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.






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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







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      up vote
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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







      share|cite|improve this question















      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




      sequences-and-series geometry triangle






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      edited 15 hours ago









      KReiser

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      asked 15 hours ago









      priyanka kumari

      1007




      1007






















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          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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            up vote
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            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

























              up vote
              1
              down vote













              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























                up vote
                1
                down vote










                up vote
                1
                down vote









                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












                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










                answered 15 hours ago









                Crazy for maths

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