1979 AHSME Problems/Problem 1

Revision as of 13:17, 3 January 2017 by E power pi times i (talk | contribs) (Solution)

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

1999 AHSME (ProblemsAnswer KeyResources)
Preceded by
First question
Followed by
Problem 2
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