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1985 AJHSME Problems/Problem 4: Difference between revisions

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New page: ==Solution== Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we '''do''' know how to find the area of.<br><br>If we continue se...
 
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==Problem==
The area of polygon <math>ABCDEF</math>, in square units, is
<math>\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74</math>
<asy>
draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle);
label("A",(0,9),NW);
label("B",(6,9),NE);
label("C",(6,0),SE);
label("D",(2,0),SW);
label("E",(2,4),NE);
label("F",(0,4),SW);
label("6",(3,9),N);
label("9",(6,4.5),E);
label("4",(4,0),S);
label("5",(0,6.5),W);
</asy>
==Solution==
==Solution==
Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we '''do''' know how to find the area of.<br><br>If we continue segment <math>FE</math> until it reaches the right side, we create two rectangles - one on the top and one on the bottom.<br><br>We know how to find the area of a rectangle, and we're given the sides! We can easily find that <math>6\times5 = 30</math>. For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other, so we'll call it <math>x</math>. And we'll also color in the part of the right side that it does cover.<br><br>We realize that the space that <math>x</math> doesn't cover is 5, so <math>x</math> must be <math>9 - 5 = 4</math>. So the area of the bottom rectangle is <math>4\times4</math>, or <math>16</math>.<br><br>Finally, we just add the area of the rectangles together to get <math>16 + 30 = 46</math>.


<math>C</math>
===Solution 1===
 
<asy>
draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle);
draw((2,4)--(6,4),dashed);
label("A",(0,9),NW);
label("B",(6,9),NE);
label("C",(6,0),SE);
label("D",(2,0),SW);
label("E",(2,4),NE);
label("F",(0,4),SW);
label("G",(6,4),SW);
label("6",(3,9),N);
label("9",(6,4.5),E);
label("4",(4,0),S);
label("5",(0,6.5),W);
</asy>
 
Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we '''do''' know how to find the area of.
 
If we continue segment <math>\overline{FE}</math> until it reaches the right side at <math>G</math>, we create two rectangles - one on the top and one on the bottom.
 
We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of <math>ABGF</math> is <math>6\times5 = 30</math>. For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.
 
Note that <math>GC+GB=9</math>, and <math>GB=AF=5</math>, so we must have <cmath>GC+5=9\Rightarrow GC=4</cmath>
 
The area of the bottom rectangle is then <cmath>(DC)(GC)=4\times 4=16</cmath>
 
Finally, we just add the areas of the rectangles together to get <math>16 + 30 = 46</math>.
 
<math>\boxed{\text{C}}</math>
 
===Solution 2===
 
<asy>
draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle);
draw((0,4)--(0,0),dashed);
draw((0,0)--(2,0),dashed);
label("A",(0,9),NW);
label("B",(6,9),NE);
label("C",(6,0),SE);
label("D",(2,0),SW);
label("E",(2,4),NE);
label("F",(0,4),SW);
label("G",(0,0),SW);
label("6",(3,9),N);
label("9",(6,4.5),E);
label("4",(4,0),S);
label("5",(0,6.5),W);
</asy>
 
Let <math>\langle ABCDEF \rangle</math> be the area of polygon <math>ABCDEF</math>. Also, let <math>G</math> be the intersection of <math>DC</math> and <math>AF</math> when both are extended.
 
Clearly, <cmath>\langle ABCDEF \rangle = \langle ABCG \rangle - \langle GFED \rangle</cmath>
 
Since <math>AB=6</math> and <math>BC=9</math>, <math>\langle ABCG \rangle =6\times 9=54</math>.
 
To compute the area of <math>GFED</math>, note that <cmath>AB=GD+DC</cmath> <cmath>BC=GF+FA</cmath>
 
We know that <math>AB=6</math>, <math>DC=4</math>, <math>BC=9</math>, and <math>FA=5</math>, so <cmath>6=GD+4\Rightarrow GD=2</cmath> <cmath>9=GF+5\Rightarrow GF=4</cmath>
 
Thus <math>\langle GFED \rangle = 4\times 2=8</math>
 
Finally, we have
<cmath>\begin{align*}
\langle ABCDEF \rangle &= \langle ABCG \rangle - \langle GFED \rangle \\
&= 54-8 \\
&= 46 \\
\end{align*}</cmath>
 
This is answer choice <math>\boxed{\text{C}}</math>
 
==See Also==
 
[[1985 AJHSME Problems]]

Revision as of 17:41, 12 January 2009

Problem

The area of polygon $ABCDEF$, in square units, is

$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Solution

Solution 1

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((2,4)--(6,4),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(6,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of.

If we continue segment $\overline{FE}$ until it reaches the right side at $G$, we create two rectangles - one on the top and one on the bottom.

We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\times5 = 30$. For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.

Note that $GC+GB=9$, and $GB=AF=5$, so we must have \[GC+5=9\Rightarrow GC=4\]

The area of the bottom rectangle is then \[(DC)(GC)=4\times 4=16\]

Finally, we just add the areas of the rectangles together to get $16 + 30 = 46$.

$\boxed{\text{C}}$

Solution 2

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(0,0),dashed); draw((0,0)--(2,0),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(0,0),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]

Let $\langle ABCDEF \rangle$ be the area of polygon $ABCDEF$. Also, let $G$ be the intersection of $DC$ and $AF$ when both are extended.

Clearly, \[\langle ABCDEF \rangle = \langle ABCG \rangle - \langle GFED \rangle\]

Since $AB=6$ and $BC=9$, $\langle ABCG \rangle =6\times 9=54$.

To compute the area of $GFED$, note that \[AB=GD+DC\] \[BC=GF+FA\]

We know that $AB=6$, $DC=4$, $BC=9$, and $FA=5$, so \[6=GD+4\Rightarrow GD=2\] \[9=GF+5\Rightarrow GF=4\]

Thus $\langle GFED \rangle = 4\times 2=8$

Finally, we have \begin{align*} \langle ABCDEF \rangle &= \langle ABCG \rangle - \langle GFED \rangle \\ &= 54-8 \\ &= 46 \\ \end{align*}

This is answer choice $\boxed{\text{C}}$

See Also

1985 AJHSME Problems

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