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Difference between revisions of "1978 AHSME Problems/Problem 29"

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==Problem==
 
==Problem==
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Sides <math>AB,~ BC, ~CD</math> and <math>DA</math>, respectively, of convex quadrilateral <math>ABCD</math> are extended past <math>B,~ C ,~ D</math> and <math>A</math> to points <math>B',~C',~ D'</math> and <math>A'</math>. Also, <math>AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8</math> and <math>DA = AA' = 9</math>; and the area of <math>ABCD</math> is <math>10</math>. The area of <math>A 'B 'C'D'</math> is
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Sides <math>AB,~ BC, ~CD</math> and <math>DA</math>, respectively, of convex quadrilateral <math>ABCD</math> are extended past <math>B,~ C ,~ D</math> and <math>A</math> to points <math>B',~C',~ D'</math> and <math>A'</math>.  
 +
Also, <math>AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8</math> and <math>DA = AA' = 9</math>; and the area of <math>ABCD</math> is <math>10</math>. The area of <math>A 'B 'C'D'</math> is
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<math>\textbf{(A) }20\qquad
 +
\textbf{(B) }40\qquad
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\textbf{(C) }45\qquad
 +
\textbf{(D) }50\qquad
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\textbf{(E) }60    </math>
  
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<math>\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad  \textbf{(E) }60</math>
 
 
==Solution==
 
==Solution==
  

Latest revision as of 12:44, 26 September 2024

Problem

Sides $AB,~ BC, ~CD$ and $DA$, respectively, of convex quadrilateral $ABCD$ are extended past $B,~ C ,~ D$ and $A$ to points $B',~C',~ D'$ and $A'$. Also, $AB = BB' = 6,~ BC = CC' = 7, ~CD = DD' = 8$ and $DA = AA' = 9$; and the area of $ABCD$ is $10$. The area of $A 'B 'C'D'$ is

$\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad \textbf{(E) }60$

Solution

Notice that the area of $\triangle$ $DAB$ is the same as that of $\triangle$ $A'AB$ (same base, same height). Thus, the area of $\triangle$ $A'AB$ is twice that (same height, twice the base). Similarly, [$\triangle$ $BB'C$] = 2 $\cdot$ [$\triangle$ $ABC$], and so on.

Adding all of these, we see that the area the four triangles around $ABCD$ is twice [$\triangle$ $DAB$] + [$\triangle$ $ABC$] + [$\triangle$ $BCD$] + [$\triangle$ $CDA$], which is itself twice the area of the quadrilateral $ABCD$. Finally, [$A'B'C'D'$] = [$ABCD$] + 4 $\cdot$ [$ABCD$] = 5 $\cdot$ [$ABCD$] = $\fbox{50}$.

~ Mathavi

Note: Anyone with a diagram would be of great help (still new to LaTex).

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