A question based on triangles and sequence and series.












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The sides of a right angle triangle are in arithmetic progression if the triangle has area $24$. What is the length of the smallest side? I try to solve this problem by taking $c^2=a^2+b^2$ and $2b=a+c$ but was unable to proceed. This question had come in my country's JEE advanced examination for the year 2017.






sequences-and-series geometry triangle arithmetic-progressions







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  • $begingroup$
    This part question might be useful: Arithmetic progression and right angled triangle.
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 8:47
















0












$begingroup$


The sides of a right angle triangle are in arithmetic progression if the triangle has area $24$. What is the length of the smallest side? I try to solve this problem by taking $c^2=a^2+b^2$ and $2b=a+c$ but was unable to proceed. This question had come in my country's JEE advanced examination for the year 2017.






sequences-and-series geometry triangle arithmetic-progressions







share|cite|improve this question











$endgroup$












  • $begingroup$
    This part question might be useful: Arithmetic progression and right angled triangle.
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 8:47














0












0








0





$begingroup$


The sides of a right angle triangle are in arithmetic progression if the triangle has area $24$. What is the length of the smallest side? I try to solve this problem by taking $c^2=a^2+b^2$ and $2b=a+c$ but was unable to proceed. This question had come in my country's JEE advanced examination for the year 2017.






sequences-and-series geometry triangle arithmetic-progressions







share|cite|improve this question











$endgroup$




The sides of a right angle triangle are in arithmetic progression if the triangle has area $24$. What is the length of the smallest side? I try to solve this problem by taking $c^2=a^2+b^2$ and $2b=a+c$ but was unable to proceed. This question had come in my country's JEE advanced examination for the year 2017.






sequences-and-series geometry triangle arithmetic-progressions




sequences-and-series geometry triangle arithmetic-progressions






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share|cite|improve this question













share|cite|improve this question




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edited Dec 1 '18 at 8:45









Martin Sleziak

44.7k8117272




44.7k8117272










asked Nov 13 '18 at 7:27









priyanka kumaripriyanka kumari

1297




1297












  • $begingroup$
    This part question might be useful: Arithmetic progression and right angled triangle.
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 8:47


















  • $begingroup$
    This part question might be useful: Arithmetic progression and right angled triangle.
    $endgroup$
    – Martin Sleziak
    Dec 1 '18 at 8:47
















$begingroup$
This part question might be useful: Arithmetic progression and right angled triangle.
$endgroup$
– Martin Sleziak
Dec 1 '18 at 8:47




$begingroup$
This part question might be useful: Arithmetic progression and right angled triangle.
$endgroup$
– Martin Sleziak
Dec 1 '18 at 8:47










1 Answer
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$begingroup$

Take the sides of the triangle to be $x+y$, $x$, $x-y$ (where $x$ and $y$ are positive numbers). Apply Pythagoras theorem, $(x+y)^2 = x^2+(x-y)^2$



$Longrightarrow(x+y)^2-(x-y)^2=x^2$



$Longrightarrow 4xy = x^2$



$Longrightarrow x=4y$



$therefore$ sides are in the ratio 3:4:5, let them be 3k,4k and 5k and use the area.



Hope it helps:)






share|cite|improve this answer











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    $begingroup$

    Take the sides of the triangle to be $x+y$, $x$, $x-y$ (where $x$ and $y$ are positive numbers). Apply Pythagoras theorem, $(x+y)^2 = x^2+(x-y)^2$



    $Longrightarrow(x+y)^2-(x-y)^2=x^2$



    $Longrightarrow 4xy = x^2$



    $Longrightarrow x=4y$



    $therefore$ sides are in the ratio 3:4:5, let them be 3k,4k and 5k and use the area.



    Hope it helps:)






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Take the sides of the triangle to be $x+y$, $x$, $x-y$ (where $x$ and $y$ are positive numbers). Apply Pythagoras theorem, $(x+y)^2 = x^2+(x-y)^2$



      $Longrightarrow(x+y)^2-(x-y)^2=x^2$



      $Longrightarrow 4xy = x^2$



      $Longrightarrow x=4y$



      $therefore$ sides are in the ratio 3:4:5, let them be 3k,4k and 5k and use the area.



      Hope it helps:)






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Take the sides of the triangle to be $x+y$, $x$, $x-y$ (where $x$ and $y$ are positive numbers). Apply Pythagoras theorem, $(x+y)^2 = x^2+(x-y)^2$



        $Longrightarrow(x+y)^2-(x-y)^2=x^2$



        $Longrightarrow 4xy = x^2$



        $Longrightarrow x=4y$



        $therefore$ sides are in the ratio 3:4:5, let them be 3k,4k and 5k and use the area.



        Hope it helps:)






        share|cite|improve this answer











        $endgroup$



        Take the sides of the triangle to be $x+y$, $x$, $x-y$ (where $x$ and $y$ are positive numbers). Apply Pythagoras theorem, $(x+y)^2 = x^2+(x-y)^2$



        $Longrightarrow(x+y)^2-(x-y)^2=x^2$



        $Longrightarrow 4xy = x^2$



        $Longrightarrow x=4y$



        $therefore$ sides are in the ratio 3:4:5, let them be 3k,4k and 5k and use the area.



        Hope it helps:)







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 1 '18 at 8:46









        Martin Sleziak

        44.7k8117272




        44.7k8117272










        answered Nov 13 '18 at 7:36









        MartundMartund

        1,415212




        1,415212






























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