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mathslover

Hey friends. This is not a question but a tutorial on Heron's formula .. please see the attachment

  • one year ago
  • one year ago

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  1. mathslover
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    • one year ago
  2. mathslover
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    any suggestions and feedbacks will be welcomed. ..

    • one year ago
  3. rebeccaskell94
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    I can't get it to download right, sorry :c

    • one year ago
  4. mathslover
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    is their any problem with file or downloading speed is low @rebeccaskell94 ?

    • one year ago
  5. rebeccaskell94
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    The file. Some files when I convert them/download them only upload as symbols and stuff.

    • one year ago
  6. mathslover
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    wait i am going to type that all soon

    • one year ago
  7. rebeccaskell94
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    Okay :) Just tag me again and I'll come back. I must go study now c:

    • one year ago
  8. lgbasallote
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    you made this maths?

    • one year ago
  9. lgbasallote
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    interesting

    • one year ago
  10. mathslover
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    yes @lgbasallote ..

    • one year ago
  11. mukushla
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    one of the most useful formulas in geometry thank u @mathslover very useful tutorial

    • one year ago
  12. mathslover
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    gr8 to know @mukushla thanks a lot

    • one year ago
  13. UnkleRhaukus
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    why does herons formula work?

    • one year ago
  14. kritima
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    great job mathslover ! i think it's going to help me a lot

    • one year ago
  15. mathslover
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    gr8 to know @kritima @UnkleRhaukus do u mean for proof

    • one year ago
  16. UnkleRhaukus
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    yeah,

    • one year ago
  17. mathslover
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    it is very long .. can u just wait for some time i will upload soon

    • one year ago
  18. mathslover
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    i have got it upto very nearer ... for the proof

    • one year ago
  19. mathslover
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    HERONS' FORMULA : Basically Herons' formula is : Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) where s =\(\frac{a+b+c}{2}\) and a , b and c are the sides of a triangle s can also be said as : semi perimeter as a + b + c = perimeter of a triangle and when we half it ..then it becomes semi perimeter . I should also introduce you all with : a basic formula for the triangle area : (base*corresponding height) / 2 In some cases we are not able to find the height .. but we are given with all sides of the triangle.. Hence in that case we generally use : heron's formula to find the area of a triangle For example : Find the area of a triangle having sides : 5 cm , 6 cm and 10 cm . In this case we are unable to find the height .. Hence we will be going to use : herons' formula \[\large{s=\frac{5 cm + 6 cm + 10 cm}{2}=\frac{21 cm }{2}}\] now applying the formula : area of the triangle : \(\sqrt{\frac{21}{2}(\frac{21}{2}-5)(\frac{21}{2}-6)(\frac{21}{2}-10)}\) \[\large{\sqrt{\frac {21}{2}*\frac{11}{2}*\frac{9}{2}*\frac{1}{2}}}\] \[\large{\sqrt{\frac{21*11*9*1}{2^4}}}\] \[\large{\sqrt{\frac{21*11*3^2*1^2}{(2^2)^2}}}\] \[\large{\frac{3}{4}\sqrt{231}}\] hence the area of the triangle with the given information will be \(\frac{3}{4}\sqrt{231}\) Now coming to the main point : area of an equilateral triangle : (base*corresponding height)/2 Since this is an equilateral triangle : having all sides equal ( let it be : a ) \[\large{\frac{a*h}{2}}\] Now we will calculate h ( height ) as we know that whenever we draw a perpendicual bisector on a base of an equilateral triangle , it will divide the base into 2 equal parts . hence the equal divided lengths of the base = \(\frac{a}{2}\) as per pythagoras theorem : \[\large{h^2+\frac{a^2}{4}=a^2}\] \[\large{h^2=a^2-\frac{a^2}{4}}\] \[\large{h^2=\frac{3a^2}{4}}\] \[\large{h=\sqrt{\frac{3a^2}{4}}}\] \[\large{h=\frac{\sqrt{3}}{2}a}\] \[\large{h=\frac{\sqrt{3}a}{2}}\] \[\large{\textbf{Area of the equilateral triangle}=\frac{a*h}{2}}\] \[\large{\textbf{Area of the equilateral triangle}=\frac{a*\frac{\sqrt{3}a}{2}}{2}}\] \[\large{\textbf{Area of the equilateral triangle}=\frac{\sqrt{3}a^2}{4}}\] Now prooving this formula by heron's formula \[\sqrt{s(s-a)(s-b)(s-c)}=\textbf{Area of the equilateral triangle}\] \[\sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}\] \[\sqrt{\frac{3a}{2}*\frac{a}{2}*\frac{a}{2}*\frac{a}{2}}\] \[\sqrt{\frac{3a^4}{2^4}}\] \[\frac{a^2}{4}\sqrt{3}\]

    • one year ago
  20. mathslover
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    @estudier @Diyadiya @annas @rebeccaskell94 @Romero @satellite73 @ujjwal

    • one year ago
  21. mathslover
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    @ParthKohli

    • one year ago
  22. rebeccaskell94
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    You really wrote this?

