How to prove $ sqrt { frac {s(c-a-b)}{2∆} } $ is equal to i?












0












$begingroup$


Consider a right angles triangle. Let, c be it's hypotenuse and a,b it's other sides.



Then prove,



$ sqrt { frac {s(c-a-b)}{2∆} } $ is complex.
Where ∆= area of the triangle and s is semi perimeter.



I am not able to tackle this questions second part which is the above fraction can be proven equal to i.



Proving it is complex is easy by the triangle equality. But how to prove that it is equal to i. Also, is there any other way to prove that the above fraction is complex.



Edit- Including the other method, there are basically 2 ways for proving it is complex. Is there any other way?






inequality triangle







share|cite|improve this question











$endgroup$












  • $begingroup$
    sorry, what is s?
    $endgroup$
    – gimusi
    Dec 12 '18 at 13:14










  • $begingroup$
    What is $s$ in the square root?
    $endgroup$
    – fantasie
    Dec 12 '18 at 13:14










  • $begingroup$
    I am so sorry. S is semi perimeter
    $endgroup$
    – Rajdeep Singh
    Dec 12 '18 at 13:14
















0












$begingroup$


Consider a right angles triangle. Let, c be it's hypotenuse and a,b it's other sides.



Then prove,



$ sqrt { frac {s(c-a-b)}{2∆} } $ is complex.
Where ∆= area of the triangle and s is semi perimeter.



I am not able to tackle this questions second part which is the above fraction can be proven equal to i.



Proving it is complex is easy by the triangle equality. But how to prove that it is equal to i. Also, is there any other way to prove that the above fraction is complex.



Edit- Including the other method, there are basically 2 ways for proving it is complex. Is there any other way?






inequality triangle







share|cite|improve this question











$endgroup$












  • $begingroup$
    sorry, what is s?
    $endgroup$
    – gimusi
    Dec 12 '18 at 13:14










  • $begingroup$
    What is $s$ in the square root?
    $endgroup$
    – fantasie
    Dec 12 '18 at 13:14










  • $begingroup$
    I am so sorry. S is semi perimeter
    $endgroup$
    – Rajdeep Singh
    Dec 12 '18 at 13:14














0












0








0





$begingroup$


Consider a right angles triangle. Let, c be it's hypotenuse and a,b it's other sides.



Then prove,



$ sqrt { frac {s(c-a-b)}{2∆} } $ is complex.
Where ∆= area of the triangle and s is semi perimeter.



I am not able to tackle this questions second part which is the above fraction can be proven equal to i.



Proving it is complex is easy by the triangle equality. But how to prove that it is equal to i. Also, is there any other way to prove that the above fraction is complex.



Edit- Including the other method, there are basically 2 ways for proving it is complex. Is there any other way?






inequality triangle







share|cite|improve this question











$endgroup$




Consider a right angles triangle. Let, c be it's hypotenuse and a,b it's other sides.



Then prove,



$ sqrt { frac {s(c-a-b)}{2∆} } $ is complex.
Where ∆= area of the triangle and s is semi perimeter.



I am not able to tackle this questions second part which is the above fraction can be proven equal to i.



Proving it is complex is easy by the triangle equality. But how to prove that it is equal to i. Also, is there any other way to prove that the above fraction is complex.



Edit- Including the other method, there are basically 2 ways for proving it is complex. Is there any other way?






inequality triangle




inequality triangle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 13:19







Rajdeep Singh

















asked Dec 12 '18 at 13:09









Rajdeep SinghRajdeep Singh

184




184












  • $begingroup$
    sorry, what is s?
    $endgroup$
    – gimusi
    Dec 12 '18 at 13:14










  • $begingroup$
    What is $s$ in the square root?
    $endgroup$
    – fantasie
    Dec 12 '18 at 13:14










  • $begingroup$
    I am so sorry. S is semi perimeter
    $endgroup$
    – Rajdeep Singh
    Dec 12 '18 at 13:14


















  • $begingroup$
    sorry, what is s?
    $endgroup$
    – gimusi
    Dec 12 '18 at 13:14










  • $begingroup$
    What is $s$ in the square root?
    $endgroup$
    – fantasie
    Dec 12 '18 at 13:14










  • $begingroup$
    I am so sorry. S is semi perimeter
    $endgroup$
    – Rajdeep Singh
    Dec 12 '18 at 13:14
















