New formulas in area

Document Type : Original Article

Author

Cairo University

Abstract

This formula develops old theories to facilitate the solution The use of geometric spaces is very important in our live. In ancient times it took many steps to resolve the area of the triangle, but in this research, it will be easy to solve the area of triangle with one side and two angles only and this saves time , effort and thinking. In Old times, solving rhombus area took a number of difficult steps, but in this research it will be easily solved..

Generally, rhombus area is calculated by half multiplied the lengths of the two diagonals, but in but in this research area is calculated in terms of one diagonal length and an angle only Parallelogram area is equal to multiplication of the base length by the perpendicular , but the new method uses diagonal length and angles of the diagonal only Area of the rectangle equal multiplication of length and width, Using the new method diagonal length and angles of the diagonal are used only Uses

1- triangle area in a new method 2- parallelogram area a new method 3 – rectangle area in a new method 4 - rhombus area in a new method 5 - square area in a new method

Keywords


Article Title [العربیة]

صیغ جدیدة فی المساحه

Author [العربیة]

  • ایهاب احمد
جامعة القاهرة
Abstract [العربیة]

الملخص عربى

هذه الصیغ تطویر لنظریات قدیمة لتسهیل الحل استخدام المساحات الهندسیة هام جدا فى حیاتنا قدیما کان یستخدم خطوات کثیرة لحل المساحة للمثلث ولکن هذا البحث یسهل حل المثلث ومثال ذلک مساحة المثلث بضلع واحد فقط وزاویتین وذلک یوفر الوقت والجهد والتفکیر قدیما یتم حل مساحة المعین بطریق صعبة ولکن فى هذا البحث یتم الحل بسهولة مساحة المعین نصف حاصل ضرب القطرین ولکن فى القانون الجدید المساحة بدلالة قطر وزاویة فقط مساحة متوازى الاضلاع تساوى القاعدة فى الارتفاع ولکن القانون الجدید بدلالة قطر وزاویتى القطر فقط مساحة المستطیل تساوى الطول فى العرض ولکن القانون الجدید بدلالة قطر وزاویتى القطر فقط وبذلک یسهل الاجابة فى ایجاد مساحة المثلث والمعین والمستطیل والمربع بطرق جدیدة وهذا یفید الطلاب فى هندسة مدنى کلیة الهندسة

 

 

New formulas in area

 


Ehab esmail ahmed

 

Bank el taslef  Street, Tahta, Sohag, egypt

 

mathehab75@yahoo.com  E-mail

 

 

Abstract

This formula develops old theories to facilitate the solution The use of geometric spaces is very important in our live. In ancient times it took many steps to resolve the area of the triangle, but in this research, it will be easy to solve the area of triangle  with one side and  two angles only and this saves time , effort and thinking.

In Old times, solving  rhombus area took a number of difficult steps, but in this research it will be easily solved..

 

Generally, rhombus area is calculated by half multiplied the lengths  of the two diagonals, but in but in this research area is calculated in terms of one  diagonal length and an angle only

Parallelogram area is equal to multiplication of the base length  by the perpendicular , but the new method uses diagonal length and angles  of the diagonal only

Area of the rectangle equal multiplication of length and width, Using the  new method diagonal length and angles  of the diagonal are used  only

Uses


1- triangle area in a new method
2- parallelogram area a new method
3 – rectangle area in a new method
4 - rhombus area in a new method
5 - square area in a new method

 

 

Keyword

 

The area; Triangle;  rhombus (diamond); Parallelogram; rectang

 

 

 

 

 

 

 

الملخص عربى

 

هذه الصیغ  تطویر  لنظریات  قدیمة  لتسهیل  الحل

استخدام المساحات الهندسیة هام جدا فى حیاتنا قدیما کان یستخدم خطوات کثیرة لحل المساحة للمثلث ولکن هذا البحث یسهل حل المثلث ومثال ذلک مساحة المثلث بضلع واحد فقط وزاویتین  وذلک یوفر الوقت  والجهد  والتفکیر قدیما یتم حل مساحة المعین بطریق صعبة ولکن فى هذا البحث یتم الحل بسهولة  مساحة المعین نصف حاصل ضرب القطرین ولکن فى القانون الجدید المساحة بدلالة قطر وزاویة فقط  مساحة متوازى الاضلاع  تساوى  القاعدة فى الارتفاع  ولکن القانون الجدید  بدلالة قطر وزاویتى القطر فقط  مساحة المستطیل تساوى الطول فى العرض ولکن القانون الجدید  بدلالة قطر وزاویتى القطر فقط  وبذلک یسهل الاجابة فى ایجاد مساحة المثلث والمعین والمستطیل والمربع بطرق جدیدة   وهذا یفید الطلاب  فى هندسة مدنى کلیة الهندسة

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Introduction

 

Calculating the area of a triangle  using the length of   one side and  two angles is  better than  using the old solution solving  rhombus area took a number of difficult steps, but in this research it will be easily solved.

