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Difficulty:
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In a certain right triangle, the hypotenuse equals (x+z) while the other two sides are (x+y) and (y+z) for positive integers x, y, z.
If y+4=z and z>x, then compute x+y+z.
A. 12
B. 16
C. 20
D. 24
E. 28
If y+4=z and z>x, then compute x+y+z.
A. 12
B. 16
C. 20
D. 24
E. 28
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3c28c027fc.2d8b1a221a.oiHyH7.png [ 8.51 KiB | Viewed 2177 times ]
09 Sep 2019, 02:42
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Using Pythagoras theorem and substituting y+4=z:
(x+y+4)^2 = (2y+4)^2 + (x+y)^2
Solving above equation, we will get:
y^2 = 2x-2y
y^2+2y =2x
y(y+2) = 2x
Now here y should be even
so substituting y=2,4,6.... and checking
So here only y= 2 will satisfy the condition of z>x
So we get y=2,x=4,z=6, sum is 12
Posted from my mobile device
(x+y+4)^2 = (2y+4)^2 + (x+y)^2
Solving above equation, we will get:
y^2 = 2x-2y
y^2+2y =2x
y(y+2) = 2x
Now here y should be even
so substituting y=2,4,6.... and checking
So here only y= 2 will satisfy the condition of z>x
So we get y=2,x=4,z=6, sum is 12
Posted from my mobile device
rocky620
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13 Sep 2019, 07:18
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looking for officail / algebraic method
