogram with base 8 cm and altitude 4.5 cm.

Given base = 8 cm and altitude = 4.5 cm

We know that area of the parallelogram = Base x Altitude

= 8 cm x 4.5 cm

Therefore, area of parallelogram = 36 cm 2

2. Find the area in square meters of the parallelogram whose base and altitudes are as under

(i) Base =15 dm, altitude = 6.4 dm

(ii) Base =1 m 40 cm, altitude = 60 cm

(i) Given base =15 dm, altitude = 6.4 dm

By converting these to standard form we get,

Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m

Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m

= 1.5 m x 0.64 m

Area of parallelogram = 0.96 m 2

(ii) Given base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]

Altitude = 60 cm = 0.6 m [Since 100 cm = 1 m]

= 1.4 m x 0.6 m

3.  Find the altitude of a parallelogram whose area is 54 d m 2  and base is 12 dm.

Given area of the given parallelogram = 54 d m 2

Base of the given parallelogram = 12 dm

Therefore altitude of the given parallelogram = Area/Base

4. The area of a rhombus is 28 m 2 . If its perimeter be 28 m, find its altitude.

Given perimeter of a rhombus = 28 m

But we know that perimeter of a rhombus = 4 (Side)

4(Side) = 28 m

Side = 28/4

Now, Area of the rhombus = 28 m 2

But we know that area of rhombus = Side x Altitude

(Side x Altitude) = 28 m 2

(7 x Altitude) = 28 m 2

Altitude = 28/7 = 4 m

5. In Fig. 20, ABCD is a parallelogram, DL ⊥ AB and DM ⊥ BC. If AB = 18 cm, BC =12 cm and DM= 9.3 cm, find DL.

Given DL ⊥ AB and DM ⊥ BC

Taking BC as the base, BC = 12 cm and altitude DM = 9.3 cm

We know that area of parallelogram ABCD = Base x Altitude

= (12 cm x 9.3 cm)

= 111.6 c m 2  ….. Equation (i)

Now, by taking AB as the base,

We have, Area of the parallelogram ABCD = Base x Altitude

= (18 cm x DL) ….. Equation (ii)

From (i) and (ii), we have

18 cm x DL = 111.6 c m 2

DL = 111.6/18

6. The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm. If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter side.

Let ABCD is a parallelogram with the longer side AB = 54 cm and corresponding altitude AE = 16 cm.

The shorter side is BC and the corresponding altitude is CF = 24 cm.

We know that area of a parallelogram = base x height.

We have two altitudes and two corresponding bases.

By equating them we get,

½ x BC x CF = ½ x AB x AE

On simplifying, we get

BC x CF = AB x AE

BC x 24 = 54 x 16

BC = (54 × 16)/24

Hence, the length of the shorter side BC = AD = 36 cm.

7. In Fig. 21, ABCD is a parallelogram, DL ⊥ AB. If AB = 20 cm, AD = 13 cm and area of the parallelogram is 100 c m 2 , find AL.

From the figure we have ABCD is a parallelogram with base AB = 20 cm and corresponding altitude DL.

It is given that the area of the parallelogram ABCD = 100 c m 2

We know that the area of a parallelogram = Base x Height

100 = AB x DL

100 = 20 x DL

DL = 100/20 = 5 cm

By observing the picture it is clear that we have to apply the Pythagoras theorem,

Therefore by Pythagoras theorem, we have,

(AD) 2  = (AL) 2  + (DL) 2

(13) 2  = (AL) 2 + (5) 2

(AL) 2 = (13) 2  – (5) 2

(AL) 2 = 169 – 25

We know that 12 2 = 144

(AL) 2  = (12) 2

Hence, length of AL is 12 cm.

8. In Fig. 21, if AB = 35 cm, AD= 20 cm and area of the parallelogram is 560 cm 2 , find LB.

From the figure, ABCD is a parallelogram with base AB = 35 cm and corresponding altitude DL.

The adjacent side of the parallelogram AD = 20 cm.

