Difference between revisions of "2008 AMC 10A Problems/Problem 20"
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− | + | ==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>? | ||
+ | |||
+ | <math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math> | ||
+ | |||
+ | ==Solution== | ||
+ | ==Solution 1== | ||
+ | <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>. | ||
+ | ==Solution 2== | ||
+ | We denote <math>KA</math> with length <math>x</math> and <math>KD</math> with length <math>\frac{4x}{3}</math> (which follows from similar triangles), and we denote <math>\angle{AKD}=\theta</math>. Note that <math>\frac{4x^2}{3}\cdot \sin\theta=48\implies 4x^2\cdot \sin\theta=36</math>. The areas of triangles <math>ABK</math> and <math>CDK</math> combined are <math>\frac{x^2\cdot\sin\theta+\frac{16x^2}{9}\cdot\sin\theta}{2}=\frac{25x^2}{18}\cdot\sin\theta=36\cdot\frac{25}{18}=50</math>. Thus, <math>[ABCD]=[ABK]+[BCK]+[CDK]+[ADK]=48+50=98\ \mathrm{(D)}</math>, as desired. | ||
+ | -mop | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Revision as of 02:53, 27 January 2022
Problem
Trapezoid has bases and and diagonals intersecting at Suppose that , , and the area of is What is the area of trapezoid ?
Solution
Solution 1
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 .
Solution 2
We denote with length and with length (which follows from similar triangles), and we denote . Note that . The areas of triangles and combined are . Thus, , as desired. -mop
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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