Difference between revisions of "2007 AMC 10A Problems/Problem 18"
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Consider the <math>12</math>-sided polygon <math>ABCDEFGHIJKL</math>, as shown. Each of its sides has length <math>4</math>, and each two consecutive sides form a right angle. Suppose that <math>\overline{AG}</math> and <math>\overline{CH}</math> meet at <math>M</math>. What is the area of quadrilateral <math>ABCM</math>? | Consider the <math>12</math>-sided polygon <math>ABCDEFGHIJKL</math>, as shown. Each of its sides has length <math>4</math>, and each two consecutive sides form a right angle. Suppose that <math>\overline{AG}</math> and <math>\overline{CH}</math> meet at <math>M</math>. What is the area of quadrilateral <math>ABCM</math>? | ||
− | [ | + | <asy> |
+ | unitsize(13mm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); | ||
+ | pair M=intersectionpoints(A--G,H--C)[0]; | ||
+ | |||
+ | draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); | ||
+ | draw(A--G); | ||
+ | draw(H--C); | ||
+ | dot(M); | ||
+ | |||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",Ep,SE); | ||
+ | label("$F$",F,SE); | ||
+ | label("$G$",G,SE); | ||
+ | label("$H$",H,SW); | ||
+ | label("$I$",I,SW); | ||
+ | label("$J$",J,SW); | ||
+ | label("$K$",K,NW); | ||
+ | label("$L$",L,NW); | ||
+ | label("$M$",M,W); | ||
+ | </asy> | ||
<math>\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}</math> | <math>\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}</math> | ||
==Solution== | ==Solution== | ||
+ | |||
+ | === Solution 1 === | ||
We can obtain the solution by calculating the area of rectangle <math>ABGH</math> minus the combined area of triangles <math>\triangle AHG</math> and <math>\triangle CGM</math>. | We can obtain the solution by calculating the area of rectangle <math>ABGH</math> minus the combined area of triangles <math>\triangle AHG</math> and <math>\triangle CGM</math>. | ||
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Taking the area of rectangle <math>ABGH</math> and subtracting the combined area of <math>\triangle AHG</math> and <math>\triangle CGM</math> yields <math>48 - (24 + \frac{32}{5}) = \boxed{\frac{88}{5}}\ \text{(C)}</math>. | Taking the area of rectangle <math>ABGH</math> and subtracting the combined area of <math>\triangle AHG</math> and <math>\triangle CGM</math> yields <math>48 - (24 + \frac{32}{5}) = \boxed{\frac{88}{5}}\ \text{(C)}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | |||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | dotfactor=4; | ||
+ | |||
+ | pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); | ||
+ | pair M=intersectionpoints(A--G,H--C)[0]; | ||
+ | pair Z=(2.5,3); | ||
+ | |||
+ | draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); | ||
+ | draw(A--G); | ||
+ | draw(H--C); | ||
+ | draw(B--Z--C); | ||
+ | draw(C--F); | ||
+ | dot(M); | ||
+ | |||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,N); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,NE); | ||
+ | label("$E$",Ep,SE); | ||
+ | label("$F$",F,SE); | ||
+ | label("$G$",G,SE); | ||
+ | label("$H$",H,SW); | ||
+ | label("$I$",I,SW); | ||
+ | label("$J$",J,SW); | ||
+ | label("$K$",K,NW); | ||
+ | label("$L$",L,NW); | ||
+ | label("$M$",M,W); | ||
+ | label("$N$",Z,NE); | ||
+ | </asy> | ||
+ | |||
+ | Extend <math>AB</math> and <math>CH</math> and call their intersection <math>N</math>. | ||
+ | |||
+ | The triangles <math>CBN</math> and <math>CGH</math> are clearly similar with ratio <math>1:2</math>, hence <math>BN=2</math> and thus <math>AN=6</math>. The area of the triangle <math>BCN</math> is <math>\frac{2\cdot 4}2 = 4</math>. | ||
+ | |||
+ | The triangles <math>MAN</math> and <math>MGH</math> are similar as well, and we now know that the ratio of their dimensions is <math>AN:GH = 6:4 = 3:2</math>. | ||
+ | |||
+ | Draw altitudes from <math>M</math> onto <math>AN</math> and <math>GH</math>, let their feet be <math>M_1</math> and <math>M_2</math>. We get that <math>MM_1 : MM_2 = 3:2</math>. Hence <math>MM1 = \frac 35 \cdot 12 = \frac {36}5 </math>. | ||
+ | |||
+ | Then the area of <math>AMN</math> is <math>\frac 12 \cdot AN \cdot MM_1 = \frac{108}5</math>, and the area of <math>ABCM</math> can be obtained by subtracting the area of <math>BCN</math>, which is <math>4</math>. Hence the answer is <math>\frac{108}5 - 4 = \boxed{\frac{88}5}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | |||
==See also== | ==See also== |
Revision as of 15:00, 1 March 2009
Problem
Consider the -sided polygon , as shown. Each of its sides has length , and each two consecutive sides form a right angle. Suppose that and meet at . What is the area of quadrilateral ?
Solution
Solution 1
We can obtain the solution by calculating the area of rectangle minus the combined area of triangles and .
We know that triangles and are similar because . Also, since , the ratio of the distance from to to the distance from to is also . Solving with the fact that the distance from to is 4, we see that the distance from to is .
The area of is simply , the area of is , and the area of rectangle is .
Taking the area of rectangle and subtracting the combined area of and yields .
Solution 2
Extend and and call their intersection .
The triangles and are clearly similar with ratio , hence and thus . The area of the triangle is .
The triangles and are similar as well, and we now know that the ratio of their dimensions is .
Draw altitudes from onto and , let their feet be and . We get that . Hence .
Then the area of is , and the area of can be obtained by subtracting the area of , which is . Hence the answer is .
See also
2007 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 AMC 10 Problems and Solutions |