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

This Question is Closed

mathslover Group TitleBest ResponseYou've already chosen the best response.43
any suggestions and feedbacks will be welcomed. ..
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.2
I can't get it to download right, sorry :c
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
is their any problem with file or downloading speed is low @rebeccaskell94 ?
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.2
The file. Some files when I convert them/download them only upload as symbols and stuff.
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
wait i am going to type that all soon
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.2
Okay :) Just tag me again and I'll come back. I must go study now c:
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
you made this maths?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
interesting
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
yes @lgbasallote ..
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
one of the most useful formulas in geometry thank u @mathslover very useful tutorial
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
gr8 to know @mukushla thanks a lot
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
why does herons formula work?
 2 years ago

kritima Group TitleBest ResponseYou've already chosen the best response.0
great job mathslover ! i think it's going to help me a lot
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
gr8 to know @kritima @UnkleRhaukus do u mean for proof
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
it is very long .. can u just wait for some time i will upload soon
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
i have got it upto very nearer ... for the proof
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
HERONS' FORMULA : Basically Herons' formula is : Area of a triangle = \(\sqrt{s(sa)(sb)(sc)}\) 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 \[\sqrt{s(sa)(sb)(sc)}=\textbf{Area of the equilateral triangle}\] \[\sqrt{\frac{3a}{2}(\frac{3a}{2}a)(\frac{3a}{2}a)(\frac{3a}{2}a)}\] \[\sqrt{\frac{3a}{2}*\frac{a}{2}*\frac{a}{2}*\frac{a}{2}}\] \[\sqrt{\frac{3a^4}{2^4}}\] \[\frac{a^2}{4}\sqrt{3}\]
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
@estudier @Diyadiya @annas @rebeccaskell94 @Romero @satellite73 @ujjwal
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
@ParthKohli
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.2
You really wrote this?
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
yes ..r u talking about that pdf file or this. . latex file ?
 2 years ago

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.2
This Latex one. Your English is really good for this, so I was kinda surprised! Good job :)
 2 years ago

annas Group TitleBest ResponseYou've already chosen the best response.1
awesome @mathslover your work is clearly appreciable ... keep up the good work and god bless you bro!!
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
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

annas Group TitleBest ResponseYou've already chosen the best response.1
@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

rebeccaskell94 Group TitleBest ResponseYou've already chosen the best response.2
Well it has that, but sometimes it downloads weird. It's not really a big deal, it's just frustrating.
 2 years ago

annas Group TitleBest ResponseYou've already chosen the best response.1
sometimes your system cant identify some symbols because there ASCII codes are unknown to CPU ... btw .pdf files never create problems
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
@lalaly @jiteshmeghwal9 @TheViper @maheshmeghwal9 @ash2326 @goformit100 @robtobey @waterineyes @CarlosGP
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
@amistre64 sir please have a look
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
nice work:) latex one is a very very nice one :D
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
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

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
yeah it is really better.
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
& i think the best.
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
thanks @jiteshmeghwal9 ...more comments and suggestions will be appreciated and welcomed
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
draw more picture , less words
 2 years ago

goformit100 Group TitleBest ResponseYou've already chosen the best response.0
Thanks @mathslover
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
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

lalaly Group TitleBest ResponseYou've already chosen the best response.0
Amazing:D thanks for sharing @mathslover
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
dw:1342456608856:dwdw:1342456626544:dw
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
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

mathslover Group TitleBest ResponseYou've already chosen the best response.43
We can calculate this also by using heron's formula : Let the diagonal be x : hence : 1) \(Ar(ACD)=\sqrt{\frac{(a+c+d)}{2}(a+c+dd)(a+c+dc)(a+c+da)}\) \(Ar(ACD)=\sqrt{\frac{(a+c+d)}{2}(a+c)(a+d)(c+d)}\) 2) \(Ar(ABC)=\sqrt{\frac{(a+b+x)}{2}(a+b+xa)(a+b+xx)(a+bb+x)}\) \(Ar(ABC)=\sqrt{\frac{(a+b+x)}{2}(b+x)(a+b)(a+x)}\) 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

mathslover Group TitleBest ResponseYou've already chosen the best response.43
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

mathslover Group TitleBest ResponseYou've already chosen the best response.43
Hope it helps .. thats all thanks mathslover
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1342458885083:dw
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
r'_a is a radius of External surrounded circle and r related to inner circle.
 2 years ago

mahmit2012 Group TitleBest ResponseYou've already chosen the best response.0
dw:1342459347799:dw
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Would have been a great video tutorial if the feature existed.
 2 years ago

Vaidehi09 Group TitleBest ResponseYou've already chosen the best response.0
great job @mathslover. its quite detailed with explanations for each step. plus the examples! i'd say job well done!
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
nice work!!!!! well that's the tutorial i would love to say:D
 2 years ago

maheshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.0
u deservedxD
 2 years ago

sami21 Group TitleBest ResponseYou've already chosen the best response.0
nice ineresting..u did all of this . awsum.
 2 years ago

kropot72 Group TitleBest ResponseYou've already chosen the best response.0
A very interesting topic. This formula goes way back in history. Thanks for your good work mathslover.
 2 years ago

Ron.mystery Group TitleBest ResponseYou've already chosen the best response.0
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

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
brilliant work!!  your name is appropriate mathslover
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
gr8 to know thanks a lot @Ron.mystery and @cwrw238 .... I hope this will be useful for all @estudier
 2 years ago

Calcmathlete Group TitleBest ResponseYou've already chosen the best response.0
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

anjali_pant Group TitleBest ResponseYou've already chosen the best response.0
nice tutorial ! Great work @mathslover ! :)
 2 years ago

Master.RohanChakraborty Group TitleBest ResponseYou've already chosen the best response.0
good job my bro!! love u!!
 2 years ago

across Group TitleBest ResponseYou've already chosen the best response.0
Thanks for referring me to this. I'll give it a look later.
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Record number of medals?
 2 years ago

sami21 Group TitleBest ResponseYou've already chosen the best response.0
i think so this is a record!
 2 years ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
Hey everybody, go back to my post, lol
 2 years ago

jiteshmeghwal9 Group TitleBest ResponseYou've already chosen the best response.1
\[\Huge{\color{gold}{\star \star \star}\color{red}{breacked \space the \space record \space of \space lgbasallote}}\]
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
FINALLY! SOMEONE BEATS MY RECORD!! i can peacefully retire form tutorials now ^_^
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
@RahulZ @rajathsbhat
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
@RahulZ please comment
 2 years ago

RahulZ Group TitleBest ResponseYou've already chosen the best response.0
Hey u made the tutorial , it was cool ,.... really cool
 2 years ago

RahulZ Group TitleBest ResponseYou've already chosen the best response.0
i am taking a printout
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
Nice to hear. .. . .. well u in which class ?
 2 years ago

RahulZ Group TitleBest ResponseYou've already chosen the best response.0
I am an University student. :)
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
Oh !! nice.
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
@Preetha mam
 2 years ago

mathslover Group TitleBest ResponseYou've already chosen the best response.43
@mayankdevnani
 one year ago
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