Difference between revisions of "2003 AMC 10B Problems/Problem 20"
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Then calculate the area of trapezoid <math>FGBA</math> and triangle <math>EGF</math> separately and add them. The area of the trapezoid is <math>\frac {2+5}{2}\cdot 3 = \frac {21}{2}</math> and the area of the triangle is <math>\frac{1}{2}\cdot 2 \cdot 2 = 2</math>. <math>\frac{21}{2}+2=\boxed{\textbf{(D)}\ \frac{25}{2}}</math> | Then calculate the area of trapezoid <math>FGBA</math> and triangle <math>EGF</math> separately and add them. The area of the trapezoid is <math>\frac {2+5}{2}\cdot 3 = \frac {21}{2}</math> and the area of the triangle is <math>\frac{1}{2}\cdot 2 \cdot 2 = 2</math>. <math>\frac{21}{2}+2=\boxed{\textbf{(D)}\ \frac{25}{2}}</math> | ||
+ | |||
+ | ==Solution 4== | ||
+ | Since <math>\Delta{ABE}\sim{\Delta{FGE}}</math> then <math>[AFGB]\sim{[FXYG]}</math>, where <math>X</math> and <math>Y</math> are ponts on <math>EF</math> and <math>EG</math> respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of <math>\frac{FG}{AB}=\frac{2}{5}</math>, or something like this | ||
+ | <cmath>[AEB]=[AFGB]+[FXYZ]+...</cmath><cmath>[AEB]=[AFGB]+\frac{2}{5}[AFGB]+(\frac{2}{5})^2[AFGB]+...</cmath>we have to find the ratio of the areas when the sides have shrunk by length <math>\frac{2}{5}l</math> | ||
+ | [asy] | ||
+ | unitsize(0.6 cm); | ||
+ | |||
+ | pair A, B, C, D, E, F, G; | ||
+ | |||
+ | A = (0,0); | ||
+ | B = (5,0); | ||
+ | C = (5,3); | ||
+ | D = (0,3); | ||
+ | F = (1,3); | ||
+ | G = (3,3); | ||
+ | E = extension(A,F,B,G); | ||
+ | |||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(A--E--B); | ||
+ | |||
+ | label("<math>A</math>", A, SW); | ||
+ | label("<math>B</math>", B, SE); | ||
+ | label("<math>C</math>", C, NE); | ||
+ | label("<math>D</math>", D, NW); | ||
+ | label("<math>E</math>", E, N); | ||
+ | label("<math>F</math>", F, SE); | ||
+ | label("<math>G</math>", G, SW); | ||
+ | label("<math>2/5</math>", (D + F)/2, N); | ||
+ | label("<math>4/5</math>", (G + C)/2, N); | ||
+ | label("<math>6/5</math>", (B + C)/2, dir(0)); | ||
+ | label("<math>6/5</math>", (A + D)/2, W); | ||
+ | label("<math>2</math>", (A + B)/2, S); | ||
+ | [/asy] | ||
+ | Let <math>[AFGB]'</math> be the area of the shape whose length is <math>\frac{2}{5}l</math> | ||
+ | <cmath>[AFGB]'=[ADCB]-[ADF]-[BCG]</cmath><cmath>[AFGB]'=12/5-6/25-12/25</cmath><cmath>[AFGB]'=42/25</cmath>Now comparing the ratios of <math>[AFGB]'</math> to <math>[AFGB]</math> we get | ||
+ | <cmath>\frac{[AFGB]'}{[AFGB]}=\frac{42}{25}/\frac{21}{2}\implies \frac{[AFGB]'}{[AFGB]}=\frac{4}{25}</cmath>By applying an infinite summation | ||
+ | <cmath>[AEB]=\sum_{n=0}^{\infty} \frac{21}{2}\cdot{(\frac{4}{25})^n}</cmath><cmath>S=\frac{a_1}{1-r}</cmath><cmath>\boxed{[AEB]=\frac{25}{2}}</cmath> | ||
==See Also== | ==See Also== |
Revision as of 01:18, 15 December 2019
- The following problem is from both the 2003 AMC 12B #14 and 2003 AMC 10B #20, so both problems redirect to this page.
Problem
In rectangle and . Points and are on so that and . Lines and intersect at . Find the area of .
Solution 1
because The ratio of to is since and from subtraction. If we let be the height of
The height is so the area of is .
Solution 2
We can look at this diagram as if it were a coordinate plane with point being . This means that the equation of the line is and the equation of the line is . From this we can set of the follow equation to find the coordinate of point :
We can plug this into one of our original equations to find that the coordinate is , meaning the area of is
Solution 3
At points and , segment is 5 units from segment . At points and , the segments are 2 units from each other. This means that collectively, the two lines closed the distance between them by 3 units over a height of 3 units. Therefore, to close the next two units of distance, they will have to travel a height of 2 units.
Then calculate the area of trapezoid and triangle separately and add them. The area of the trapezoid is and the area of the triangle is .
Solution 4
Since then , where and are ponts on and respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of , or something like this we have to find the ratio of the areas when the sides have shrunk by length [asy] unitsize(0.6 cm);
pair A, B, C, D, E, F, G;
A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G);
draw(A--B--C--D--cycle); draw(A--E--B);
label("", A, SW); label("", B, SE); label("", C, NE); label("", D, NW); label("", E, N); label("", F, SE); label("", G, SW); label("", (D + F)/2, N); label("", (G + C)/2, N); label("", (B + C)/2, dir(0)); label("", (A + D)/2, W); label("", (A + B)/2, S); [/asy] Let be the area of the shape whose length is Now comparing the ratios of to we get By applying an infinite summation
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.