RENOVATION

IN

R.P.V.V

ROHINI DELHI SEC - 11

GROUP - 3

MEMBERS

LEADER

PRACHI

TANNU – 33 AANCHAL – 02

NITISH – 22 HARENDER – 11

SONU – 31 VIVEK – 35

WORK DISTRIBUTION

Tannu – Written Work

Prachi – PresentationComputer Work

Aanchal – Written Work

Nitish – Field Work Graphs

Vivek – Design Work

Sonu – Drawing Work

HARENDER – COMPUTERWORK …..

INTRODUCTIONFirstly I want to thanks mrs. Sushma

SINGH

MAM FOR THER CONSULTS AND AISO FOR GIVEN YOUS

A UNIQUE AND PARTICULAR PROJECT ARCHITECTURE

STRUCTURE OF R.P.V.V IN THIS PROJECT WQE MEASURE

THE AREA OF STAFF ROME AND AUDITIRIUM

STAFF ROOM – 2FLOOR AUDITORIUM – GROUND FLOORWE USE MATHS CHAPTER IN OUR

PROJECT ;

1 . Coordinate Geometry 2 . Statistics

3 . MensurationFor easiest understanding of the enormous data

this project is easy understand but also time saver after the all the factors . Its an educative project . We all the group members are happy to work on this project and we got success also all the topics were explain in very detail and very nicely in this

project .

Aims AndTHE AIMS OF OUR PROJECT IS TO THE ABOUT THE

ARCHITECTURE STRUCTURE OF R . P . V . V ROHINI .IN THIS PROJECT WE WILL GIVE OUR BEST AND

DO VERY HARD WORK . AND WE WILL TELL ABOUT INTRODUCTION ACKNOWLEDGEMENT , CONCLUSION ,

LIMINATION . FIRST OF ALL WE MEASURE THE ROOMS AND THEN WE MAKE THE GRAPH . IN THIS PROJECT MANY MACHENICAL TOPICS ARE

USED SUCH AS MESURATION AND CORDINATEGEOMETRY OUR GROUP MEMBERS .

THEY WILL ALSO MAKE A FINAL PRESENTATION WITHDRAWING SCALE DIAGRAME . THIS PROJECT HELPIN FUTURE TO MAKE ARCHITECTURE STRUCTURE

OF BUILDING . IT CREATE OUR INVENSTIONIN MATHEMATICAL . IT HELPS IN STUDY AND MAKE OUR

MIND SHARP . SECOND THANKS TO MRS. SUSHMA SINGH MAM .

Objective

HISTORY OF AREAMEASUREMENT

What occupied the geometers of five thousand year ago? The answer is

Earth mesurement the area problem has been studies since the baby-lonian . The method

used to calculate area range from decomposing the complicated

poiygons/poiyhedran into simpler regions , for example with parellel lines / planes or using

triangulations up to calculate Methods .

BABY ONION

• THE EARLIST RECORD FROM ANCIENTR BABY

LONION DEALS WITH THE MEASUREMENT OF CERTAIN

QUADRILATERALS . A (ABCD) = ¼( A + C ) ( B + D ) WHERE A , B , C ,

D IS THE SIDES OF A QUADRILATERAL UNFORTUNATELY

THIS GIVES THE CORRECT RESULT ONLY IN THE CASE OF THE RECTANGLE .

EGYPTIONThe visual justification method were employed

By Egyptions to calculate the area . The Area of an isosceles triangle . The isoscelesTriangle can be divided by altitude into two right

triangleThen join to form a rectangle of height equal to the

altitudeAnd base equal to one – half the base of the triangle

Area of Curved bodies A = ( ½ π r )

The formula will give us the area of a semicircle If r is the radius or a hemisphere if r be the diameter.