$begingroup$
sorry, what is s?
$endgroup$
– gimusi
Dec 12 '18 at 13:14




$begingroup$
sorry, what is s?
$endgroup$
– gimusi
Dec 12 '18 at 13:14












$begingroup$
What is $s$ in the square root?
$endgroup$
– fantasie
Dec 12 '18 at 13:14




$begingroup$
What is $s$ in the square root?
$endgroup$
– fantasie
Dec 12 '18 at 13:14












$begingroup$
I am so sorry. S is semi perimeter
$endgroup$
– Rajdeep Singh
Dec 12 '18 at 13:14




$begingroup$
I am so sorry. S is semi perimeter
$endgroup$
– Rajdeep Singh
Dec 12 '18 at 13:14










4 Answers
4






active

oldest

votes


















0












$begingroup$

We have that



$$frac {s(c-a-b)}{2∆}=frac {frac12(c+(a+b))(c-(a+b))}{2frac12ab}=frac12frac{c^2-(a+b)^2}{ab}=frac12frac{-2ab}{ab}=-1 $$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Semi-perimeter $s=frac{a+b+c}2, Delta=frac12abimpliessqrt{frac{s(c-a-b)}{2∆}}=sqrt{frac{c^2-(a+b)^2}{2ab}}=sqrt{-frac{2ab}{2ab}}because a^2+b^2=c^2$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      It is equal to $i$
      $$Longleftrightarrow s(a+b-c)=2Delta$$
      $$Longleftrightarrow (a+b+c)(a+b-c)=4Delta$$
      $$Longleftrightarrow (a+b)^2-c^2=4Delta$$
      $$Longleftrightarrow 2ab=4Delta$$
      $$LongleftrightarrowDelta=frac{ab}{2}$$
      which is obviously true.



      Hope it helps:)






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        The first thing to do is to replace $s$ with $(a+b+c)/2$ and $Delta$ with $ab/2$. This means the expression becomes



        $$sqrt{frac{(a+b+c)(c-a-b)}{2ab}}$$



        now, simplify the numerator (either by multiplying everything out or first writing it as $(c+(a+b))(c-(a+b))$ and then using the fact that $(A+B)(A-B)=A^2-B^2$. Also, use the fact that $c^2=a^2+b^2$.






        share|cite|improve this answer









        $endgroup$













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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          We have that



          $$frac {s(c-a-b)}{2∆}=frac {frac12(c+(a+b))(c-(a+b))}{2frac12ab}=frac12frac{c^2-(a+b)^2}{ab}=frac12frac{-2ab}{ab}=-1 $$






          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            We have that



            $$frac {s(c-a-b)}{2∆}=frac {frac12(c+(a+b))(c-(a+b))}{2frac12ab}=frac12frac{c^2-(a+b)^2}{ab}=frac12frac{-2ab}{ab}=-1 $$






            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              We have that



              $$frac {s(c-a-b)}{2∆}=frac {frac12(c+(a+b))(c-(a+b))}{2frac12ab}=frac12frac{c^2-(a+b)^2}{ab}=frac12frac{-2ab}{ab}=-1 $$






              share|cite|improve this answer









              $endgroup$



              We have that



              $$frac {s(c-a-b)}{2∆}=frac {frac12(c+(a+b))(c-(a+b))}{2frac12ab}=frac12frac{c^2-(a+b)^2}{ab}=frac12frac{-2ab}{ab}=-1 $$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 12 '18 at 13:15









              gimusigimusi

              92.9k84494




              92.9k84494























                  1












                  $begingroup$

                  Semi-perimeter $s=frac{a+b+c}2, Delta=frac12abimpliessqrt{frac{s(c-a-b)}{2∆}}=sqrt{frac{c^2-(a+b)^2}{2ab}}=sqrt{-frac{2ab}{2ab}}because a^2+b^2=c^2$






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Semi-perimeter $s=frac{a+b+c}2, Delta=frac12abimpliessqrt{frac{s(c-a-b)}{2∆}}=sqrt{frac{c^2-(a+b)^2}{2ab}}=sqrt{-frac{2ab}{2ab}}because a^2+b^2=c^2$






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Semi-perimeter $s=frac{a+b+c}2, Delta=frac12abimpliessqrt{frac{s(c-a-b)}{2∆}}=sqrt{frac{c^2-(a+b)^2}{2ab}}=sqrt{-frac{2ab}{2ab}}because a^2+b^2=c^2$






                      share|cite|improve this answer









                      $endgroup$



                      Semi-perimeter $s=frac{a+b+c}2, Delta=frac12abimpliessqrt{frac{s(c-a-b)}{2∆}}=sqrt{frac{c^2-(a+b)^2}{2ab}}=sqrt{-frac{2ab}{2ab}}because a^2+b^2=c^2$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 12 '18 at 13:14









