Problem: Diagonal of triangle is 19cm. One side of triangle is 4 cm longer than the other leg. Find the two missing legs.
Here is my work so far:
a^2+b^2=c^2
x^2+(x+4)^2=19^2
x^2+x^2+8x+16=361
2x^2+8x=345
This isn't a problem that can be easily factored so I am wondering if I am taking the right approach? Can someone please point me in the right direction?
Thanks!
Here is my work so far:
a^2+b^2=c^2
x^2+(x+4)^2=19^2
x^2+x^2+8x+16=361
2x^2+8x=345
This isn't a problem that can be easily factored so I am wondering if I am taking the right approach? Can someone please point me in the right direction?
Thanks!