    • one year ago
  23. mathslover
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    yes ..r u talking about that pdf file or this. . latex file ?

    • one year ago
  24. rebeccaskell94
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    This Latex one. Your English is really good for this, so I was kinda surprised! Good job :)

    • one year ago
  25. mathslover
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    thanks

    • one year ago
  26. annas
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    awesome @mathslover your work is clearly appreciable ... keep up the good work and god bless you bro!!

    • one year ago
  27. mathslover
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    thanks a lot annas ... just needed all of ur's wishes . . . that is what i got ! thanks a lot I promise that i will continue to maintain this ...

    • one year ago
  28. annas
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    @rebeccaskell94 you can download it by pressing right mouse button a box will appear with some options there is an option save as click it ... file will be downloaded as .pdf

    • one year ago
  29. rebeccaskell94
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    Well it has that, but sometimes it downloads weird. It's not really a big deal, it's just frustrating.

    • one year ago
  30. annas
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    sometimes your system cant identify some symbols because there ASCII codes are unknown to CPU ... btw .pdf files never create problems

    • one year ago
  31. mathslover
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    @lalaly @jiteshmeghwal9 @TheViper @maheshmeghwal9 @ash2326 @goformit100 @robtobey @waterineyes @CarlosGP

    • one year ago
  32. mathslover
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    @amistre64 sir please have a look

    • one year ago
  33. jiteshmeghwal9
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    nice work:) latex one is a very very nice one :D

    • one year ago
  34. mathslover
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    thanks @jiteshmeghwal9 that's why i put up latex here also .... in the place of that pdf ..so that all can view easily ...

    • one year ago
  35. jiteshmeghwal9
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    yeah it is really better.

    • one year ago
  36. jiteshmeghwal9
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    & i think the best.

    • one year ago
  37. mathslover
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    thanks @jiteshmeghwal9 ...more comments and suggestions will be appreciated and welcomed

    • one year ago
  38. Libniz
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    draw more picture , less words

    • one year ago
  39. goformit100
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    Thanks @mathslover

    • one year ago
  40. mathslover
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    So here i go with the explanation for : \[\textbf{How to find the area of a quadrilateral using heron's formula}\] |dw:1342456404056:dw| In the above diagram we have : a quadrilateral .. So how to find the area of a quadrilateral ..having sides a , b , c and d as i drew the diagonals of the quadrilateral .. we can find the area of the quadrilateral very easily .. let me show u all how .

    • one year ago
  41. lalaly
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    Amazing:D thanks for sharing @mathslover

    • one year ago
  42. mathslover
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    |dw:1342456608856:dw||dw:1342456626544:dw|

    • one year ago
  43. mathslover
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    So from this we have 2 triangles : ACD and ABC now : \(Ar(ACD)+Ar(ABC)=Ar(ABCD)\) hence first calculating \(Ar(ACD)\) as we know that the area of a triangle = \(\large{\frac{b*h}{2}}\) hence \[\large{Ar(ACD)=\frac{c*h_1}{2}}\] and similarly \(Ar(ABC)\) : \[\large{Ar(ABC)=\frac{a*h_2}{2}}\] hence now adding them we get : \[\large{Ar(ABCD)=\frac{c*h_1}{2}+\frac{a*h_2}{2}}\] \[\large{Ar(ABCD)=\frac{(c*h_1)+(a*h_2)}{2}}\]

    • one year ago
  44. mathslover
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    We can calculate this also by using heron's formula : Let the diagonal be x : hence : 1) \(Ar(ACD)=\sqrt{\frac{(a+c+d)}{2}(a+c+d-d)(a+c+d-c)(a+c+d-a)}\) \(Ar(ACD)=\sqrt{\frac{(a+c+d)}{2}(a+c)(a+d)(c+d)}\) 2) \(Ar(ABC)=\sqrt{\frac{(a+b+x)}{2}(a+b+x-a)(a+b+x-x)(a+b-b+x)}\) \(Ar(ABC)=\sqrt{\frac{(a+b+x)}{2}(b+x)(a+b)(a+x)}\) finally adding both these equations we may get the area of the quadrilateral .. This seems hard but one if we get the values of the sides then we can calculate this very easily .. I am going to explain this to you all by taking an example : Find the area of a quadrilateral having sides : a = 4 cm. b = 3 cm. c = 10 cm. d = 12 cm. diagonal ( x ) = 5 cm. |dw:1342457755184:dw|

