Difference between revisions of "2002 AMC 10A Problems/Problem 25"
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=== Solution 2 but quicker === | === Solution 2 but quicker === | ||
− | From | + | From Solution <math>2</math> we know that the the altitude of the trapezoid is <math>\frac{60}{13}</math> and the triangle's area is <math>30</math>. |
− | Note that once we remove the triangle we get a rectangle with length <math>39</math> and height <math>\frac{60}{13}</math> | + | Note that once we remove the triangle we get a rectangle with length <math>39</math> and height <math>\frac{60}{13}</math>. |
− | + | The numbers multiply nicely to get <math>180+30=\boxed{(C) 210}</math> | |
-harsha12345 | -harsha12345 | ||
Revision as of 14:29, 1 January 2020
Contents
Problem
In trapezoid with bases and , we have , , , and . The area of is
Solution
Solution 1
It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend and to meet at point :
Since we have , with the ratio of proportionality being . Thus So the sides of are , which we recognize to be a right triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),
Solution 2
Draw altitudes from points and :
Translate the triangle so that coincides with . We get the following triangle:
The length of in this triangle is equal to the length of the original , minus the length of . Thus .
Therefore is a well-known right triangle. Its area is , and therefore its altitude is .
Now the area of the original trapezoid is
Solution 3
Draw altitudes from points and :
Call the length of to be , the length of to be , and the height of the trapezoid to be . By the Pythagorean Theorem, we have:
Subtracting these two equation yields:
We also have: .
We can substitute the value of into the equation we just obtained, so we now have:
.
We can add the and the equation to find the value of , which simplifies down to be . Finally, we can plug in and use the Pythagorean theorem to find the height of the trapezoid.
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.
The median of the trapezoid is , and multiplying this and the height of the trapezoid gets us:
Solution 4
We construct a line segment parallel to from point to line and label the intersection of this segment with line as point Then quadrilateral is a parallelogram, so and Triangle is therefore a right triangle, with area
By continuing to split and into segments of length we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides and and each with area The total area is therefore
Solution 2 but quicker
From Solution we know that the the altitude of the trapezoid is and the triangle's area is . Note that once we remove the triangle we get a rectangle with length and height . The numbers multiply nicely to get -harsha12345
See also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.