 

Generally, rhombus area is calculated by half multiplied the lengths  of the two diagonals, but in but in this research area is calculated in terms of one  diagonal length and an angle only

Parallelogram area is equal to multiplication of the base length  by the perpendicular , but the new method uses diagonal length and angles  of the diagonal only

Area of the rectangle equal multiplication of length and width, Using the  new method diagonal length and angles  of the diagonal are used  only 

 

 

 

Idea of the research    

 

"The new methods"

The area of :

(1) Triangle                                          (2) rhombus (diamond)

(3) Parallelogram                    (4) rectangle

 

The area of triangle

                                                           A

 

 

 

 

                                                      b                                   c          Figure(6.1)

 

 

 

 

proof

 

 

the area of the triangle

 =  × (a b)× (a c)× sin ȃ

000   =

        

000 ac = (ab) × sin (b) / sin (c)

000 sin c = sin(180- ( a + b ))

                                = sin ( a + b )          

from (1), (2), (3)

000 The area at triangle =    

 

 

 

Example (1)

 

                                                                        c

                                         2 cm               45

                                                                         1cm

                                                45     

 

                             a                1 cm          b

 

Solution (1)

 

The Area of triangle    abc   =   a b  × b c ×   sin b

                                                      =    × 1 ×   1 × 1  =   cm2

Solution (2) by new method

The Area of triangle abc =  

 

=

=  =   cm2

                    Example (1)

a

                                    

       30          5 cm                          

     70         

                      b                                          c

The Area oF     abc   =  

 

   =    = 5.96 cm2

 

 

 

 

 

 

The area of parallelogram

 

d                                    c

         3

                2

                             a                                              b

Figure (6.2)

The proof

The area of parallelogram a b c d

= 2 × area       abd

= 2 ×                                       

         

000  ab  //  bc

000      m ( 3 ) = m ( 1 )   

from (1) , (2)

000 area    a b c d =

 

\

example (1)

                   a                                     b

 

                             5 cm                

 

        40      30

                      

                      d                                       c

 

Area of        a b c d =

                               = 8.55 cm2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The area of rhombus (diamond)

                                      b

 

                   a                                       c

 

                                  

 

 

                                         2 1

                                             d                       Figure (6.3)

 

                 Proof

 

  000 Area of        a b c d =  

    000 m(1) = m(2) = m(

            from (1), (2)

000 Area of        a b c d =

new method

example

                                      b

                  

  a              

 

             4 cm                  c

                              

                                     30  30

 

 

                                      d

Area of        a b c d =

 

                                        =     = =  cm2

 

 

 

 

 

(2.4)The area of rectangle

                                    a                                                      b

                                                                 1             2

         

                                      d                                                    c    Figure (6.1)

Proof

 

 

  =  = (bd)2 × sin(2) cos(2)

 

Because

sin(90) = 1

Sin (1) = cos(2)                       because 1+2 = 90

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

References

 

Book calculate areas and volumes of geometric shapes

 

Carson, P. B. (2012). Effects of levels of formal educational training in mathematics on teacher self-efficacy (Doctoral dissertation, Piedmont College).‏

 

Gardner, R. J. (1995). Geometric tomography (Vol. 1). Cambridge: Cambridge University Press

 

http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-triangleformulae-2009-1.pdf

 

httpfiles.books.elebda3.netdownload-pdf-ebooks.org-ku-9199.pdfhttp://www.mathcentre.ac.uk/resources/uploaded/mc-ty-triangleformulae-2009-1.pdf

.‏

 

 

 

 

 

 
 
Carson, P. B. (2012). Effects of levels of formal educational training in mathematics on teacher self-efficacy (Doctoral dissertation, Piedmont College).‏
 
Gardner, R. J. (1995). Geometric tomography (Vol. 1). Cambridge: Cambridge University Press
 
http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-triangleformulae-2009-1.pdf
 
httpfiles.books.elebda3.netdownload-pdf-ebooks.org-ku-9199.pdfhttp://www.mathcentre.ac.uk/resources/uploaded/mc-ty-triangleformulae-2009-1.pdf