finding ratio of sides in a right angled triangle when its two sides are given












-3












$begingroup$


The perimeter of a right-angled triangle is 4 times the shorter side. Find the ratio of other sides ??
I tried approaching it this way .
Let's have a triangle ABC right angled at B and BC=x .Then P=5x and on adding the three sides AB+AC=4x.But couldn't move after this as how to get ratios.






geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Whenever I see a right angled triangle I immediately assume a ratio of 3,4,5 Which often works out. Of course if you have to show your working don't do this.
    $endgroup$
    – yolo
    Dec 16 '18 at 12:39
















-3












$begingroup$


The perimeter of a right-angled triangle is 4 times the shorter side. Find the ratio of other sides ??
I tried approaching it this way .
Let's have a triangle ABC right angled at B and BC=x .Then P=5x and on adding the three sides AB+AC=4x.But couldn't move after this as how to get ratios.






geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Whenever I see a right angled triangle I immediately assume a ratio of 3,4,5 Which often works out. Of course if you have to show your working don't do this.
    $endgroup$
    – yolo
    Dec 16 '18 at 12:39














-3












-3








-3





$begingroup$


The perimeter of a right-angled triangle is 4 times the shorter side. Find the ratio of other sides ??
I tried approaching it this way .
Let's have a triangle ABC right angled at B and BC=x .Then P=5x and on adding the three sides AB+AC=4x.But couldn't move after this as how to get ratios.






geometry







share|cite|improve this question









$endgroup$




The perimeter of a right-angled triangle is 4 times the shorter side. Find the ratio of other sides ??
I tried approaching it this way .
Let's have a triangle ABC right angled at B and BC=x .Then P=5x and on adding the three sides AB+AC=4x.But couldn't move after this as how to get ratios.






geometry




geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 1 '18 at 6:41









iamredpandaiamredpanda

54




54












  • $begingroup$
    Whenever I see a right angled triangle I immediately assume a ratio of 3,4,5 Which often works out. Of course if you have to show your working don't do this.
    $endgroup$
    – yolo
    Dec 16 '18 at 12:39


















  • $begingroup$
    Whenever I see a right angled triangle I immediately assume a ratio of 3,4,5 Which often works out. Of course if you have to show your working don't do this.
    $endgroup$
    – yolo
    Dec 16 '18 at 12:39
















$begingroup$
Whenever I see a right angled triangle I immediately assume a ratio of 3,4,5 Which often works out. Of course if you have to show your working don't do this.
$endgroup$
– yolo
Dec 16 '18 at 12:39




$begingroup$
Whenever I see a right angled triangle I immediately assume a ratio of 3,4,5 Which often works out. Of course if you have to show your working don't do this.
$endgroup$
– yolo
Dec 16 '18 at 12:39










3 Answers
3






active

oldest

votes


















0












$begingroup$

Read yourself : "perimeter is 4 times the shorter side .... P = 5 x". Can you see the illogic point ? You can add "lengths" in your sentences too.



My hint is : you have not deduced anything from the specific property "right angle triangle".






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Take it easy. You are both new and unpracticed here. Your $answer$ might have been better posted as a comment. Asis, you are open to losing reputation if anyone thinks your answer is not useful.
    $endgroup$
    – poetasis
    Dec 1 '18 at 12:16



















0












$begingroup$

Let the sides be $a,b,c$ such that:
$$begin{cases}4a=a+b+c \ a^2+b^2=c^2 end{cases} Rightarrow begin{cases}9a^2=b^2+2bc+c^2 \ 9a^2=9c^2-9b^2end{cases} Rightarrow 10b^2+2bc-8c^2=0 Rightarrow \
5left(frac bcright)^2+frac bc-4=0 Rightarrow frac bc=frac{-1+9}{10}=frac45.$$






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    One approach could start with $4A=A+B+C$. Subracting $A$ from both sides, we get $3A=B+C$. Also $A^2+B^2=C^2$ so $A=sqrt{C^2-B^2}$.



    Now $$A=sqrt{C^2-B^2}=frac{B+C}{3}$$



    $$A^2=C^2-B^2=frac{C^2+2CB+B^2}{9}$$
    so
    $$9C^2-9B^2=C^2+2CB+B^2$$ subtracting the right side from both sides
    $$8C^2+2CB-10B^2=0$$
    $$4C^2+BC-5B^2=0$$



    Perhaps you could work it out algebraically using the quadratic formula to solve for either C but I noticed the smallest Pythagorean triplet has sides of $3,4,5$ and, since $$Perimeter=3+4+5=12=4*3(theshortest side)$$ I think you have your ratios.






