# Difference between revisions of "1991 AHSME Problems/Problem 5"

## Problem

$[asy] draw((0,0)--(2,2)--(2,1)--(5,1)--(5,-1)--(2,-1)--(2,-2)--cycle,dot); MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(2,1),S);MP("D",(5,1),NE);MP("E",(5,-1),SE);MP("F",(2,-1),NW);MP("G",(2,-2),S); MP("5",(2,1.5),E);MP("5",(2,-1.5),E);MP("20",(3.5,1),N);MP("20",(3.5,-1),S);MP("10",(5,0),E); [/asy]$

In the arrow-shaped polygon [see figure], the angles at vertices $A,C,D,E$ and $F$ are right angles, $BC=FG=5, CD=FE=20, DE=10$, and $AB=AG$. The area of the polygon is closest to $\text{(A) } 288\quad \text{(B) } 291\quad \text{(C) } 294\quad \text{(D) } 297\quad \text{(E) } 300$

## Solution

$\fbox{E}$ 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 $20$ by $\sqrt2$. Both legs have length $10\sqrt2$, so the area of the right triangle is $100$. The rectangle is simple, just $20\times10$, so the area is 200. Adding these areas, we get $300$ as the total area.

## See also

 1991 AHSME (Problems • Answer Key • Resources) Preceded byProblem 4 Followed byProblem 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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