The value pai comes from the formula : The area of a circle of diameter g is calculated as that

Of a square of side 8 , so the value π is = π = [ 16/9 ] 2

Greek

Heron [ 75 AD ] ‘ s formula for the area of triangle

A = Where a , b , C are the sides and s is one

half of the Diameter , i , e s = ( a + b + c ) / 2

INDIANo An extension of Heron ‘ s triangle area formula

toQuadrilaterals was discovered by Brahmagupta

( ~ 620 AD ) : A ( abcd ) =

where a, b, c are the edges of lengths of cyclical Quadrilaterals , semiperimeter s = ( a + b + c +

d ) / 2 .This formula is an amazing symetric formula . If one side is zero length , say d = o , then we

haveA triangle ( which is always cyclic ) and this

formula reduces toHeron ‘ s one .

Generalizations of formula for quadrilaterals

A generalixations of Brahmagupta ‘ s formula for general

Quadrilaterals is an follows : A ( abcd ) = ( s – a ) ( s – b )

( s – c ) ( s – d ) – abcd cos ( a + b ) where a and b are

Two non -0 adjacent angles . If the quadrilateral is inscribed in one circle and circumscribed in another , than

the areaIs A ( abcd ) = abcd

C . A . Bretschneider ( 1842 ) gives another formula :

A ( abcd ) = 1 4 p – ( b + d – a – c ) Where P and Q Are the lengths of the

diagnals .

THE MAN WHO INVENTED ANALYTIC GEOMETRY, REN ‘E

DESCARTES ( 1596 – 1650 ) NEVER GOT OUT OF BED BEFORE 11 IN THE MORNING

ANALYTIC GEOMETRY BRINGS TOGETHER THE ANALYTICAL

OF ALGEBRA AND THE VISUAL IMMEDIACY OF GEOMETRY

BY PROVIDING A WAY TO VISUALIZE ALGEBRAIC FUNCTIONS . DESCARTES , A FRENCH PHILOSOPHER

AND MATHEMATICIAN , DID THIS THROUGH THE INTRODUCTION

OF THE COORDINATE SYSTEM THAT STILL BEARS HISNAME , THE CARTESIAN COORDINATE SYSTEM .DESCARTES PUBLISHED HIS IDEAS IN 1637 IN A

TREATISE CALLED LA GEOMETRIC ( GEOMETRY ) .

CORDINATE GEOMETRY

IT IS SAID ( ALTHOUGH THE STORY ISPROBABLE A MYTH ) THAT DESCARTES

CAME UP WITH THE IDEA FOR HIS COORDINATESYSTEM WHILE LYING IN BED AND WATCHINGA FLY CRAWL ON THE CEILING OF HIS ROOM .

DESCARDS GEOMETRY WAS AN APPENDIN TO A LARGER

WORK CALLED DISCOURSE ON THE METHOD OF PROPERLY

CONDUCTING ONE ‘ S REASON AND OF SEEKING THE TRUTH

IN THE SCIENCES . IN THIS WORK , DISCARTES SET OUT TO PLACE

HUMAN KNOWLEDGE ON A NEW , FIRM FOOTING , AN IMPORTANT TASK

IN AN AGE BESET WITH DOUBTS AND CONTROVERSY AND IN WHICH

SKEPTICISM REIGNED . DESCARTES HIMSELF HAD SEEN SERVICE

IN WARFARE SPURRED BY RELIGIOUS DISAGREEMENTS AND

HE WAS NO STRANGE TO SCIENTIFIC CONTROVERSY . EITHER . IN 1633 ,

HEARING THAT GALILEO HAD BEEN CONDEMNED FOR TEACHING

THAT THE EARTH MOVED THE SUM , DESCARTES ABRUPTLY

DECIDED NOT TO PUBLISH A WORK DEFENDING THE IDEA OF A SUN – CENTERED UNIVERSE

DESCARTES FOUND A MODEL FOR PROPERREASONING IN MATHEMATICS , ESPECIALLY

IN GEOMETRY AND HIS APPENDIN ON GEOMETRYWAS MEANT TO ILLUSTRATE THE EFFECTIVENESSAND USEFULNESS OF HIS METHOD . HIS METHOD

