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

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==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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==Solution==
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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.
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Adding all of these, we see that the area the four triangles around <math>ABCD</math> is twice [<math>\triangle</math> <math>DAB</math>] + [<math>\triangle</math> <math>ABC</math>] + [<math>\triangle</math> <math>BCD</math>] + [<math>\triangle</math> <math>CDA</math>], which is itself twice the area of the quadrilateral <math>ABCD</math>. Finally, [<math>A'B'C'D'</math>] = [<math>ABCD</math>] + 4 <math>\cdot</math> [<math>ABCD</math>] = 5 <math>\cdot</math> [<math>ABCD</math>] = <math>\fbox{50}</math>.
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~ Mathavi
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Note: Anyone with a diagram would be of great help (still new to LaTex).

Latest revision as of 14:24, 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

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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