It is given that the area of the parallelogram ABCD = 560 cm 2

Now, Area of the parallelogram = Base x Height

560 cm 2  = AB x DL

560 cm 2  = 35 cm x DL

DL = 560/35

Again by Pythagoras theorem, we have, (AD) 2  = (AL) 2  + (DL) 2

(20) 2 = (AL) 2  + (16) 2

(AL) 2  = (20) 2  – (16) 2

= 400 – 256

From the figure, AB = AL + LB

35 = 12 + LB

LB = 35 – 12 = 23 cm

Hence, length of LB is 23 cm.

9. The adjacent sides of a parallelogram are 10 m and 8 m. If the distance between the longer sides is 4 m, find the distance between the shorter sides

Let ABCD is a parallelogram with side AB = 10 m and corresponding altitude AE = 4 m.

The adjacent side AD = 8 m and the corresponding altitude is CF.

We know that area of a parallelogram = Base x Height

So, equating them we get

AD x CF = AB x AE

8 x CF = 10 x 4

CF = (10 x 4)/8 = 5 m

Hence, the distance between the shorter sides is 5 m.

10. The base of a parallelogram is twice its height. If the area of the parallelogram is 512 cm 2 , find the base and height.

Let the height of the parallelogram be x cm.

Then the base of the parallelogram is 2x cm. [from given data]

Given that the area of the parallelogram = 512 cm 2

512 = (2x) (x)

x 2  = 512/2

x 2  = (16) 2

Hence height of parallelogram = x = 16cm

And base of the parallelogram = 2x = 2 x 16 = 32 cm

11. Find the area of a rhombus having each side equal to 15 cm and one of whose diagonals is 24 cm.

Let ABCD be the rhombus where diagonals intersect at 0 as given in the figure.

Then AB = 15 cm and AC = 24 cm.

The diagonals of a rhombus bisect each other at right angles.

Therefore, from the figure triangle A0B is a right-angled triangle, right angled at O such that

Therefore, OA = ½(AC) = ½ (24) = 12 cm and AB = 15 cm.

By applying Pythagoras theorem, we get,

(AB) 2  = (OA) 2  + (OB) 2

(15) 2  = (12) 2  + (OB) 2

(OB) 2 = (15) 2 – (12) 2

(OB) 2  = 225 – 144

(OB) 2  = (9) 2

BD = 2 x OB

Hence, Area of the rhombus ABCD = (½ x AC x BD)

= (1/2 x 24 x 18)

12. Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.

Let ABCD be the rhombus whose diagonals intersect at 0.

Then AB = 20 cm and AC = 24 cm by given data

Therefore Triangle AOB is a right-angled triangle, right angled at O

OA = ½ (AC) = ½ (24) = 12 cm and AB =20 cm

By Pythagoras theorem, we have,

(AB) 2  = (OA) 2 + (OB)  2

(20) 2  = (12) 2  + (OB)  2

(OB)  2 = (20)  2 – (12)  2

(OB) 2  = 400 – 144

(OB)  2  = (16)  2

= 2 x 16 cm

Hence, Area of the rhombus ABCD = ½ x AC x BD

= ½ x 24 x 32

Area of rhombus ABCD = 384 cm 2

13. The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonals is 2 m?

Given the length of a side of a square field is 4m

Also given that,

Area of the rhombus = Area of the square of side

½ x AC x BD = (4m) 2

½ x AC x 2 = 16 m 2

AC = 16 m 2

We know that the diagonals of a rhombus are perpendicular bisectors of each other.

AO = ½ (AC) = ½ (16) = 8 m and BO = ½ (BD) = ½ (2) = 1 m

By Pythagoras theorem, we can write as

AO 2  + BO 2  = AB 2

AB 2  = (8 m)  2  + (1 m)  2

= 64 m 2  + 1 m 2  = 65 m 2

Side of a rhombus = AB = √65 m.

Let DX be the altitude.

Area of the rhombus = AB x DX

16 m 2 = √65m × DX

DX = 16/ (√65) m

Hence, the altitude of the rhombus will be 16/√65 m.

