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

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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>
 
<math>\textbf{(A) }20\qquad \textbf{(B) }40\qquad \textbf{(C) }45\qquad \textbf{(D) }50\qquad  \textbf{(E) }60</math>
 
==Solution==
 
==Solution==
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\usepackage{asymptote}
 
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\begin{asy} [asy]
 
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unitsize(1 cm);
 
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pair A, Ap, B, C, P, Q;
 
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A = 3*dir(60);
 
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Ap = (1,0);
 
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B = (0,0);
 
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C = (3,0);
 
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P = 8/5*dir(60);
 
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Q = C + 5/4*dir(120);
 
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draw(B--P--Q--C--cycle);
 
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draw(P--Ap--Q);
 
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draw(P--A--Q,dashed);
 
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label("<math>A</math>", A, N);
 
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label("<math>A'</math>", Ap, S);
 
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label("<math>B</math>", B, SW);
 
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label("<math>C</math>", C, SE);
 
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label("<math>P</math>", P, NW);
 
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label("<math>Q</math>", Q, NE);
 
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[/asy]
 
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\end{asy}
 
  
 
Notice that the area of <math>\triangle</math> <math>DAB</math> is the same as that of <math>\triangle</math> <math>A'AB</math> (same base, same height). Thus, the area of <math>\triangle</math> <math>A'AB</math> is twice that (same height, twice the base). Similarly, [<math>\triangle</math> <math>BB'C</math>] = 2 <math>\cdot</math> [<math>\triangle</math> <math>ABC</math>], and so on.
 
Notice that the area of <math>\triangle</math> <math>DAB</math> is the same as that of <math>\triangle</math> <math>A'AB</math> (same base, same height). Thus, the area of <math>\triangle</math> <math>A'AB</math> is twice that (same height, twice the base). Similarly, [<math>\triangle</math> <math>BB'C</math>] = 2 <math>\cdot</math> [<math>\triangle</math> <math>ABC</math>], and so on.

Revision as of 13:14, 23 August 2022

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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