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Manager
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triangle [#permalink] New post 04 May 2006, 08:03
Can someone please help me with the solution and explanation to this problem?

For right triangle RST, RS and RT are legs and RS+RT=12. ST is hypotenuse w/length of 10. What is the area of the triangle?
Manager
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 [#permalink] New post 04 May 2006, 08:13
let the arms be x and y respectively say x as base and y as height
so x +y =12 eqn 1

also its a right angle traingle so

x^2 +y^2 = 10^2 (since 10 is hyptoneous) eqn 2

but we know

(x+y)^2 = x^2 +y^2 +2*x*y
i.e 144 =100 +2xy

implies xy =22

area or right angle traingle = (1/2) * base * height
= (1/2) *x *y
= (1/2) *22
=11

Hope the above explanation was clear enough to comprehend.
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 [#permalink] New post 04 May 2006, 08:20
Thank you, you made it very easy. I was trying to work out a complex quadratic, forgetting that I could plug in 100.
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 [#permalink] New post 05 May 2006, 03:03
Area of the triangle is RT*RS/2
RS^2+RT^2=100
RS^2+2RS*RT+RT^2=144
Subtract and get 2 RS*RT=44 RS*RT/2= 11
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 [#permalink] New post 07 May 2006, 13:22
Agree Area = 11 :wink:
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 [#permalink] New post 08 May 2006, 02:53
Agreed 11.. Nicely explained!
  [#permalink] 08 May 2006, 02:53
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