Difference between revisions of "2008 AMC 10A Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | Trapezoid <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K</math>. Suppose that <math>AB = 9</math>, <math>DC = 12</math>, and the area of <math>\triangle AKD</math> is <math>24</math>. What is the area of trapezoid <math>ABCD</math>? | + | [[Trapezoid]] <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K</math>. Suppose that <math>AB = 9</math>, <math>DC = 12</math>, and the area of <math>\triangle AKD</math> is <math>24</math>. What is the area of trapezoid <math>ABCD</math>? |
− | <math>\ | + | <math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math> |
==Solution== | ==Solution== | ||
− | {{ | + | <center><asy> |
+ | 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); | ||
+ | </asy></center> | ||
+ | Since <math>\overline{AB} \parallel \overline{DC}</math> it follows that <math>\triangle ABK \sim \triangle CDK</math>. Thus <math>\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}</math>. | ||
+ | |||
+ | 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 <math>\triangle AKB, \triangle AKD</math> share a common [[altitude]] to <math>\overline{BD}</math>, it follows that (we let <math>[\triangle \ldots]</math> denote the area of the triangle) <math>\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}</math>, so <math>[\triangle AKB] = \frac{3}{4}(24) = 18</math>. Similarly, we find <math>[\triangle DKC] = \frac{4}{3}(24) = 32</math> and <math>[\triangle BKC] = 24</math>. | ||
+ | |||
+ | Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>. | ||
==See also== | ==See also== | ||
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[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:38, 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
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 |
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