Difference between revisions of "2006 AMC 10A Problems/Problem 17"
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<math>\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{\sqrt{2}}{2}\qquad\textbf{(C) } \frac{\sqrt{3}}{2}\qquad\textbf{(D) } \sqrt{2} \qquad\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad</math> | <math>\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{\sqrt{2}}{2}\qquad\textbf{(C) } \frac{\sqrt{3}}{2}\qquad\textbf{(D) } \sqrt{2} \qquad\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad</math> | ||
− | + | == Solution 1 == | |
− | + | By [[symmetry]], <math>WXYZ</math> is a square. | |
− | By [[symmetry]], <math>WXYZ</math> is a | ||
<asy> | <asy> | ||
Line 33: | Line 32: | ||
</asy> | </asy> | ||
− | Draw <math>\overline{BZ}</math>. <math>BZ = \frac 12AH = 1</math>, so <math>\triangle BWZ</math> is a <math>45-45-90 \triangle</math>. Hence <math>WZ = \frac{1}{\sqrt{2}}</math>, and <math>[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 = \ | + | Draw <math>\overline{BZ}</math>. <math>BZ = \frac 12AH = 1</math>, so <math>\triangle BWZ</math> is a <math>45-45-90 \triangle</math>. Hence <math>WZ = \frac{1}{\sqrt{2}}</math>, and <math>[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 =\boxed{\textbf{(A) }\frac 12}</math>. |
There are many different similar ways to come to the same conclusion using different [[right triangle|45-45-90 triangles]]. | There are many different similar ways to come to the same conclusion using different [[right triangle|45-45-90 triangles]]. | ||
Line 51: | Line 50: | ||
</asy> | </asy> | ||
− | Drawing lines as shown above and piecing together the triangles, we see that <math>ABCD</math> is made up of 12 squares congruent to <math>WXYZ</math>. Hence <math>[WXYZ] = \frac{2\cdot 3}{12} = \frac 12</math>. | + | Drawing lines as shown above and piecing together the triangles, we see that <math>ABCD</math> is made up of <math>12</math> squares congruent to <math>WXYZ</math>. Hence <math>[WXYZ] = \frac{2\cdot 3}{12} =\boxed{\textbf{(A) }\frac 12} </math>. |
=== Solution 3 === | === Solution 3 === | ||
− | We see that if we draw a line to <math>BZ</math> it is half the width of the rectangle so that length would be <math>1</math>, and the resulting triangle is a <math>45-45-90</math> so using the Pythagorean Theorem we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\frac{1}{2}</math> which is our answer. | + | We see that if we draw a line to <math>BZ</math> it is half the width of the rectangle so that length would be <math>1</math>, and the resulting triangle is a <math>45-45-90</math> so using the [[Pythagorean Theorem]] we can get that each side is <math>\sqrt{\frac{1^2}{2}}</math> so the area of the middle square would be <math>(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\frac{1}{2}</math> which is our answer. |
=== Solution 4 === | === Solution 4 === |
Revision as of 12:10, 17 December 2021
Contents
Problem
In rectangle , points and trisect , and points and trisect . In addition, , and . What is the area of quadrilateral shown in the figure?
Solution 1
By symmetry, is a square.
Draw . , so is a . Hence , and .
There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
Solution 2
Drawing lines as shown above and piecing together the triangles, we see that is made up of squares congruent to . Hence .
Solution 3
We see that if we draw a line to it is half the width of the rectangle so that length would be , and the resulting triangle is a so using the Pythagorean Theorem we can get that each side is so the area of the middle square would be which is our answer.
Solution 4
Since and are trisection points and , we see that . Also, , so triangle is a right isosceles triangle, i.e. . By symmetry, triangles , , and are also right isosceles triangles. Therefore, , which means triangle is also a right isosceles triangle. Also, triangle is a right isosceles triangle.
Then , and . Hence, .
By symmetry, quadrilateral is a square, so its area is
~made by AoPS (somewhere) -put here by qkddud~
Video Solution by the Beauty of Math
https://www.youtube.com/watch?v=GX33rxlJz7s
~IceMatrix
See Also
2006 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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