                      Shubham JohriShubham Johri

                      5,189718




                      5,189718























                          0












                          $begingroup$

                          It is equal to $i$
                          $$Longleftrightarrow s(a+b-c)=2Delta$$
                          $$Longleftrightarrow (a+b+c)(a+b-c)=4Delta$$
                          $$Longleftrightarrow (a+b)^2-c^2=4Delta$$
                          $$Longleftrightarrow 2ab=4Delta$$
                          $$LongleftrightarrowDelta=frac{ab}{2}$$
                          which is obviously true.



                          Hope it helps:)






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            It is equal to $i$
                            $$Longleftrightarrow s(a+b-c)=2Delta$$
                            $$Longleftrightarrow (a+b+c)(a+b-c)=4Delta$$
                            $$Longleftrightarrow (a+b)^2-c^2=4Delta$$
                            $$Longleftrightarrow 2ab=4Delta$$
                            $$LongleftrightarrowDelta=frac{ab}{2}$$
                            which is obviously true.



                            Hope it helps:)






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              It is equal to $i$
                              $$Longleftrightarrow s(a+b-c)=2Delta$$
                              $$Longleftrightarrow (a+b+c)(a+b-c)=4Delta$$
                              $$Longleftrightarrow (a+b)^2-c^2=4Delta$$
                              $$Longleftrightarrow 2ab=4Delta$$
                              $$LongleftrightarrowDelta=frac{ab}{2}$$
                              which is obviously true.



                              Hope it helps:)






                              share|cite|improve this answer









                              $endgroup$



                              It is equal to $i$
                              $$Longleftrightarrow s(a+b-c)=2Delta$$
                              $$Longleftrightarrow (a+b+c)(a+b-c)=4Delta$$
                              $$Longleftrightarrow (a+b)^2-c^2=4Delta$$
                              $$Longleftrightarrow 2ab=4Delta$$
                              $$LongleftrightarrowDelta=frac{ab}{2}$$
                              which is obviously true.



                              Hope it helps:)







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Dec 12 '18 at 13:15









                              MartundMartund

                              1,645213




                              1,645213























                                  0












                                  $begingroup$

                                  The first thing to do is to replace $s$ with $(a+b+c)/2$ and $Delta$ with $ab/2$. This means the expression becomes



                                  $$sqrt{frac{(a+b+c)(c-a-b)}{2ab}}$$



                                  now, simplify the numerator (either by multiplying everything out or first writing it as $(c+(a+b))(c-(a+b))$ and then using the fact that $(A+B)(A-B)=A^2-B^2$. Also, use the fact that $c^2=a^2+b^2$.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    The first thing to do is to replace $s$ with $(a+b+c)/2$ and $Delta$ with $ab/2$. This means the expression becomes



                                    $$sqrt{frac{(a+b+c)(c-a-b)}{2ab}}$$



                                    now, simplify the numerator (either by multiplying everything out or first writing it as $(c+(a+b))(c-(a+b))$ and then using the fact that $(A+B)(A-B)=A^2-B^2$. Also, use the fact that $c^2=a^2+b^2$.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      The first thing to do is to replace $s$ with $(a+b+c)/2$ and $Delta$ with $ab/2$. This means the expression becomes



                                      $$sqrt{frac{(a+b+c)(c-a-b)}{2ab}}$$



                                      now, simplify the numerator (either by multiplying everything out or first writing it as $(c+(a+b))(c-(a+b))$ and then using the fact that $(A+B)(A-B)=A^2-B^2$. Also, use the fact that $c^2=a^2+b^2$.






                                      share|cite|improve this answer









                                      $endgroup$



                                      The first thing to do is to replace $s$ with $(a+b+c)/2$ and $Delta$ with $ab/2$. This means the expression becomes



                                      $$sqrt{frac{(a+b+c)(c-a-b)}{2ab}}$$



                                      now, simplify the numerator (either by multiplying everything out or first writing it as $(c+(a+b))(c-(a+b))$ and then using the fact that $(A+B)(A-B)=A^2-B^2$. Also, use the fact that $c^2=a^2+b^2$.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 12 '18 at 13:17









                                      5xum5xum

                                      91.1k394161




                                      91.1k394161






























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