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









        $endgroup$













          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036668%2fhow-to-prove-sqrt-frac-sc-a-b2%25e2%2588%2586-is-equal-to-i%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          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






























                                          draft saved

                                          draft discarded




















































                                          Thanks for contributing an answer to Mathematics Stack Exchange!


                                          • Please be sure to answer the question. Provide details and share your research!

                                          But avoid …



                                          • Asking for help, clarification, or responding to other answers.

                                          • Making statements based on opinion; back them up with references or personal experience.


                                          Use MathJax to format equations. MathJax reference.


                                          To learn more, see our tips on writing great answers.




                                          draft saved


                                          draft discarded














                                          StackExchange.ready(
                                          function () {
                                          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3036668%2fhow-to-prove-sqrt-frac-sc-a-b2%25e2%2588%2586-is-equal-to-i%23new-answer', 'question_page');
                                          }
                                          );

                                          Post as a guest















                                          Required, but never shown





















































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown

































                                          Required, but never shown














                                          Required, but never shown












                                          Required, but never shown







                                          Required, but never shown







                                          -

                                          Popular posts from this blog

                                          Bundesstraße 106

                                          Image
                                          Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                                          Read more

                                          Verónica Boquete

                                          Image
                                          Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                                          Read more

                                          Ida-Boy-Ed-Garten

                                          Image
                                          Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                                          Read more