A right triangle has legs of length 6 and b, and a hypotenuse of length c . The perimeter of the triangle is 24 . Compute c.
+9674 Since the triangle is a right triangle, by Pythagorean theorem,
\(6^2 + b^2 = c^2\\ c^2 - b^2 = 36\quad\text{---}(1)\)
Now, since the perimeter is 24,
\(6 + b + c = 24\\ b + c = 18\quad \text{---}(2)\)
From (1), we can factorize and substitute (2) into the result.
\(\begin{array}{rcl} c^2 - b^2 &=& 36\\ (c - b)(b + c) &=& 36\\ (c - b)(18) &=& 36\\ c - b &=& 2 \end{array}\)
Now, solving the system \(\begin{cases}b + c = 18\\c - b = 2\end{cases}\) isn't too hard. We have b = 8 and c = 10.
+9674 Since the triangle is a right triangle, by Pythagorean theorem,
\(6^2 + b^2 = c^2\\ c^2 - b^2 = 36\quad\text{---}(1)\)
Now, since the perimeter is 24,
\(6 + b + c = 24\\ b + c = 18\quad \text{---}(2)\)
From (1), we can factorize and substitute (2) into the result.
\(\begin{array}{rcl} c^2 - b^2 &=& 36\\ (c - b)(b + c) &=& 36\\ (c - b)(18) &=& 36\\ c - b &=& 2 \end{array}\)
Now, solving the system \(\begin{cases}b + c = 18\\c - b = 2\end{cases}\) isn't too hard. We have b = 8 and c = 10.
