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

I got my questions answered at in under 10 minutes. Go to now for free help!
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any suggestions and feedbacks will be welcomed. ..
I can't get it to download right, sorry :c

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Other answers:

is their any problem with file or downloading speed is low @rebeccaskell94 ?
The file. Some files when I convert them/download them only upload as symbols and stuff.
wait i am going to type that all soon
Okay :) Just tag me again and I'll come back. I must go study now c:
you made this maths?
one of the most useful formulas in geometry thank u @mathslover very useful tutorial
gr8 to know @mukushla thanks a lot
why does herons formula work?
great job mathslover ! i think it's going to help me a lot
gr8 to know @kritima @UnkleRhaukus do u mean for proof
it is very long .. can u just wait for some time i will upload soon
i have got it upto very nearer ... for the proof
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 \[\sqrt{s(s-a)(s-b)(s-c)}=\textbf{Area of the equilateral triangle}\] \[\sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}\] \[\sqrt{\frac{3a}{2}*\frac{a}{2}*\frac{a}{2}*\frac{a}{2}}\] \[\sqrt{\frac{3a^4}{2^4}}\] \[\frac{a^2}{4}\sqrt{3}\]
You really wrote this?
yes ..r u talking about that pdf file or this. . latex file ?
This Latex one. Your English is really good for this, so I was kinda surprised! Good job :)
awesome @mathslover your work is clearly appreciable ... keep up the good work and god bless you bro!!
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 ...
@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
Well it has that, but sometimes it downloads weird. It's not really a big deal, it's just frustrating.
sometimes your system cant identify some symbols because there ASCII codes are unknown to CPU ... btw .pdf files never create problems
@amistre64 sir please have a look
nice work:) latex one is a very very nice one :D
thanks @jiteshmeghwal9 that's why i put up latex here also .... in the place of that pdf that all can view easily ...
yeah it is really better.
& i think the best.
thanks @jiteshmeghwal9 ...more comments and suggestions will be appreciated and welcomed
draw more picture , less words
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 .
Amazing:D thanks for sharing @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 : \[\large{Ar(ABCD)=\frac{c*h_1}{2}+\frac{a*h_2}{2}}\] \[\large{Ar(ABCD)=\frac{(c*h_1)+(a*h_2)}{2}}\]
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|
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 : \[Ar(Quadrilateral)=6 cm^2+\frac{9}{4}\sqrt{119} cm^2\]
Hope it helps .. thats all thanks mathslover
  • hba
yes ml
r'_a is a radius of External surrounded circle and r related to inner circle.
Would have been a great video tutorial if the feature existed.
great job @mathslover. its quite detailed with explanations for each step. plus the examples! i'd say job well done!
nice work!!!!! well that's the tutorial i would love to say:D
u deservedxD
nice ineresting..u did all of this . awsum.
A very interesting topic. This formula goes way back in history. Thanks for your good work mathslover.
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
brilliant work!! - your name is appropriate mathslover
gr8 to know thanks a lot @Ron.mystery and @cwrw238 .... I hope this will be useful for all @estudier
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 :)
nice tutorial ! Great work @mathslover ! :-)
good job my bro!! love u!!
Thanks for referring me to this. I'll give it a look later.
Record number of medals?
i think so this is a record!
Hey everybody, go back to my post, lol
\[\Huge{\color{gold}{\star \star \star}\color{red}{breacked \space the \space record \space of \space lgbasallote}}\]
FINALLY! SOMEONE BEATS MY RECORD!! i can peacefully retire form tutorials now ^_^
@RahulZ please comment
Hey u made the tutorial , it was cool ,.... really cool
i am taking a printout
Nice to hear. .. . .. well u in which class ?
I am an University student. :)
Oh !! nice.
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