    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%2f3021069%2ffinding-ratio-of-sides-in-a-right-angled-triangle-when-its-two-sides-are-given%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      Read yourself : "perimeter is 4 times the shorter side .... P = 5 x". Can you see the illogic point ? You can add "lengths" in your sentences too.



      My hint is : you have not deduced anything from the specific property "right angle triangle".






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Take it easy. You are both new and unpracticed here. Your $answer$ might have been better posted as a comment. Asis, you are open to losing reputation if anyone thinks your answer is not useful.
        $endgroup$
        – poetasis
        Dec 1 '18 at 12:16
















      0












      $begingroup$

      Read yourself : "perimeter is 4 times the shorter side .... P = 5 x". Can you see the illogic point ? You can add "lengths" in your sentences too.



      My hint is : you have not deduced anything from the specific property "right angle triangle".






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Take it easy. You are both new and unpracticed here. Your $answer$ might have been better posted as a comment. Asis, you are open to losing reputation if anyone thinks your answer is not useful.
        $endgroup$
        – poetasis
        Dec 1 '18 at 12:16














      0












      0








      0





      $begingroup$

      Read yourself : "perimeter is 4 times the shorter side .... P = 5 x". Can you see the illogic point ? You can add "lengths" in your sentences too.



      My hint is : you have not deduced anything from the specific property "right angle triangle".






      share|cite|improve this answer









      $endgroup$



      Read yourself : "perimeter is 4 times the shorter side .... P = 5 x". Can you see the illogic point ? You can add "lengths" in your sentences too.



      My hint is : you have not deduced anything from the specific property "right angle triangle".







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Dec 1 '18 at 9:45









      Dominique LaurainDominique Laurain

      342




      342












      • $begingroup$
        Take it easy. You are both new and unpracticed here. Your $answer$ might have been better posted as a comment. Asis, you are open to losing reputation if anyone thinks your answer is not useful.
        $endgroup$
        – poetasis
        Dec 1 '18 at 12:16


















      • $begingroup$
        Take it easy. You are both new and unpracticed here. Your $answer$ might have been better posted as a comment. Asis, you are open to losing reputation if anyone thinks your answer is not useful.
        $endgroup$
        – poetasis
        Dec 1 '18 at 12:16
















      $begingroup$
      Take it easy. You are both new and unpracticed here. Your $answer$ might have been better posted as a comment. Asis, you are open to losing reputation if anyone thinks your answer is not useful.
      $endgroup$
      – poetasis
      Dec 1 '18 at 12:16




      $begingroup$
      Take it easy. You are both new and unpracticed here. Your $answer$ might have been better posted as a comment. Asis, you are open to losing reputation if anyone thinks your answer is not useful.
      $endgroup$
      – poetasis
      Dec 1 '18 at 12:16











      0












      $begingroup$

      Let the sides be $a,b,c$ such that:
      $$begin{cases}4a=a+b+c \ a^2+b^2=c^2 end{cases} Rightarrow begin{cases}9a^2=b^2+2bc+c^2 \ 9a^2=9c^2-9b^2end{cases} Rightarrow 10b^2+2bc-8c^2=0 Rightarrow \
      5left(frac bcright)^2+frac bc-4=0 Rightarrow frac bc=frac{-1+9}{10}=frac45.$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Let the sides be $a,b,c$ such that:
        $$begin{cases}4a=a+b+c \ a^2+b^2=c^2 end{cases} Rightarrow begin{cases}9a^2=b^2+2bc+c^2 \ 9a^2=9c^2-9b^2end{cases} Rightarrow 10b^2+2bc-8c^2=0 Rightarrow \
        5left(frac bcright)^2+frac bc-4=0 Rightarrow frac bc=frac{-1+9}{10}=frac45.$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Let the sides be $a,b,c$ such that:
          $$begin{cases}4a=a+b+c \ a^2+b^2=c^2 end{cases} Rightarrow begin{cases}9a^2=b^2+2bc+c^2 \ 9a^2=9c^2-9b^2end{cases} Rightarrow 10b^2+2bc-8c^2=0 Rightarrow \
          5left(frac bcright)^2+frac bc-4=0 Rightarrow frac bc=frac{-1+9}{10}=frac45.$$






