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mathslover

  • 2 years ago

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

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

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

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

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

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

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

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

  9. lgbasallote
    • 2 years ago
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    interesting

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

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

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

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

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

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

  16. UnkleRhaukus
    • 2 years ago
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    yeah,

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

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

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

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

  21. mathslover
    • 2 years ago
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    @ParthKohli

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

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

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

  25. mathslover
    • 2 years ago
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    thanks

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

  27. mathslover
    • 2 years ago
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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 ...

  28. annas
    • 2 years ago
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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

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

  30. annas
    • 2 years ago
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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

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

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

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

  34. mathslover
    • 2 years ago
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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 ...

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

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

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

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

  39. goformit100
    • 2 years ago
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    Thanks @mathslover

  40. mathslover
    • 2 years ago
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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 .

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

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

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

  44. mathslover
    • 2 years ago
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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|

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

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

  47. hba
    • 2 years ago
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    yes ml

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

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

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

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

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

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

  54. maheshmeghwal9
    • 2 years ago
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    u deservedxD

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

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

  57. Ron.mystery
    • 2 years ago
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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

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

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

  60. Calcmathlete
    • 2 years ago
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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 :)

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

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

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

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

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

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

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

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

  69. mathslover
    • 2 years ago
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    @RahulZ @rajathsbhat

  70. mathslover
    • 2 years ago
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    @RahulZ please comment

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

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

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

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

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

  76. mathslover
    • 2 years ago
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    @Preetha mam

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

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