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 ==
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{{AHSME box|year=1979|before=First question|num-a=2}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 11:48, 5 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]

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

$\textbf{(A) }8\qquad \textbf{(B) }9\qquad \textbf{(C) }12\qquad \textbf{(D) }18\qquad \textbf{(E) }24$

Solution

Solution by e_power_pi_times_i

Since the dimensions of $DEFG$ are half of the dimensions of $ABCD$, the area of $DEFG$ is $\dfrac{1}{2}\cdot\dfrac{1}{2}$ of $ABCD$, so the area of $ABCD$ is $\dfrac{1}{4}\cdot72 = \boxed{\textbf{(D) } 18}$.

See also

1979 AHSME (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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All AHSME Problems and Solutions

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