    • one year ago
  45. mathslover
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    given x = 5 cm .. Hence we have all sides in 1st part of the triangle : calculating the area of the first part of the quadrilateral : 1) \(Ar(fig.1)=\sqrt{6(2)(3)(1)cm^4}\) \(Ar(fig.1)=6 cm^2\) 2) \(Ar(fig.2)=\sqrt{\frac{27}{2}*\frac{7}{2}*\frac{3}{2}*\frac{17}{2}}\) \(Ar(fig.2)=\sqrt{\frac{9*9*17*7}{2^4}}\) \(Ar(fig.2)=\frac{9}{4}\sqrt{119}\) Hence adding them we get : \[Ar(Quadrilateral)=6 cm^2+\frac{9}{4}\sqrt{119} cm^2\]

    • one year ago
  46. mathslover
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    Hope it helps .. thats all thanks mathslover

    • one year ago
  47. hba
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    yes ml

    • one year ago
  48. mahmit2012
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    |dw:1342458885083:dw|

    • one year ago
  49. mahmit2012
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    r'_a is a radius of External surrounded circle and r related to inner circle.

    • one year ago
  50. mahmit2012
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    |dw:1342459347799:dw|

    • one year ago
  51. Hero
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    Would have been a great video tutorial if the feature existed.

    • one year ago
  52. Vaidehi09
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    great job @mathslover. its quite detailed with explanations for each step. plus the examples! i'd say job well done!

    • one year ago
  53. maheshmeghwal9
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    nice work!!!!! well that's the tutorial i would love to say:D

    • one year ago
  54. maheshmeghwal9
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    u deservedxD

    • one year ago
  55. sami-21
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    nice ineresting..u did all of this . awsum.

    • one year ago
  56. kropot72
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    A very interesting topic. This formula goes way back in history. Thanks for your good work mathslover.

    • one year ago
  57. Ron.mystery
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    got the medal dude!! but nice work as better as LGBA and he's the best on this so a fab job is been done!! @mathslover

    • one year ago
  58. cwrw238
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    brilliant work!! - your name is appropriate mathslover

    • one year ago
  59. mathslover
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    gr8 to know thanks a lot @Ron.mystery and @cwrw238 .... I hope this will be useful for all @estudier

    • one year ago
  60. Calcmathlete
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    Great job! Heron's formula really can be very useful. It's a bit unfortunate you usually don't learn it until higher levels of math though...Nevertheless, awesome work :)

    • one year ago
  61. anjali_pant
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    nice tutorial ! Great work @mathslover ! :-)

    • one year ago
  62. Master.RohanChakraborty
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    good job my bro!! love u!!

    • one year ago
  63. across
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    Thanks for referring me to this. I'll give it a look later.

    • one year ago
  64. Hero
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    Record number of medals?

    • one year ago
  65. sami-21
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    i think so this is a record!

    • one year ago
  66. Hero
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    Hey everybody, go back to my post, lol

    • one year ago
  67. jiteshmeghwal9
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    \[\Huge{\color{gold}{\star \star \star}\color{red}{breacked \space the \space record \space of \space lgbasallote}}\]

    • one year ago
  68. lgbasallote
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    FINALLY! SOMEONE BEATS MY RECORD!! i can peacefully retire form tutorials now ^_^

    • one year ago
  69. mathslover
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    @RahulZ @rajathsbhat

    • one year ago
  70. mathslover
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    @RahulZ please comment

    • one year ago
  71. RahulZ
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    Hey u made the tutorial , it was cool ,.... really cool

    • one year ago
  72. RahulZ
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    i am taking a printout

    • one year ago
  73. mathslover
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    Nice to hear. .. . .. well u in which class ?

    • one year ago
  74. RahulZ
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    I am an University student. :)

    • one year ago
  75. mathslover
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    Oh !! nice.

    • one year ago
  76. mathslover
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    @Preetha mam

    • one year ago
  77. mathslover
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    • one year ago
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  78. mathslover
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    @mayankdevnani

    • one year ago
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