Difference between revisions of "1985 AJHSME Problems/Problem 4"
(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== |
− | |||
− | <math>C</math> | + | 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 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== | ||
+ | |||
+ | {{AJHSME box|year=1985|num-b=3|num-a=5}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | |||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 02:13, 16 February 2021
Contents
Problem
The area of polygon , in square units, is
Solution 1
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 until it reaches the right side at , 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 is . 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 , and , so we must have
The area of the bottom rectangle is then
Finally, we just add the areas of the rectangles together to get .
Solution 2
Let be the area of polygon . Also, let be the intersection of and when both are extended.
Clearly,
Since and , .
To compute the area of , note that
We know that , , , and , so
Thus
Finally, we have
This is answer choice
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.