SOLUTION: The area for a triangle is (4x squared + 10) squared. the base is 4x-4 and the height is x-3 and the hypotinus is 2x+5. what is x? how do you find what x is?

Algebra ->  Triangles -> SOLUTION: The area for a triangle is (4x squared + 10) squared. the base is 4x-4 and the height is x-3 and the hypotinus is 2x+5. what is x? how do you find what x is?      Log On


   

Question 8615: The area for a triangle is (4x squared + 10) squared. the base is 4x-4 and the height is x-3 and the hypotinus is 2x+5. what is x? how do you find what x is?
Found 2 solutions by jennifer, rapaljer:
Answer by jennifer(9) About Me  (Show Source):
You can put this solution on YOUR website!
Ok, A=(4x-sqrt+10)sqrt
(4x-4)+(x-3)= 2x+5 You solve for x to figure out what base, height, and hypotenuse is.
(4x+x)+ (-4-3)=2x+5
5x-7=2x+5
5x=2x+12
3x=12
x=4

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Jennifer's solution is NOT correct! When you add the base and to the height you do NOT get the hypotenuse, as she suggested!!

Area of a triangle:
A+=+%281%2F2%29%2A+%28bh%29
+%281%2F2%29%2A%284x-4%29%2A%28x-3%29=+%284x%5E2%2B10%29%5E2
+%282x-2%29%2A%28x-3%29=+%284x%5E2%2B10%29%5E2
2x%5E2+-6x-2x+%2B+6+=+16x%5E4+%2B+80x%5E2+%2B+100+
0+=+16x%5E4+%2B+78x%5E2+%2B+8x+%2B+94+

Since this seems unreasonable to solve, perhaps you stated the problem wrong!! Recheck the problem. I was thinking that maybe the area is just 4x%5E2+%2B+10, without the "squared."

+%282x-2%29%2A%28x-3%29=+%284x%5E2%2B10%29+
2x%5E2+-6x-2x+%2B+6+=+4x%5E2+%2B+10+
0+=+2x%5E2+%2B+8x+%2B+4+
0+=+2%28x%5E2+%2B+4x+%2B+2%29+
Solve by quadratic formula, and it just comes out ugly?? Recheck your copying of the problem to make sure that is correct.

R^2 at SCC