catheter? I don't think so.
The legs are x and x-7, and the hypotenuse is x+6. So,
x^2 + (x-7)^2 = (x+6)^2
Solve for x, and then of course, the area is 1/2 x(x-7)
A right-angled triangle has a catheter 7 cm shorter than the other catheter and a hypotenuse 6 cm longer. Determine the area of the triangle.
2 answers
Ok so.
x^2+(x-7)^2 = (x+6)^2
it eventually left me with:
=13±2√39
Is that right? Do I now have to that into 1/2 x(x-7) Or did I muck it up?
x^2+(x-7)^2 = (x+6)^2
it eventually left me with:
=13±2√39
Is that right? Do I now have to that into 1/2 x(x-7) Or did I muck it up?