WAS BASED ON FOUR BASIC TRULES FOR DEDUCINGKNOWLEDGE IN THE MANNER OF A GEOMETRY

PROOF . HIS FAMOUS LINE , ‘’ I THINK THERFOREI AM , ‘’ REVEALS THE FIRST FIRM PIECE OF KNOWLEDGE

UPON WHICH HIS SUBSEQUENT DEDUCTIONS WERE BASED .IN 1649 , DESCARTES ACCEPTED AN INVITATION FROM

QUEEN CHRISTINE OF SWEDON TO MOVE TO THESWEDISH COURT IN STOKHLOM AND BECOME HER

PRIVATE TUTOR . HIS NEW EMPLOYER , HOWEVER , FORCED

HIM TO BEIGN LESSONAT 5 A. M .F FOR A MAN WHO

HADSTAYED IN BED TILL 11 SINCE

CHILDHOOD , THIS EARLY RISING , COMBINED WITH THE COLD ,

CLIMATE AND PROVED ….

STATISTICS (History)

IN OUR DAILY LIFE , WE HAVE TO COLLECT FACTSWHICH HELP US IN ANSWERING MOST OF THE QUESTIONS

CONCERNING THE WORLD IN WHICH WE LIVE . THE FACTS WE COLLECTARE OFTEN NUMBER FACTORS SUCH AS THE NUMBER OF RUN SCORED BY INDIAN TEAM

AGAINST PAKISTAN . THE METHOD OF TECHNIQUESOF COLLECTION , PRESENTATION , ANLYSES OFNUMERICAL DATA IN A LOGICAL SYSTEMATIC

MANNEER SO AS TO SERVE A PURPOSEIS KNOWN AS STATICS .

STATISTICS IS A MATHEMATICAL SCIENCEPERTAINING TO THE COLLECTION OF

PRESENTATION OF DATA .

Meaning of statistics

STATISTICS IS CONCERNED WITH

SCIENTIFIC METHOD FOR COLLECTING

AND PRESENTING , ORGANICING DATD

AS WELL AS DERIVING VALID CONCULSIONS

AND MAKING REASONABLE DECISION

ON THE BASIS OF THIS ANALYSIS …

ORIGIN AND GROWTH OF STATISTICS ..

THE WORD ‘’ STATISTICS ‘’ AND ‘’ STASTICAL ‘’

ARE DERIVED FROM THE LATIN WORDS STATUS MEANS

POLITICAL STATES . THE GERMAN STATISTICS , FIRST

INTRODUCED BYGOLTIFRIED ACHE WALL (1749) ,

ORIGINALLYDESIGNATED THE DATA ANLYSIS OF

STATE . IT WAS USED BY THE BRITISH MAINLY

FORADMINISTRATIVE AND GOVERMENTAL

BODIES . IN PARTICULAR CENSUS PROVIDE

REGULARINFORMATION ABOUT THE POPULATION

…..

FUNDAMENTAL CHARACTERSTICSOF STAISTICS ….

They are related to each other and are comparable . They are aggregate of facts and not a single

observationStatics' data are numerically expressed .

Statics are collection of data in a systematic mannerStatics are collected for a predetermine purpose

Statics deals with group and doesnot study individually .

Statics laws are not exact , they are true only on averages .

The data collected by someone else , other than the investigate

Are known as SECONDARY DATA . The data obtained in the original from

Are called UNGROUP DATA or RAW DATA . Today however statics and broadened

For beyond the service of a state or government , it includes areas such as :

Business Natural and social science

Medicines Before 3000 B . C the Babylonian used small

Clay tablets to record of tabulation of agricultural yields

And of commodities bartered or sold . The Egyptian analyzed the population and

metrialWealth of their Kingdoms .

