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

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(Solution)
 
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== Solution ==
 
== Solution ==
 
<math>\fbox{E}</math>
 
<math>\fbox{E}</math>
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.
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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> (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 18:52, 21 November 2014

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 $20$ (the hypotenuse) 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 (ProblemsAnswer KeyResources)
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
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