14. Two sides of a parallelogram are 20 cm and 25 cm. If the altitude corresponding to the sides of length 25 cm is 10 cm, find the altitude corresponding to the other pair of sides.

Let ABCD is a parallelogram with longer side AB = 25 cm and altitude AE = 10 cm.

Therefore AB = CD (opposite sides of parallelogram are equal).

The shorter side is AD = 20 cm and the corresponding altitude is CF.

So, by equating them

AD x CF = CD x AE

20 x CF = 25 x 10

CF = 12.5 cm

Hence, the altitude corresponding to the other pair of the side AD is 12.5 cm.

15. The base and corresponding altitude of a parallelogram are 10 cm and 12 cm respectively. If the other altitude is 8 cm, find the length of the other pair of parallel sides.

Let ABCD is a parallelogram with side AB = CD = 10 cm (since opposite sides of parallelogram are equal) and corresponding altitude AM = 12 cm.

The other side is AD and the corresponding altitude is CN = 8 cm.

So, by equating them we get

AD x CN = CD x AM

AD x 8 = 10 x 12

AD = (10×12)/8 = 15 cm

Hence, the length of the other pair of the parallel sides = 15 cm.

16. A floral design on the floor of a building consists of 280 tiles. Each tile is in the shape of a parallelogram of altitude 3 cm and base 5 cm. Find the cost of polishing the design at the rate of 50 paise per cm 2 .

Given altitude of a tile = 3 cm

Base of a tile = 5 cm

Area of one tile = Attitude x Base

= 5 cm x 3 cm

Area of 280 tiles = 280 x 15 c m 2  = 4200 c m 2

Rate of polishing the tile = Rs. 0.5 per cm 2

Thus, Total cost of polishing the design = 4200 x 0.5

Exercise 20.4 Page No: 20.26

1. Find the area in square centimetres of a triangle whose base and altitude are as under:

(i) Base =18 cm, altitude = 3.5 cm

(ii) Base = 8 dm, altitude =15 cm

(i) Given base = 18 cm and height = 3.5 cm

We know that the area of a triangle = ½ (Base x Height)

Therefore area of the triangle = ½ x 18 x 3.5

= 31.5 cm 2

(ii) Given base = 8 dm = (8 x 10) cm = 80 cm [Since 1 dm = 10 cm]

And height = 15 cm

Therefore area of the triangle = ½ x 80 x 15

2. Find the altitude of a triangle whose area is 42 cm 2  and base is 12 cm.

Given base = 12 cm and area = 42 cm 2

Therefore altitude of a triangle = (2 x Area)/Base

Altitude = (2 x 42)/12

3. The area of a triangle is 50 cm 2 . If the altitude is 8 cm, what is its base?

Given, altitude = 8 cm and area = 50 cm 2

Therefore base of a triangle = (2 x Area)/ Altitude

Base = (2 x 50)/ 8

4. Find the area of a right angled triangle whose sides containing the right angle are of lengths 20.8 m and 14.7 m.

In a right-angled triangle,

The sides containing the right angles are of lengths 20.8 m and 14.7 m.

Let the base be 20.8 m and the height be 14.7 m.

Area of a triangle = 1/2 (Base x Height)

= 1/2 (20.8 × 14.7)

= 152.88 m 2

5. The area of a triangle, whose base and the corresponding altitude are 15 cm and 7 cm, is equal to area of a right triangle whose one of the sides containing the right angle is 10.5 cm. Find the other side of this triangle.

For the first triangle, given that

Base = 15 cm and altitude = 7 cm

We know that area of a triangle = ½ (Base x Altitude)

= ½ (15 x 7)

= 52.5 cm 2

It is also given that the area of the first triangle and the second triangle are equal.

Area of the second triangle = 52.5 c m 2

One side of the second triangle = 10.5 cm

Therefore, The other side of the second triangle = (2 x Area)/One side of a triangle

= (2x 52.5)/10.5

Hence, the other side of the second triangle will be 10 cm.