The roma empire was the first government to gather extensive

Data about population , area and wealth of territaries that they

Controlled . An arrangement of raw numerical data in ascending

Or descending order of magnitude is called ARRAY .

SUB TOPICS

SOME RELATED DIFFINATION

This figure are in the ascendingOrder 4 ,5 , 8 , 18 , 28 , 29 , 31,

40 , 40 , 43 , 43 , 46 , 46 , 47 , 47 ,50 , 50 , 55 , 55 , 70 , 71 , 75 , 75 ,

80 , 90 …

MARKS

MARKS NO OF STUDENT

4 146 3

5 1 47 2

8 1 50 2

18 2 55 3

28 1 70 1

29 2 71 1

31 1 75 2

The above data in the ascendingOder is called an arrayed and the way of

Arrangement is called an ‘array’The way of arrangement of data is the

Table is known as “frequency distribution”Make are called “variants“ the no of student who

Secured a particular no of marks are Frequency of variants is called “frequency of the variates”the number of times a number has been Repeated is called the “frequency of the variant”

continuous: quantities which can take are numerical values within a certain intervals

Discontinuous: quantities or variable which can take only a finite set of values . Each group into

which a Data is condensed is called a

The size of class is known as the class interval For Example – 10 is the class interval of class 0 – 10

Each class is bounded by 2 figure , which are called the ‘’Lower limits’’ and 20 is the ‘’ upper limit ‘’. The different

Between the upper limit of the class and the lower limitOf class is called as the class size .

The Value which lies midway between lower and upperLimits of class is Known as it ‘’ Mid Value ‘’ or ‘’ Class

Mark ‘’ . Class Mark = upper limits + lower limit

The Difference between the two extreme observationIs Known as the ‘’ RANGE ‘’ .

THREE MEASURES OF CENTRAL TRANDENCY ARE :A. MEANB. MODE

C. MEDIAN

MEAN

Median of Grouped data : if x1 , x 2 , x3 ………. Xn

Are veriable of variable x , then the arithmetic mean

Or simply mean of these values is denoted by x and

Is defined as x = x1+ x2 + x3+ …….. Xn

X = 1 = 1 n x x 1/ n ……..

ALGORITHMSTEPS – 1 = Prepare the frequency table in such a wayThat its column consist of the values of the variabe and

The second column the ∑ STEPS 2 = multiply the frequency of each row with

The corresponding values of variable to obtain Third column containing fix :

STEPS 3 = Find the sum of all the entries in column I. to obtain .

STEPS 4 = Find the Sum of all the frequencies in column2 to obtain .

STEPS 5 = Use the Formula x =For Example = Find the Missing frequency in the following

Frequency disturbution if it is Known that the meanOf the distribution is 1 : 46 No of accidents

(x) : 012345Frequency (f):

? 25105

Computation of arithmetic mean ∑ f = N = 86 + F1+ f2∑f1x1 =

140 + F1 + 2F2N = 200

200= 86 + F 1 + F 2F1 + F2 = 114 + f2 + f2

Also mean = 1461.46 = ∑f1 x 1n

1.46 = 140 + f1 + f2/200Sloving ½

F1=76 and f2 = 38

Steps deviation Method ….

STEPS – 1 – Obtain the FrequencyDistribution and prepare the frquency

Table in such away that first column consistsOf the values of the variable and the second

column Corresponding frequencies .

STEPS – 2 – Choose a number ‘’A’’ ( generally Known

As the assumed mean ) and take deviation d 1= x1 -4

About A . Write these deviation against theCorresponding in the 4 column .

DATA OFPRESENTATION

RENOVATION IN STAFF ROOM ...

Staff Room has 4 walls .1. Let the name of wall of staff room ‘’A’’ Stay in east side

Length of wall = 431 cmBreadth of wall = 90 cm

Area of rectangle = length x breadthArea of wall ‘’A’’ = 431 X 90 = 38,790 cm 2

2 . Let the name of wall of staff room ‘’B’’ stay in north side .Length of the wall = 628 cmBreadth of the wall = 90 cm

Area of rectangle = length x breadthArea of wall ‘’B’’ = 628 x 90 = 56, 520 cm 2 .

