User:Pepper2831

Solution 3 (Pythagorean Theorem) Assign ZA as $x$ , then AY as $4 - x$ . Assign XM as $y$ and MY as $8 - y$ . Since triangles WXM and WZA are together, we can say $4x = 8y$ , so $y = 2x$ . Then therefore, XM is $2x$ and MY has length $8 - 2x$ . We can use the Pythagorean theorem to find WM, which is actually $\sqrt{(2x)^2 + 4^2)} = \sqrt{4x^2 + 16}$ . We don't factor it yet - we are going to find $x$ again using the Pythagorean Theorem. Similarly, finding MA is just the square root of the squares of AY and MY individually, or $\sqrt{(8 - 2x)^2 + (4 - x)^2} = \sqrt{64 - 32x + 4x^2 + 16 - 8x + x^2} = \sqrt{5x^2 - 40x + 80}$ . Then simply, WA is really $\sqrt{x^2 + 64}$ .

Now we have the three sides of the right triangle: $\sqrt{4x^2 + 16}$ , $\sqrt{5x^2 - 40x + 80}$ , and $\sqrt{x^2 + 64}$ . Per the Pythagorean theorem again, we can see $(4x^2 + 16) + (5x^2 - 40x + 80) = (x^2 + 64)$ . Combining like terms gives us $8x^2 - 40x + 32 = 0$ , then dividing by 8 gives $x^2 - 5x + 4 = 0$ . As this elementary and well-known quadratic gives us the roots of $1$ and $4$ , we can see it is a bit weird to have $x = 4$ , as then point Z is point A. So we'll assume $x = 1$ . We have two legs of the triangle by plugging in the sides with x in them, given that $x = 1$ : $\sqrt{20}$ and $\sqrt{45}$ . We should know that $20 \cdot 45 = 900$ , and $\sqrt{900} = 30.$ Dividing by 2 reveals us our answer: $\boxed{\textbf{(C) }15}$

~pepper2831

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User:Pepper2831