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mathslover Group Title

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

  • 2 years ago
  • 2 years ago

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  1. mathslover Group Title
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    • 2 years ago
  2. mathslover Group Title
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    any suggestions and feedbacks will be welcomed. ..

    • 2 years ago
  3. rebeccaskell94 Group Title
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    I can't get it to download right, sorry :c

    • 2 years ago
  4. mathslover Group Title
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    is their any problem with file or downloading speed is low @rebeccaskell94 ?

    • 2 years ago
  5. rebeccaskell94 Group Title
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    The file. Some files when I convert them/download them only upload as symbols and stuff.

    • 2 years ago
  6. mathslover Group Title
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    wait i am going to type that all soon

    • 2 years ago
  7. rebeccaskell94 Group Title
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    Okay :) Just tag me again and I'll come back. I must go study now c:

    • 2 years ago
  8. lgbasallote Group Title
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    you made this maths?

    • 2 years ago
  9. lgbasallote Group Title
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    interesting

    • 2 years ago
  10. mathslover Group Title
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    yes @lgbasallote ..

    • 2 years ago
  11. mukushla Group Title
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    one of the most useful formulas in geometry thank u @mathslover very useful tutorial

    • 2 years ago
  12. mathslover Group Title
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    gr8 to know @mukushla thanks a lot

    • 2 years ago
  13. UnkleRhaukus Group Title
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    why does herons formula work?

    • 2 years ago
  14. kritima Group Title
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    great job mathslover ! i think it's going to help me a lot

    • 2 years ago
  15. mathslover Group Title
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    gr8 to know @kritima @UnkleRhaukus do u mean for proof

    • 2 years ago
  16. UnkleRhaukus Group Title
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    yeah,

    • 2 years ago
  17. mathslover Group Title
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    it is very long .. can u just wait for some time i will upload soon

    • 2 years ago
  18. mathslover Group Title
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    i have got it upto very nearer ... for the proof

    • 2 years ago
  19. mathslover Group Title
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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 \[\sqrt{s(s-a)(s-b)(s-c)}=\textbf{Area of the equilateral triangle}\] \[\sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}\] \[\sqrt{\frac{3a}{2}*\frac{a}{2}*\frac{a}{2}*\frac{a}{2}}\] \[\sqrt{\frac{3a^4}{2^4}}\] \[\frac{a^2}{4}\sqrt{3}\]

    • 2 years ago
  20. mathslover Group Title
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    @estudier @Diyadiya @annas @rebeccaskell94 @Romero @satellite73 @ujjwal

    • 2 years ago
  21. mathslover Group Title
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    @ParthKohli

    • 2 years ago
  22. rebeccaskell94 Group Title
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    You really wrote this?

    • 2 years ago
  23. mathslover Group Title
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    yes ..r u talking about that pdf file or this. . latex file ?

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

    • 2 years ago
  25. mathslover Group Title
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    thanks

    • 2 years ago
  26. annas Group Title
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    awesome @mathslover your work is clearly appreciable ... keep up the good work and god bless you bro!!

    • 2 years ago
  27. mathslover Group Title
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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 ...

    • 2 years ago
  28. annas Group Title
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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

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

    • 2 years ago
  30. annas Group Title
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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

    • 2 years ago
  31. mathslover Group Title
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    @lalaly @jiteshmeghwal9 @TheViper @maheshmeghwal9 @ash2326 @goformit100 @robtobey @waterineyes @CarlosGP

    • 2 years ago
  32. mathslover Group Title
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    @amistre64 sir please have a look

    • 2 years ago
  33. jiteshmeghwal9 Group Title
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    nice work:) latex one is a very very nice one :D

    • 2 years ago
  34. mathslover Group Title
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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 ...

    • 2 years ago
  35. jiteshmeghwal9 Group Title
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    yeah it is really better.

    • 2 years ago
  36. jiteshmeghwal9 Group Title
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    & i think the best.

    • 2 years ago
  37. mathslover Group Title
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    thanks @jiteshmeghwal9 ...more comments and suggestions will be appreciated and welcomed

    • 2 years ago
  38. Libniz Group Title
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    draw more picture , less words

    • 2 years ago
  39. goformit100 Group Title
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    Thanks @mathslover

    • 2 years ago
  40. mathslover Group Title
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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 .