3 . Let the name of wallOf staff room ‘’C’’ stay in

West side .Area of rectangle = l x b

Length of Almirah = 157 cmBreadth of Almirah = 80 cmArea of Almirah = 157 x 80

= 12,560 cm 2 .Area 0f 3 Almirah = 12,560 X 3

= 37,680 cm 2 .

AREA OF ONE RECTANGLE BOXESSIDE WALL = LENGTH X BREADTH

= 80 X 53 = 4240 CM 2 .AREA OF SIX RECTANGLE BOXES =

4240 X 6 = 25,440 CM 2 AREA OF ONE PILLAR = L X B = 90X29

= 2610 CM 2AREA OF FOUR PILLAR = 2610 X 4

= 10,440 CMAREA OF 3 BOXES BOUNDARY =

157 X 10 = 1570 X 3 = 4,710 CM 2TOTAL AREA OF WALL ‘’C’’

= 37, 680 + 25,440 + 10,440+4,710= 78,270 CM 2 .

Lets the name of wall of staff room‘’D’’ stay in south side .Length of wall = 628 cmBreadth of wall = 90 cmArea of wall ‘’D’’ = L X B

= 628 X 90 = 56,520 cm 2Total Area of wall in staff room

Area of ‘’A’’ wall = 38790Area of ‘’B’’ wall = 56520Area of ‘’C’’ wall = 78270

Area of ‘’D’’ wall = 56520Total = 230100

Total area of wall in Staff room = 2,30,100 cm 2 .

Area of one tile = L X BLength of one tile = 46 cmBreadth of one tile = 30 cm

Area of one tile = ?Total area of total tile = 2,30,100 cm

Total tile = 166.73 tiles= 167 approx .

Rate of one tile = 46.66 rupeesCost of 167 tile = 167 x 46.66 = Rs . 27843 . 91 approx

Area of floor in staff room Length of floor = 628 cmBreadth of floor = 618 cmArea of Rectangle = L X B

Area of floor = 628 x 618 = 3,88,104 cm 2Area of floor boundary :628 x 10 = 6280 cm 2

618 x 10 = 6180 cm 2 .

628 X 10 = 6280 CM 2431 X 10 = 4310 CM 2

TOTAL AREA OF FLOOR BOUNDARY :6280 + 6180 + 6280 + 4310 = 23,050 CM 2

AREA OF 1 BOX FLOOR = 157 X 53 = 8,321 CM 2AREA OF 3 BOX FLOOR = 3 X 8231 = 24963 CM2

TOTAL AREA OF FLOOR TILES = 388104 + 23050 + 24963 = 436117 CM 2

AREA OF ONE FLOOR TILES = LX B

Length of one tile = 61 cm Breadth of one tile = 61 cm = 3,721 cm

Total floor tiles = 436117 upon 3721 = 117.20 tiles= 117 tiles approx .

Total floor tiles = 117 tiles approxCost of one tile = 135 rupees

Cost of 117 tiles = 117 x 135 = Rs . 15,795 approxTotal Cost of floor tiles = Rs . 15,795 approx .

Total renovation cost in staff room= Rs . 27843 . 91 + Rs . 15795 = 43,638.91

Total renovation cost in staff room = Rs , 43,640 approx …………..

Cost of Renovation

Inauditorium

The Wall of Auditorium stage Length of Wall = 1256 cm

Breadth of Wall = L x B = 1256 x 75 = 94200 cm Stair First = 58 x 21 = 1218 cm

Stair Second = 40 x 21 = 840 cm Stair third = 23 x 21 = 483 cmStair Fourth = 5 x 21 = 105 cm

Total area of Stair = 1218 + 840 + 483 + 105 = 2646 ..

