Answer:
9
Step-by-step explanation:
→You can use the Pythagorean Theorem to solve this, by plugging in the numbers, like so:
\begin{gathered}a^2+b^2=c^2\\x^2+(x-3)^2=(x+3)^2\end{gathered}
a
2
+b
2
=c
2
x
2
+(x−3)
2
=(x+3)
2
x^2+x^2-6x+9=x^2+6x+9x
2
+x
2
−6x+9=x
2
+6x+9
→Subtract x^2-6x+9x
2
−6x+9 from both sides:
x^2 = 12xx
2
=12x
→Subtract 12x from both sides:
x^2 -12x=0x
2
−12x=0
→Factor out x:
x(x-12)=0x(x−12)=0
→Separate, set = to 0, and solve:
\begin{gathered}x = 0\\x -12=0\end{gathered}
x=0
x−12=0
→ Add 12 to both sides: x = 12x=12
→So we have 0 and 12, as our answers. However, we cannot have 0 as a side length, since this would not be possible.
→All we need to do is take 12, and plug it into the equation for the shortest leg.
\begin{gathered}x - 3=?\\12-3=9\end{gathered}
x−3=?
12−3=9
