Calculate the ratio of the sides of a given triangle given the ratio of areas.












3












$begingroup$


Given a triangle $triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $frac{|AM|}{|MB|} = frac{|BN|}{|NC|}= frac{|PC|}{|PA|}=k$, where $k>0$.




Calculate $k$, given that the area of the triangle $triangle MNP$ and the area of the triangle $ triangle ABC$ are in the following ratio: $Area_{triangle MNP} = frac{7}{25} times Area_{triangle ABC}$.




I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
I was also looking for similar triangles.



I would appreciate some hint.






algebra-precalculus geometry euclidean-geometry triangle area







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    Given a triangle $triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $frac{|AM|}{|MB|} = frac{|BN|}{|NC|}= frac{|PC|}{|PA|}=k$, where $k>0$.




    Calculate $k$, given that the area of the triangle $triangle MNP$ and the area of the triangle $ triangle ABC$ are in the following ratio: $Area_{triangle MNP} = frac{7}{25} times Area_{triangle ABC}$.




    I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
    I was also looking for similar triangles.



    I would appreciate some hint.






    algebra-precalculus geometry euclidean-geometry triangle area







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      1



      $begingroup$


      Given a triangle $triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $frac{|AM|}{|MB|} = frac{|BN|}{|NC|}= frac{|PC|}{|PA|}=k$, where $k>0$.




      Calculate $k$, given that the area of the triangle $triangle MNP$ and the area of the triangle $ triangle ABC$ are in the following ratio: $Area_{triangle MNP} = frac{7}{25} times Area_{triangle ABC}$.




      I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
      I was also looking for similar triangles.



      I would appreciate some hint.






      algebra-precalculus geometry euclidean-geometry triangle area







      share|cite|improve this question











      $endgroup$




      Given a triangle $triangle ABC$, points $M$, $N$, $P$ are drawn on the sides of the triangle in a way that $frac{|AM|}{|MB|} = frac{|BN|}{|NC|}= frac{|PC|}{|PA|}=k$, where $k>0$.




      Calculate $k$, given that the area of the triangle $triangle MNP$ and the area of the triangle $ triangle ABC$ are in the following ratio: $Area_{triangle MNP} = frac{7}{25} times Area_{triangle ABC}$.




      I have tried Heron formula, bot the calculations seem to be too complicated, I have also tried to simplify the problem and assume that k is equal to 1 and then calculate the ratio of the triangles area but it also doesn't help.
      I was also looking for similar triangles.



      I would appreciate some hint.






      algebra-precalculus geometry euclidean-geometry triangle area




      algebra-precalculus geometry euclidean-geometry triangle area






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 14:49







      user593746

















      asked Dec 17 '18 at 12:44









      Jack BlackwellJack Blackwell

      1249




      1249






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Draw $BB1$ is perpendicular to $AC$; $MM1//BC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55











          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%2f3043884%2fcalculate-the-ratio-of-the-sides-of-a-given-triangle-given-the-ratio-of-areas%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Draw $BB1$ is perpendicular to $AC$; $MM1//BC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55
















          3












          $begingroup$

          Draw $BB1$ is perpendicular to $AC$; $MM1//BC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55














          3












          3








          3





          $begingroup$

          Draw $BB1$ is perpendicular to $AC$; $MM1//BC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$






          share|cite|improve this answer











          $endgroup$



          Draw $BB1$ is perpendicular to $AC$; $MM1//BC$ and perpendicular to $AC$, then we have $frac{AM}{AB}=frac{MM1}{BB1}$



          So $$frac{AMP}{ABC}=frac{1/2 cdot APcdot MM1}{1/2cdot AC cdot BB1} =frac{APcdot AM}{ACcdot AB}$$



          $$Rightarrow frac{AMP}{ABC} = frac{CPN}{ABC} = frac{BMN}{ABC} = frac{AM}{AB}. frac{AP}{AC} = frac{k}{(k+1)^2}$$



          Or $$frac{MNP}{ABC} = 1 - frac{AMP}{ABC} - frac{CPN}{ABC} - frac{BMN}{ABC} = 1 - frac{3k}{(k+1)^2}=frac{7}{25}$$



          It is not difficult to find $k=frac{2}{3}$ or $k=frac{3}{2}$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 15:54

























          answered Dec 17 '18 at 13:58









          Word ShallowWord Shallow

          768518




          768518












          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55


















          • $begingroup$
            $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 13:59












          • $begingroup$
            $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
            $endgroup$
            – Quang Hoang
            Dec 17 '18 at 15:42










          • $begingroup$
            Got it. I edited something good in my solution, see now.
            $endgroup$
            – Word Shallow
            Dec 17 '18 at 15:55
















          $begingroup$
          $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
          $endgroup$
          – Word Shallow
          Dec 17 '18 at 13:59






          $begingroup$
          $frac{MNP}{ABC}=frac{Area_{triangle MNP}}{ Area_{triangle ABC}}$
          $endgroup$
          – Word Shallow
          Dec 17 '18 at 13:59














          $begingroup$
          $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
          $endgroup$
          – Quang Hoang
          Dec 17 '18 at 15:42




          $begingroup$
          $triangle AMP$ and $triangle ABC$ are not, in general, similar, because as you pointed out $AM:AB = k:(k+1)$ while $AP:AC = 1:(k+1)$. However, your others arguments still apply.
          $endgroup$
          – Quang Hoang
          Dec 17 '18 at 15:42












          $begingroup$
          Got it. I edited something good in my solution, see now.
          $endgroup$
          – Word Shallow
          Dec 17 '18 at 15:55




          $begingroup$
          Got it. I edited something good in my solution, see now.
          $endgroup$
          – Word Shallow
          Dec 17 '18 at 15:55


















          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%2f3043884%2fcalculate-the-ratio-of-the-sides-of-a-given-triangle-given-the-ratio-of-areas%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

          Bressuire

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more

          Cabo Verde

          Image
            Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
          Read more

          Gyllenstierna

          Image
          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
          Read more