Difference between revisions of "2012 AMC 10A Problems/Problem 15"
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Since <math>AC=2BC</math>, <math>BC=\frac{1}{\sqrt{5}}</math>. <math>\triangle ABC</math> is a right triangle so the area is just <math>\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}</math> | Since <math>AC=2BC</math>, <math>BC=\frac{1}{\sqrt{5}}</math>. <math>\triangle ABC</math> is a right triangle so the area is just <math>\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{\textbf{(B)}\ \frac15}</math> | ||
− | == Solution 2 == | + | == Solution 2 (coordbash)== |
<center><asy> | <center><asy> |
Revision as of 09:39, 11 September 2020
Problem
Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of ?
Solution 1
intersects at a right angle, so . The hypotenuse of right triangle BED is .
Since , . is a right triangle so the area is just
Solution 2 (coordbash)
Let be the origin. Then,
can be represented by the line Also, can be represented by the line
Subtracting the second equation from the first gives us . Thus, . Plugging this into the first equation gives us .
Since , is ,
and .
Thus, . The answer is .
Solution 3
Triangle is similar to triangle ; line
Triangle is similar to triangle and the ratio of line to line .
Based on similarity the length of the height of is thus .
Thus, . The answer is
Solution 4
Let be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let be the top right corner of the top right unit square, where segment is 2 units in length.
Because of the Pythagorean Theorem, since and = 1, the diagonal of triangle is .
Triangle is clearly a similar triangle to triangle . Segment is the hypotenuse of triangle . So, we can write down:
, which is equal to: Solving this equation yields:
By Pythagorean theorem, we can now find segment Solving this yields:
, so
So then we can use So
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.