TOTAL TILE GAP = 1570 CMTOTAL AREA OF WALL OF

AUDITIRIUM STAGE = 94200 + 2646 – 1570

= 95276 CMAREA OF 1 TILE = 46X 30

= 1380 CM …

Number of tiles = 95276 upon 1380 = 69 tilesRate of one tile = Rs. 46.66

Rate of 69 tile = 69 x 46.66 = Rs . 3219.54 approx

BOUNDARY WALL

Let the name of boundary wall stayIn north side is ‘’A’’

Length of Boundary Wall = 1509 cm

( I ) BREADTH OF BOUNDARY WALL = 15 CMAREA OF WALL ‘’A’’ = L X B = 1509 X 15 = 22635 CM

(II ) LET THE NAME OF BOUNDARY WALL IN STAY WEST SIDEIS ‘’B’’ LENGTH OF BOUNDARY WALL ‘’B’’ = 1536 CM

BREADTH OF BOUNDARY WALL ‘’B’’ = 15 CMAREA OF WALL ‘’B’’ = L X B = 1536 X 15 = 23040 CM

(III) LET THE NAME OF BOUNDARY WALL IN STAY SOUTH SIDE ‘’C’’ = LENGTH OF WALL ‘’C’’ = 1509 CM

BREADTH OF WALL ‘’C’’ = 15 CM AREA OF WALL ‘’C’’ == L X B = 1509 X15 = 22635 CM

(iv) LET THE NAME OF BOUNDARY WALL STAY IN EAST SIDE IS ‘’D’’ .

Length of wall ‘’d’’ = 1414 cmBreadth of wall ‘’d’’ = 10 cm

Area of wall ‘’d’’ = l x b = 14140 cmTotal area of boundary wall in

Auditorium =Area of wall ‘’a’’ = 22635 cmArea of wall ‘’b’’ = 23040 cmArea of wall ‘’c’’ = 22635 cmArea of wall ‘’d’’ = 14140 cm

Total area = 82450 cmArea of one tile = 3721 cm

Total tile = 82450 upon 3721 = 22.15

FLOOR

Length of floor = 1938 cmBreadth of floor = 1766 cm

Area of floor = l x b = 1938 x 1766= 3422508 cm

Area of one tile = 61 x 61 = 3721 cmNumber of floor tile

= 3422508 upon 3721Number of tile = 919.78 = 920 approx

Cost of one tile = 135 Cost of 920 tile = 135 x 920 =

124200Total renovation cost auditorium

Cost of stage = 3219.54Cost of boundary wall = 2990 . 25

Cost of floor = 124200 Cost of renovation in auditorium

= 130409 . 79 approx …

BIBLIOGRAPHY

We use N.C.E.R.P Maths bookOf our class 9 th . We use internet.

We need the help of our GuideTeacher Mrs . Sushma Singh she

Help us . We take help of ourClassmates and friends as well

as Parents ………

LIMITATIONS

WE FACE MANY DIFFICULTIES AND FIND VERYLIMITATIONS TO MAKE THIS PROJECT . WHEN

WE MAKE THIS PROJECT WE MISS OUR PERIODS LIKE – YOGA , GAMES AND LIBRARYBECAUSE WE HAVE NO ABSENTY PERIOD TOMAKE THIS PROJECTS LIKE SANSKRIT AND

SCIENCE . WE DO THIS PROJECT THEREARE NO TIME TO REVISION OF OUR EXAMS ..

SUGGESTIONWE WOULD LIKE TO GIVE OURSUGGESTION TO OUR GUIDE

TEACHER Mrs. SUSHMA SINGHTHAT SHE GIVE EXTRAORDINARY

PERIOD TO MAKE THIS PROJECT EDUCATEDAND PERFECT . SHE SEPARATE BOYS &

GIRLS . AND GIVE SEPRATE WORK TO EACHSTUDENT . SO , STUDENT FEEL NOT ANY

BURDEN TO THEMSELVES .

PRACHI