Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
 
Line 15: Line 15:
 
== 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.
+
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

[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 (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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AHSME_Problems/Problem_5&oldid=65998"