          share|cite|improve this answer









          $endgroup$



          Let the sides be $a,b,c$ such that:
          $$begin{cases}4a=a+b+c \ a^2+b^2=c^2 end{cases} Rightarrow begin{cases}9a^2=b^2+2bc+c^2 \ 9a^2=9c^2-9b^2end{cases} Rightarrow 10b^2+2bc-8c^2=0 Rightarrow \
          5left(frac bcright)^2+frac bc-4=0 Rightarrow frac bc=frac{-1+9}{10}=frac45.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 1 '18 at 16:35









          farruhotafarruhota

          19.6k2738




          19.6k2738























              -1












              $begingroup$

              One approach could start with $4A=A+B+C$. Subracting $A$ from both sides, we get $3A=B+C$. Also $A^2+B^2=C^2$ so $A=sqrt{C^2-B^2}$.



              Now $$A=sqrt{C^2-B^2}=frac{B+C}{3}$$



              $$A^2=C^2-B^2=frac{C^2+2CB+B^2}{9}$$
              so
              $$9C^2-9B^2=C^2+2CB+B^2$$ subtracting the right side from both sides
              $$8C^2+2CB-10B^2=0$$
              $$4C^2+BC-5B^2=0$$



              Perhaps you could work it out algebraically using the quadratic formula to solve for either C but I noticed the smallest Pythagorean triplet has sides of $3,4,5$ and, since $$Perimeter=3+4+5=12=4*3(theshortest side)$$ I think you have your ratios.






              share|cite|improve this answer











              $endgroup$


















                -1












                $begingroup$

                One approach could start with $4A=A+B+C$. Subracting $A$ from both sides, we get $3A=B+C$. Also $A^2+B^2=C^2$ so $A=sqrt{C^2-B^2}$.



                Now $$A=sqrt{C^2-B^2}=frac{B+C}{3}$$



                $$A^2=C^2-B^2=frac{C^2+2CB+B^2}{9}$$
                so
                $$9C^2-9B^2=C^2+2CB+B^2$$ subtracting the right side from both sides
                $$8C^2+2CB-10B^2=0$$
                $$4C^2+BC-5B^2=0$$



                Perhaps you could work it out algebraically using the quadratic formula to solve for either C but I noticed the smallest Pythagorean triplet has sides of $3,4,5$ and, since $$Perimeter=3+4+5=12=4*3(theshortest side)$$ I think you have your ratios.






                share|cite|improve this answer











                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  One approach could start with $4A=A+B+C$. Subracting $A$ from both sides, we get $3A=B+C$. Also $A^2+B^2=C^2$ so $A=sqrt{C^2-B^2}$.



                  Now $$A=sqrt{C^2-B^2}=frac{B+C}{3}$$



                  $$A^2=C^2-B^2=frac{C^2+2CB+B^2}{9}$$
                  so
                  $$9C^2-9B^2=C^2+2CB+B^2$$ subtracting the right side from both sides
                  $$8C^2+2CB-10B^2=0$$
                  $$4C^2+BC-5B^2=0$$



                  Perhaps you could work it out algebraically using the quadratic formula to solve for either C but I noticed the smallest Pythagorean triplet has sides of $3,4,5$ and, since $$Perimeter=3+4+5=12=4*3(theshortest side)$$ I think you have your ratios.






                  share|cite|improve this answer











                  $endgroup$



                  One approach could start with $4A=A+B+C$. Subracting $A$ from both sides, we get $3A=B+C$. Also $A^2+B^2=C^2$ so $A=sqrt{C^2-B^2}$.



                  Now $$A=sqrt{C^2-B^2}=frac{B+C}{3}$$



                  $$A^2=C^2-B^2=frac{C^2+2CB+B^2}{9}$$
                  so
                  $$9C^2-9B^2=C^2+2CB+B^2$$ subtracting the right side from both sides
                  $$8C^2+2CB-10B^2=0$$
                  $$4C^2+BC-5B^2=0$$



                  Perhaps you could work it out algebraically using the quadratic formula to solve for either C but I noticed the smallest Pythagorean triplet has sides of $3,4,5$ and, since $$Perimeter=3+4+5=12=4*3(theshortest side)$$ I think you have your ratios.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 2 '18 at 14:32

























                  answered Dec 1 '18 at 11:15









                  poetasispoetasis

                  409117




                  409117






























                      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%2f3021069%2ffinding-ratio-of-sides-in-a-right-angled-triangle-when-its-two-sides-are-given%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

                      Willebadessen

                      Image
                      Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
                      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

                      Residenzschloss Arolsen

                      Image
                      Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
                      Read more