Difference between revisions of "1979 AHSME Problems/Problem 1"
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Since the dimensions of <math>DEFG</math> are half of the dimensions of <math>ABCD</math>, the area of <math>DEFG</math> is <math>\dfrac{1}{2}\cdot\dfrac{1}{2}</math> of <math>ABCD</math>, so the area of <math>ABCD</math> is <math>\dfrac{1}{4}\cdot72 = \boxed{\textbf{(D) } 18}</math>. | Since the dimensions of <math>DEFG</math> are half of the dimensions of <math>ABCD</math>, the area of <math>DEFG</math> is <math>\dfrac{1}{2}\cdot\dfrac{1}{2}</math> of <math>ABCD</math>, so the area of <math>ABCD</math> is <math>\dfrac{1}{4}\cdot72 = \boxed{\textbf{(D) } 18}</math>. | ||
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| + | == See also == | ||
| + | {{AHSME box|year=1999|before=First question|num-a=2}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Revision as of 12:17, 3 January 2017
Problem 1
![[asy] draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle); draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle); label("$F$",(0,0),E); label("$A$",(-2,1),W); label("$B$",(2,1),E); label("$C$", (2,-1),E); label("$D$",(-2,-1),WSW); label("$E$",(-2,0),W); label("$G$",(0,-1),S); //Credit to TheMaskedMagician for the diagram [/asy]](http://latex.artofproblemsolving.com/e/6/f/e6f1f519c16bc795c83c94655c1012fb77e77c86.png)
If rectangle ABCD has area 72 square meters and E and G are the midpoints of sides AD and CD, respectively, then the area of rectangle DEFG in square meters is
Solution
Solution by e_power_pi_times_i
Since the dimensions of 
are half of the dimensions of 
, the area of is 
of , so the area of is 
.
See also
| 1999 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 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. 
