Difference between revisions of "2006 AMC 10A Problems/Problem 17"
m (minor edit) |
Icematrix2 (talk | contribs) |
||
(6 intermediate revisions by one other user not shown) | |||
Line 14: | Line 14: | ||
</asy> | </asy> | ||
− | <math>\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D) | + | <math>\mathrm{(A) \ } \frac{1}{2}\qquad\mathrm{(B) \ } \frac{\sqrt{2}}{2}\qquad\mathrm{(C) \ } \frac{\sqrt{3}}{2}\qquad\mathrm{(D)} \sqrt{2} \qquad\mathrm{(E) \ } \frac{2\sqrt{3}}{3}\qquad</math> |
== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | + | By [[symmetry]], <math>WXYZ</math> is a [[square]]. | |
<asy> | <asy> | ||
Line 33: | Line 33: | ||
</asy> | </asy> | ||
− | Draw <math>\overline{BZ}</math>. | + | 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 = \frac 12\ \mathrm{(A)}</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 51: | ||
</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>. | |
=== Solution 3 === | === Solution 3 === |
Revision as of 23:18, 22 October 2020
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
Solution 1
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 12 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~
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.