1991 AHSME Problems/Problem 5
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 | |
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