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− | == Problem ==
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− | The area of polygon <math> ABCDEF</math> is 52 with <math> AB=8</math>, <math>BC=9</math> and <math>FA=5</math>. What is <math> DE+EF</math>?
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− | <asy>
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− | pair a=(0,9), b=(8,9), c=(8,0), d=(4,0), e=(4,4), f=(0,4);
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− | draw(a--b--c--d--e--f--cycle);
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− | draw(shift(0,-.25)*a--shift(.25,-.25)*a--shift(.25,0)*a);
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− | draw(shift(-.25,0)*b--shift(-.25,-.25)*b--shift(0,-.25)*b);
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− | draw(shift(-.25,0)*c--shift(-.25,.25)*c--shift(0,.25)*c);
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− | draw(shift(.25,0)*d--shift(.25,.25)*d--shift(0,.25)*d);
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− | draw(shift(.25,0)*f--shift(.25,.25)*f--shift(0,.25)*f);
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− | label("$A$", a, NW);
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− | label("$B$", b, NE);
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− | label("$C$", c, SE);
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− | label("$D$", d, SW);
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− | label("$E$", e, SW);
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− | label("$F$", f, SW);
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− | label("5", (0,6.5), W);
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− | label("8", (4,9), N);
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− | label("9", (8, 4.5), E);
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− | </asy>
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− | <math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11 </math>
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− | == Solution ==
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− | Notice that <math>AF + DE = BC</math>, so <math>DE=4</math>. Let <math>O</math> be the intersection of the extensions of <math>AF</math> and <math>DC</math>, which makes rectangle <math>ABCO</math>. The area of the polygon is the area of <math>FEDO</math> subtracted from the area of <math>ABCO</math>.
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− | <cmath>\text{Area} = 52 = 8 \cdot 9- EF \cdot 4</cmath>
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− | Solving for the unknown, <math>EF=5</math>, therefore <math>DE+EF=4+5=\boxed{\textbf{(C)}\ 9}</math>.
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− | == Solution 2 ==
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− | Similar to solution 1, <math>AF + DE = BC</math>, so <math>DE=4</math>. Split the polygon into two rectangles by extending the <math>DE</math> so it intersects <math>AB</math>. Let's say the length of <math>FE</math> is equal to <math>x</math>. We can form the equation: <cmath>5x + 9(8-x) = 52</cmath>. We see that <math>x = 5</math>, so we can add <math>5 + 4 = \boxed{\textbf{(C)}\ 9}</math>.
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− | - ArjunBhumula
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− | ==Video Solution by OmegaLearn==
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− | https://youtu.be/abSgjn4Qs34?t=1908
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− | ~ pi_is_3.14
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− | == See Also ==
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− | {{AMC8 box|year=2005|num-b=12|num-a=14}}
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− | {{MAA Notice}}
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