6. A rectangular field is 48 m long and 20 m wide. How many right triangular flower beds, whose sides containing the right angle measure 12 m and 5 m can be laid in this field?

Given length of the rectangular field = 48 m

Breadth of the rectangular field = 20 m

= 48 m x 20 m

Area of one right triangular flower bed = ½ (12 x 5) = 30 m 2

Therefore, required number of right triangular flower beds = area of the rectangular field/ area of one right triangular flower bed.

Number of right triangular flower beds = 32

7. In Fig. 29, ABCD is a quadrilateral in which diagonal AC = 84 cm; DL ⊥AC, BM ⊥ AC, DL = 16.5 cm and BM = 12 cm. Find the area of quadrilateral ABCD.

Given AC = 84 cm, DL = 16.5 cm and BM = 12 cm

We know that area of triangle = ½ x base x height

Area of triangle ADC = ½ (AC x DL)

= ½ (84 x 16.5)

Area of triangle ABC = ½ (AC x BM)

= ½ (84 x 12) = 504 cm 2

Hence, Area of quadrilateral ABCD = Area of triangle ADC + Area of triangle ABC

= (693 + 504) cm 2

= 1197 cm 2

8. Find the area of the quadrilateral ABCD given in Fig. 30. The diagonals AC and BD measure 48 m and 32 m respectively and are perpendicular to each other.

Given diagonal AC = 48 m and diagonal BD = 32 m

Area of a quadrilateral = ½ (Product of diagonals)

= ½ (AC x BD)

= ½ (48 x 32) m 2

9.  In Fig 31, ABCD is a rectangle with dimensions 32 m by 18 m. ADE is a triangle such that EF⊥ AD and EF= 14 cm. Calculate the area of the shaded region.

Given length of rectangle = 32m and breadth = 18m

Therefore area of the rectangle = AB x BC

= 32 m x 18 m

Also given that base of triangle = 18m and height = 14m and EF⊥ AD

Area of the triangle = ½ (AD x FE)

= ½ (BC x FE) [Since AD = BC]

= ½ (18 m x 14 m)

Area of the shaded region = Area of the rectangle – Area of the triangle

= (576 – 126) m 2

10. In Fig. 32, ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm. If P, Q, R, S be the mid-points of the sides AB, BC, CD and DA respectively, find the area of the shaded region.

Given ABCD is a rectangle of length AB = 40 cm and breadth BC = 25 cm.

Join PR and SQ so that these two lines bisect each other at point O

Also OP = OR = RP/2

From the given figure it is clear that,

Area of Triangle SPQ = Area of Triangle SRQ

Hence, area of the shaded region = 2 x (Area of SPQ)

= 2 x (1/2 (SQ x OP))

= 2 x (1/2 (40 x 12.5))

11.  Calculate the area of the quadrilateral ABCD as shown in Fig.33, given that BD = 42 cm, AC = 28 cm, OD = 12 cm and AC ⊥ BO.

BD = 42 cm, AC = 28 cm, OD= 12 cm

Area of Triangle ABC = 1/2 (AC x OB)

= 1/2 (AC x (BD – OD))

= 1/2 (28 cm x (42 cm – 12 cm))

= 1/2 (28 cm x 30 cm)

= 14 cm x 30 cm

Area of Triangle ADC = 1/2 (AC x OD) = 1/2 (28 cm x 12 cm)

= 14 cm x 12 cm

Hence, Area of the quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC

= (420 + 168) cm 2

12. Find the area of figure formed by a square of side 8 cm and an isosceles triangle with base as one side of the square and perimeter as 18 cm.

Let x cm be one of the equal sides of an isosceles triangle.

