Difference between revisions of "1991 AHSME Problems/Problem 5"
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== Solution == | == Solution == | ||
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− | Since they tell us that AB=AG and that angle A is a right angle, triangle ABG is a 45-45-90 triangle. Thus we can find the legs of the triangle by dividing <math>20</math> by <math>\sqrt2</math>. Both legs have length <math>10\sqrt2</math>, so the area of the right triangle is <math>100</math>. The rectangle is simple, just <math>20\times10</math>, so the area is 200. Adding these areas, we get <math>300</math> as the total area. | + | Since they tell us that AB=AG and that angle A is a right angle, triangle ABG is a 45-45-90 triangle. Thus we can find the legs of the triangle by dividing <math>20</math> (the hypotenuse) by <math>\sqrt2</math>. Both legs have length <math>10\sqrt2</math>, so the area of the right triangle is <math>100</math>. The rectangle is simple, just <math>20\times10</math>, so the area is 200. Adding these areas, we get <math>300</math> as the total area. |
== See also == | == See also == |
Latest revision as of 17:52, 21 November 2014
Problem
In the arrow-shaped polygon [see figure], the angles at vertices and are right angles, , and . The area of the polygon is closest to
Solution
Since they tell us that AB=AG and that angle A is a right angle, triangle ABG is a 45-45-90 triangle. Thus we can find the legs of the triangle by dividing (the hypotenuse) by . Both legs have length , so the area of the right triangle is . The rectangle is simple, just , so the area is 200. Adding these areas, we get as the total area.
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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