SOLUTION: In right triangleABC , the length of the hypotenuse is 25 and the lengths of the legs are x and 3x + 3. What is the perimeter of the triangle?

Question 1007682: In right triangleABC , the length of the hypotenuse is 25 and the lengths of the legs are x and 3x + 3. What is
the perimeter of the triangle?
Found 3 solutions by solver91311, stanbon, mananth:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

Since it is a right triangle, call upon our old friend Pythagoras to help:

Solve for , then calculate , then add the measures of the three sides to get the perimeter.

John

My calculator said it, I believe it, that settles it

Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
In right triangle ABC , the length of the hypotenuse is 25 and the lengths of the legs are x and 3x + 3. What is
the perimeter of the triangle?
-----
Solve for "x"::
x^2 + (3x+3)^2 = 25^2
------
x^2 + 9x^2 + 18x + 9 - 625 = 0
-----
10x^2 + 18x - 616 = 0
---
5x^2 + 9x - 308 = 0
---
(x-7)(5x+44) = 0
---
Positive solution::
x = 7
3x+3 = 24
----
Ans: Perimeter = 25 + 24 + 7 = 56
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Cheers,
Stan H.
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Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!

Using Pythagoras theorem
x^2+(3x+3)^2 = 625

x^2+9x^2+18x +9 =625
10x^2+18x -616=0
/2
5x^2+9x-308=0
5x^2-35x+44x-308=0
5x(x-7)+44(x-7)=0
(5x+4)(x-7)=0
5x+4 = 0 OR x-7=0
x= -4/5 Or 7
x cannot be negative
Therefore x =7
Check
x^2= 49
(3x+3)^2 = 576
576 + 49 = 625
x= 7
(3x+3)= 24
Hypotenuse = 25
Perimeter = 7+24+25 = 56

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