Given that the perimeter of the isosceles triangle = 18 cm

Then, x + x + 8 =18

2x = (18 – 8) = 10 cm

Area of the figure formed = Area of the square + Area of the isosceles triangle

= (side of square) 2 + ½ (base x √[(equal side) 2 – ¼ x (base) 2 ]

= 8 2 + ½ (8) x √ [5 2 – ¼ x 8 2 ]

= 64 + 4 x √ [25 – ¼ x 64]

= 64 + 4 x √ (25 – 16)

= 64 + 4 x √9

= 64 + 4 x 3

13. Find the area of Fig. 34, in the following ways: (i) Sum of the areas of three triangles (ii) Area of a rectangle — sum of the areas of five triangles

(i) From the figure, P is the midpoint of AD.

Thus AP = PD = 25 cm and AB = CD = 20 cm

From the figure, we observed that,

Area of Triangle APB = Area of Triangle PDC

Area of Triangle APB = ½ (AB x AP)

= ½ (20 x 25)

Area of Triangle PDC = Area of Triangle APB = 250 c m 2

Area of Triangle RPQ = ½ (Base x Height)

= ½ (25 cm x 10 cm)

Hence, Sum of the three triangles = (250 + 250 + 125) cm 2

(ii) From the figure, area of the rectangle ABCD = 50 cm x 20 cm

= 1000 cm 2

Thus, Area of the rectangle – Sum of the areas of three triangles

= (1000 – 625) cm 2

14. Calculate the area of quadrilateral field ABCD as shown in Fig.35, by dividing it into a rectangle and a triangle.

Join CE, so that which intersect AD at point E.

Given AE = ED = BC = 25 m and EC = AB = 30 m

Area of rectangle = length x breadth

Area of the rectangle ABCE = AB x BC

= 30 m x 25 m

Area of triangle = ½ x base x height

Area of Triangle CED = ½ (EC x ED)

= ½ (30 m x 25 m)

Hence, Area of the quadrilateral ABCD = (750 + 375) m 2  = 1125 m 2

15. Calculate the area of the pentagon ABCDE, where AB = AE and with dimensions as shown in Fig. 36.

Join BE so that we can get rectangle and triangle.

Area of the rectangle BCDE = CD x DE

= 10 cm x 12 cm

= 120 c m 2

Area of Triangle ABE = 1/2 (BE x height of the triangle)

= ½ (10 cm x (20 – 12) cm)

= ½ (10 cm x 8 cm)

Hence, Area of the pentagon ABCDE = area of rectangle + area of triangle

= (120 + 40) cm 2

16. The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10, find its base and height.

Let altitude of the triangular field be h m

Then base of the triangular field is 3h m.

We know that area of triangle = ½ x b x h

Area of the triangular field = ½ (h x 3h) = 3h 2 /2 m 2 ……. (i)

The rate of cultivating the field is Rs 24.60 per hectare.

Area of the triangular field = 332.10 /24.60

= 13.5 hectare

= 135000 m 2  [Since 1 hectare = 10000 m 2 ] …… (ii)

From equation (i) and (ii) we have,

3h 2 /2 = 135000 m 2

3h 2  = 135000 x 2 = 270000 m 2

h 2  = 270000/3

= 90000 m 2

Hence, Height of the triangular field = 300 m and

Base of the triangular field = 3 x 300 m = 900 m

17. A wall is 4.5 m long and 3 m high. It has two equal windows, each having form and dimensions as shown in Fig. 37. Find the cost of painting the wall (leaving windows) at the rate of Rs 15 per m 2 .

Given length of a wall = 4.5 m

Breadth of the wall = 3 m

We know that the area of triangle = length x Breadth

Area of the wall = Length x Breadth

= 4.5 m x 3 m = 13.5 m 2

From the figure we observed that,

Area of the window = Area of the rectangle + Area of the triangle

= (0.8 m x 0.5 m) + (1/2 x 0.8 m x 0.2 m) [Since 1 m = 100 cm]

= 0.4 m 2  + 0.08 m 2

Area of two windows = 2 x 0.48 = 0.96 m 2

Area of the remaining wall (leaving windows) = (13.5 – 0.96) m 2

= 12.54 m 2

Cost of painting the wall per m 2  = Rs. 15

Hence, the cost of painting the wall = Rs. (15 x 12.54)

= Rs. 188.1

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

• Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.