## mathslover 4 years ago Hey friends. This is not a question but a tutorial on Heron's formula .. please see the attachment

1. mathslover

2. mathslover

any suggestions and feedbacks will be welcomed. ..

3. anonymous

4. mathslover

5. anonymous

6. mathslover

wait i am going to type that all soon

7. anonymous

Okay :) Just tag me again and I'll come back. I must go study now c:

8. lgbasallote

9. lgbasallote

interesting

10. mathslover

yes @lgbasallote ..

11. anonymous

one of the most useful formulas in geometry thank u @mathslover very useful tutorial

12. mathslover

gr8 to know @mukushla thanks a lot

13. UnkleRhaukus

why does herons formula work?

14. anonymous

great job mathslover ! i think it's going to help me a lot

15. mathslover

gr8 to know @kritima @UnkleRhaukus do u mean for proof

16. UnkleRhaukus

yeah,

17. mathslover

it is very long .. can u just wait for some time i will upload soon

18. mathslover

i have got it upto very nearer ... for the proof

19. mathslover

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

21. mathslover

@ParthKohli

22. anonymous

You really wrote this?

23. mathslover

yes ..r u talking about that pdf file or this. . latex file ?

24. anonymous

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

25. mathslover

thanks

26. anonymous

awesome @mathslover your work is clearly appreciable ... keep up the good work and god bless you bro!!

27. mathslover

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. anonymous

@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. anonymous

Well it has that, but sometimes it downloads weird. It's not really a big deal, it's just frustrating.

30. anonymous

sometimes your system cant identify some symbols because there ASCII codes are unknown to CPU ... btw .pdf files never create problems

31. mathslover

@lalaly @jiteshmeghwal9 @TheViper @maheshmeghwal9 @ash2326 @goformit100 @robtobey @waterineyes @CarlosGP

32. mathslover

@amistre64 sir please have a look

33. jiteshmeghwal9

nice work:) latex one is a very very nice one :D

34. mathslover

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

yeah it is really better.

36. jiteshmeghwal9

& i think the best.

37. mathslover

thanks @jiteshmeghwal9 ...more comments and suggestions will be appreciated and welcomed

38. anonymous

draw more picture , less words

39. goformit100

Thanks @mathslover

40. mathslover

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

Amazing:D thanks for sharing @mathslover

42. mathslover

|dw:1342456608856:dw||dw:1342456626544:dw|

43. mathslover

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

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

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

Hope it helps .. thats all thanks mathslover

47. hba

yes ml

48. anonymous

|dw:1342458885083:dw|

49. anonymous

r'_a is a radius of External surrounded circle and r related to inner circle.

50. anonymous

|dw:1342459347799:dw|

51. Hero

Would have been a great video tutorial if the feature existed.

52. anonymous

great job @mathslover. its quite detailed with explanations for each step. plus the examples! i'd say job well done!

53. maheshmeghwal9

nice work!!!!! well that's the tutorial i would love to say:D

54. maheshmeghwal9

u deservedxD

55. anonymous

nice ineresting..u did all of this . awsum.

56. kropot72

A very interesting topic. This formula goes way back in history. Thanks for your good work mathslover.

57. anonymous

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

brilliant work!! - your name is appropriate mathslover

59. mathslover

gr8 to know thanks a lot @Ron.mystery and @cwrw238 .... I hope this will be useful for all @estudier

60. anonymous

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. anonymous

nice tutorial ! Great work @mathslover ! :-)

62. anonymous

good job my bro!! love u!!

63. across

Thanks for referring me to this. I'll give it a look later.

64. Hero

Record number of medals?

65. anonymous

i think so this is a record!

66. Hero

Hey everybody, go back to my post, lol

67. jiteshmeghwal9

$\Huge{\color{gold}{\star \star \star}\color{red}{breacked \space the \space record \space of \space lgbasallote}}$

68. lgbasallote

FINALLY! SOMEONE BEATS MY RECORD!! i can peacefully retire form tutorials now ^_^

69. mathslover

@RahulZ @rajathsbhat

70. mathslover

71. anonymous

Hey u made the tutorial , it was cool ,.... really cool

72. anonymous

i am taking a printout

73. mathslover

Nice to hear. .. . .. well u in which class ?

74. anonymous

I am an University student. :)

75. mathslover

Oh !! nice.

76. mathslover

@Preetha mam

77. mathslover

78. mathslover

@mayankdevnani