Difference between revisions of "2006 AMC 10A Problems/Problem 17"
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− | Draw <math>\overline{BZ}</math>. <math>BZ = \frac 12AH = 1</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]]. |
Revision as of 17:55, 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
Draw the lines as shown above, and count the squares. There are 12, so we have .
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.