Difference between revisions of "User:Pepper2831"
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| − | + | Solution 3 (Pythagorean Theorem) | |
| + | Assign ZA as <math>x</math>, then AY as <math>4 - x</math>. Assign XM as <math>y</math> and MY as <math>8 - y</math>. Since triangles WXM and WZA are together, we can say <math>4x = 8y</math>, so <math>y = 2x</math>. Then therefore, XM is <math>2x</math> and MY has length <math>8 - 2x</math>. We can use the Pythagorean theorem to find WM, which is actually <math>\sqrt{(2x)^2 + 4^2)} = \sqrt{4x^2 + 16}</math>. We don't factor it yet - we are going to find <math>x</math> again using the Pythagorean Theorem. Similarly, finding MA is just the square root of the squares of AY and MY individually, or <math>\sqrt{(8 - 2x)^2 + (4 - x)^2} = \sqrt{64 - 32x + 4x^2 + 16 - 8x + x^2} = \sqrt{5x^2 - 40x + 80}</math>. Then simply, WA is really <math>\sqrt{x^2 + 64}</math>. | ||
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| + | Now we have the three sides of the right triangle: <math>\sqrt{4x^2 + 16}</math>, <math>\sqrt{5x^2 - 40x + 80}</math>, and <math>\sqrt{x^2 + 64}</math>. Per the Pythagorean theorem again, we can see <math>(4x^2 + 16) + (5x^2 - 40x + 80) = (x^2 + 64)</math>. Combining like terms gives us <math>8x^2 - 40x + 32 = 0</math>, then dividing by 8 gives <math>x^2 - 5x + 4 = 0</math>. As this elementary and well-known quadratic gives us the roots of <math>1</math> and <math>4</math>, we can see it is a bit weird to have <math>x = 4</math>, as then point Z is point A. So we'll assume <math>x = 1</math>. We have two legs of the triangle by plugging in the sides with x in them, given that <math>x = 1</math>: <math>\sqrt{20}</math> and <math>\sqrt{45}</math>. We should know that <math>20 \cdot 45 = 900</math>, and <math>\sqrt{900} = 30.</math> Dividing by 2 reveals us our answer: <math>\boxed{\textbf{(C) }15}</math> | ||
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| + | ~pepper2831 | ||
Latest revision as of 18:09, 15 November 2024
Solution 3 (Pythagorean Theorem)
Assign ZA as 
, then AY as 
. Assign XM as 
and MY as 
. Since triangles WXM and WZA are together, we can say 
, so 
. Then therefore, XM is 
and MY has length 
. We can use the Pythagorean theorem to find WM, which is actually 
. We don't factor it yet - we are going to find again using the Pythagorean Theorem. Similarly, finding MA is just the square root of the squares of AY and MY individually, or 
. Then simply, WA is really 
.
Now we have the three sides of the right triangle: 
, 
, and . Per the Pythagorean theorem again, we can see 
. Combining like terms gives us 
, then dividing by 8 gives 
. As this elementary and well-known quadratic gives us the roots of 
and 
, we can see it is a bit weird to have 
, as then point Z is point A. So we'll assume 
. We have two legs of the triangle by plugging in the sides with x in them, given that : 
and 
. We should know that 
, and 
Dividing by 2 reveals us our answer: 
~pepper2831
