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
edited Nov 13 at 8:07
KReiser
9,02711233
asked Nov 13 at 7:27
priyanka kumari
1007
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up vote
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favorite
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
edited Nov 13 at 8:07
KReiser
9,02711233
asked Nov 13 at 7:27
priyanka kumari
1007
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
edited Nov 13 at 8:07
KReiser
9,02711233
asked Nov 13 at 7:27
priyanka kumari
1007
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
edited Nov 13 at 8:07
KReiser
9,02711233
asked Nov 13 at 7:27
priyanka kumari
1007
edited Nov 13 at 8:07
KReiser
9,02711233
asked Nov 13 at 7:27
priyanka kumari
1007
edited Nov 13 at 8:07
KReiser
9,02711233
edited Nov 13 at 8:07
KReiser
9,02711233
edited Nov 13 at 8:07
KReiser
9,02711233
9,02711233
asked Nov 13 at 7:27
priyanka kumari
1007
asked Nov 13 at 7:27
priyanka kumari
1007
asked Nov 13 at 7:27
priyanka kumari
1007
1007
add a comment |
add a comment |
1 Answer
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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:)
answered Nov 13 at 7:36
Crazy for maths
5089
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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:)
answered Nov 13 at 7:36
Crazy for maths
5089
add a comment |
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:)
answered Nov 13 at 7:36
Crazy for maths
5089
add a comment |
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:)
answered Nov 13 at 7:36
Crazy for maths
5089
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:)
answered Nov 13 at 7:36
Crazy for maths
5089
answered Nov 13 at 7:36
Crazy for maths
5089
answered Nov 13 at 7:36
Crazy for maths
5089
answered Nov 13 at 7:36
Crazy for maths
5089
5089
add a comment |
add a comment |
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