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Right triangle ABC has sides with length x, y and z. If triangle ABC

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Right triangle ABC has sides with length x, y and z. If triangle ABC  [#permalink]

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New post Updated on: 27 Mar 2017, 05:36
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Right triangle ABC has sides with length x, y and z. If triangle ABC has perimeter 17, and x² + y² + z² = 98, then what is the area of triangle ABC?

A) 12.75
B) 13.25
C) 14
D) 14.5
E) 15.25

* kudos for all correct solutions

EDIT: Changed perimeter to 17 (sorry)

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Originally posted by GMATPrepNow on 26 Mar 2017, 05:55.
Last edited by GMATPrepNow on 27 Mar 2017, 05:36, edited 2 times in total.
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Re: Right triangle ABC has sides with length x, y and z. If triangle ABC  [#permalink]

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New post 27 Mar 2017, 05:35
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GMATPrepNow wrote:
Right triangle ABC has sides with length x, y and z. If triangle ABC has perimeter 17, and x² + y² + z² = 98, then what is the area of triangle ABC?

A) 12.75
B) 13.25
C) 14
D) 14.5
E) 15.25

* kudos for all correct solutions


Let z = length of hypotenuse
Let x and y = lengths of the sides that make up the triangle’s right angle
NOTE: since triangle area = (base)(height)/2, we know that the area of triangle ABC = [color=red]xy/2

Given: x² + y² + z² = 98
The Pythagorean Theorem tells us that x² + y² = z²
So, we can write: x² + y² + (x² + y²) = 98
Rewrite: 2(x² + y²) = 98
Or we can write: x² + y² = 49

Since we already know that x² + y² = z², we can conclude that z² = 49, which means z = 7

If the perimeter = 17, we can write: x + y + z = 17
Since z = 7, we can write: x + y + 7 = 17
Simplify to get: x + y = 10

IMPORTANT: If we square both sides of the above equation, some nice things happen.
(x + y) ² = 10²
Expand: x² + 2xy + y² = 100
Rearrange to get: (x² + y²) + 2xy = 100
Replace x² + y² with 49 to get: 49 + 2xy = 100
Simplify: 2xy = 51

NOTE: Our goal is to determine the value of xy/2
So, take 2xy = 51 and divide both sides by 4 to get: xy/2 = 51/4 = 12.75

Answer:
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Right triangle ABC has sides with length x, y and z. If triangle ABC  [#permalink]

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New post Updated on: 27 Mar 2017, 09:29
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kavyagupta wrote:
please share the solution



Sure!!!

Please follow below attached fig.

As x^2+y^2=z^2----(1)

Also given x^2+y^2+z^2=98
putting value z^2 from (1)

2(x^2+y^2)= 98
or x^2+y^2=49------(2)

again putting x^2+y^2 =z^2 from (1)
z^2 =49 or z=7

Also given that perimeter x+y+z= 17
thus x+y= 17-7 =10---------(3)
sq both side to get
x^2+y^2+2xy =100
from (2)
2xy =100-49 =51
Also area = xy/2= 51/4 = 12.75

Ans 12.75

Ans A
Attachments

Untitled.jpg
Untitled.jpg [ 6.71 KiB | Viewed 2341 times ]


Originally posted by rohit8865 on 26 Mar 2017, 18:16.
Last edited by rohit8865 on 27 Mar 2017, 09:29, edited 1 time in total.
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Re: Right triangle ABC has sides with length x, y and z. If triangle ABC  [#permalink]

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New post 26 Mar 2017, 16:56
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please share the solution
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Re: Right triangle ABC has sides with length x, y and z. If triangle ABC  [#permalink]

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New post 24 Apr 2017, 17:26
Interesting question - how would you know to square both sides? Is there a rule?
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Re: Right triangle ABC has sides with length x, y and z. If triangle ABC  [#permalink]

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New post 19 Jun 2017, 10:25
henkbdh828 wrote:
Interesting question - how would you know to square both sides? Is there a rule?


\(x^2 + y^2 + z^2 = 98\)
\(z^2 = x^2 + y^2\)
\(2*z^2 = 98\)
z = 7

given perimeter: x + y + z = 17
if z = 7, than x+y = 10

\((x+y)^2 = x^2 + 2*x*y + y^2\)
\(x^2 + 2*x*y + y^2 = (x+y)^2 = 10^2 = 100\)
\(x^2 + y^2 = 49\)
\(2*x*y = 51\)
\(x*y = 25.5\)

Area of triangle = 1/2*x*y = 12.75
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Re: Right triangle ABC has sides with length x, y and z. If triangle ABC  [#permalink]

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