    • 2 years ago
  41. lalaly Group Title
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    Amazing:D thanks for sharing @mathslover

    • 2 years ago
  42. mathslover Group Title
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    |dw:1342456608856:dw||dw:1342456626544:dw|

    • 2 years ago
  43. mathslover Group Title
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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 : \[\large{Ar(ABCD)=\frac{c*h_1}{2}+\frac{a*h_2}{2}}\] \[\large{Ar(ABCD)=\frac{(c*h_1)+(a*h_2)}{2}}\]

    • 2 years ago
  44. mathslover Group Title
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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|

    • 2 years ago
  45. mathslover Group Title
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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 : \[Ar(Quadrilateral)=6 cm^2+\frac{9}{4}\sqrt{119} cm^2\]

    • 2 years ago
  46. mathslover Group Title
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    Hope it helps .. thats all thanks mathslover

    • 2 years ago
  47. hba Group Title
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    yes ml

    • 2 years ago
  48. mahmit2012 Group Title
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    |dw:1342458885083:dw|

    • 2 years ago
  49. mahmit2012 Group Title
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    r'_a is a radius of External surrounded circle and r related to inner circle.

    • 2 years ago
  50. mahmit2012 Group Title
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    |dw:1342459347799:dw|

    • 2 years ago
  51. Hero Group Title
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    Would have been a great video tutorial if the feature existed.

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

    • 2 years ago
  53. maheshmeghwal9 Group Title
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    nice work!!!!! well that's the tutorial i would love to say:D

    • 2 years ago
  54. maheshmeghwal9 Group Title
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    u deservedxD

    • 2 years ago
  55. sami-21 Group Title
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    nice ineresting..u did all of this . awsum.

    • 2 years ago
  56. kropot72 Group Title
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    A very interesting topic. This formula goes way back in history. Thanks for your good work mathslover.

    • 2 years ago
  57. Ron.mystery Group Title
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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

    • 2 years ago
  58. cwrw238 Group Title
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    brilliant work!! - your name is appropriate mathslover

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

    • 2 years ago
  60. Calcmathlete Group Title
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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 :)

    • 2 years ago
  61. anjali_pant Group Title
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    nice tutorial ! Great work @mathslover ! :-)

    • 2 years ago
  62. Master.RohanChakraborty Group Title
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    good job my bro!! love u!!

    • 2 years ago
  63. across Group Title
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    Thanks for referring me to this. I'll give it a look later.

    • 2 years ago
  64. Hero Group Title
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    Record number of medals?

    • 2 years ago
  65. sami-21 Group Title
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    i think so this is a record!

    • 2 years ago
  66. Hero Group Title
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    Hey everybody, go back to my post, lol

    • 2 years ago
  67. jiteshmeghwal9 Group Title
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    \[\Huge{\color{gold}{\star \star \star}\color{red}{breacked \space the \space record \space of \space lgbasallote}}\]

    • 2 years ago
  68. lgbasallote Group Title
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    FINALLY! SOMEONE BEATS MY RECORD!! i can peacefully retire form tutorials now ^_^

    • 2 years ago
  69. mathslover Group Title
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    @RahulZ @rajathsbhat

    • 2 years ago
  70. mathslover Group Title
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    @RahulZ please comment

    • 2 years ago
  71. RahulZ Group Title
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    Hey u made the tutorial , it was cool ,.... really cool

    • 2 years ago
  72. RahulZ Group Title
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    i am taking a printout

    • 2 years ago
  73. mathslover Group Title
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    Nice to hear. .. . .. well u in which class ?

    • 2 years ago
  74. RahulZ Group Title
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    I am an University student. :)

    • 2 years ago
  75. mathslover Group Title
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    Oh !! nice.

    • 2 years ago
  76. mathslover Group Title
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    @Preetha mam

    • 2 years ago
  77. mathslover Group Title
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    • 2 years ago
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  78. mathslover Group Title
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    @mayankdevnani

    • 2 years ago
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