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In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK

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In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK  [#permalink]

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In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK = 20, what is the area of the triangle HEK?

A. 50
B. 75
C. 25√2
D. 100
E. 50√2

ALL THE BEST!

Originally posted by HARRY113 on 16 May 2016, 03:57.
Last edited by HARRY113 on 16 May 2016, 06:02, edited 2 times in total.
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In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK  [#permalink]

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New post 16 May 2016, 04:10
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HARRY113 wrote:
In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK = 20, what is the area of the triangle HEK?

A. 50
B. 75
C. 25√2
D. 100
E. 50√2

ALL THE BEST!

Hit Kudos if you like this post!


Hi,
This is in noway anywhere close to 700 level..
straight answer- 600 level..
1) HYP = \(10\sqrt{2}\)..
Other two sides = 20 or 10+10..
Area = 1/2 *10*10 = 50
2) Sum of square of side = hyp^2..
\((EH+HK)^2 = EH^2+HK^2+2*EH*HK = EK^2+2*EH*HK =20^2\)...
\((10\sqrt{2})^2 +2*EH*HK =400............................
2*EH*HK =400-200=200....\)
or \(EH*HK = 100..\)
area = \(\frac{1}{2}*EH*HK = \frac{1}{2}*100 = 50\)

A
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Re: In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK  [#permalink]

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New post 16 May 2016, 10:09
HARRY113 wrote:
In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK = 20, what is the area of the triangle HEK?

A. 50
B. 75
C. 25√2
D. 100
E. 50√2

ALL THE BEST!


Attachment:
Triangle.png
Triangle.png [ 3.68 KiB | Viewed 793 times ]


\((10√2)^2\) = \(x^2 + y^2\)

\(200\) = \((x + y)^2\)\(- 2xy\)

\(200\) = \((20)^2\)\(- 2xy\)

\(200\) = \(400\)\(- 2xy\)

\(- 200\) = \(- 2xy\)

\(xy = 100\)

Area of the triangle will be 1/2*xy = 50

Answer will be (A) 50 ; same opinion as chetan2u , not a 700 level question :lol:
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Re: In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK  [#permalink]

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New post 18 Mar 2018, 12:18
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Re: In right triangle HEK, HK is the hypotenuse. If HK is 10√2 and EH + EK &nbs [#permalink] 18 Mar 2018, 12:18
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