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Difference between revisions of "1985 AJHSME Problems/Problem 4"

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==Solution==
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==Solution 1==
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===Solution 1===
 
  
 
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<asy>
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<math>\boxed{\text{C}}</math>
 
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==Solution 2==
  
 
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This is answer choice <math>\boxed{\text{C}}</math>
 
This is answer choice <math>\boxed{\text{C}}</math>
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== Solution 3 (Weird Divisions) ==
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<asy>
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draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle);
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draw((0,4)--(6,9),dashed);
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draw((2,4)--(6,9),dashed);
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label("A",(0,9),NW);
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label("B",(6,9),NE);
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label("C",(6,0),SE);
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label("D",(2,0),SW);
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label("E",(2,4),SW);
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label("F",(0,4),SW);
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</asy>
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The area of <math>\Delta ABF</math> is <math>15</math>
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The area of <math>\Delta BFE</math> is <math>\frac{1}{2} \times FE \times AF</math> or <math>5</math>
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The area of quadrilateral <math>BEDC</math> is <math>\frac{1}{2} \times (DE + BC) \times DC</math> or <math>26</math>
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Thus, the area is <math>46</math> or <math>\boxed{\textbf{(C)}\ 46}</math>
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~ lovelearning999
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==Video Solution by BoundlessBrain!==
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https://youtu.be/-_r5GzafCUY
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==Video Solution==
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https://youtu.be/9OHRebmMyz4
 +
 +
~savannahsolver
  
 
==See Also==
 
==See Also==
  
−
[[1985 AJHSME Problems]]
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{{AJHSME box|year=1985|num-b=3|num-a=5}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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 +
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{{MAA Notice}}

Latest revision as of 21:39, 2 October 2024

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 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 $\boxed{\text{C}}$

Solution 3 (Weird Divisions)

[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(6,9),dashed); draw((2,4)--(6,9),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),SW); label("F",(0,4),SW); [/asy]

The area of $\Delta ABF$ is $15$

The area of $\Delta BFE$ is $\frac{1}{2} \times FE \times AF$ or $5$

The area of quadrilateral $BEDC$ is $\frac{1}{2} \times (DE + BC) \times DC$ or $26$

Thus, the area is $46$ or $\boxed{\textbf{(C)}\ 46}$

~ lovelearning999

Video Solution by BoundlessBrain!

https://youtu.be/-_r5GzafCUY

Video Solution

https://youtu.be/9OHRebmMyz4

~savannahsolver

See Also

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions


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