Difference between revisions of "2008 AMC 10A Problems/Problem 20"
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pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */ | pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */ | ||
pen sm = fontsize(10); /* small font pen */ | pen sm = fontsize(10); /* small font pen */ |
Revision as of 18:37, 3 September 2017
Problem
Trapezoid has bases and and diagonals intersecting at . Suppose that , , and the area of is . What is the area of trapezoid ?
Solution
unitsize=5cm; pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */ pen sm = fontsize(10); /* small font pen */ pair D=(0,0),C=(12,0), K=(7,16/3); /* note that K.x is arbitrary, as generator for A,B */ pair A=7*K/4-3*C/4, B=7*K/4-3*D/4; D(MP("A",A,N)--MP("B",B,N)--MP("C",C)--MP("D",D)--A--C);D(B--D);D(A--MP("K",K)--D--cycle,linewidth(0.7)); MP("9",(A+B)/2,N,sm);MP("12",(C+D)/2,sm);MP("24",(A+D)/2+(1,0),E); (Error compiling LaTeX. 3175d9566780b2a4b407ee7999b85dac0fe3c66c.asy: 3.9: cannot convert 'real' to 'void(picture pic=<default>, real x, real y=<default>, real z=<default>)' in assignment)
Since it follows that . Thus .
We now introduce the concept of area ratios: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since share a common altitude to , it follows that (we let denote the area of the triangle) , so . Similarly, we find and .
Therefore, the area of .
See also
2008 AMC 10A (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.