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Difference between revisions of "2008 AMC 10A Problems/Problem 20"
 
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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>?
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==Problem==
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[[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>?
  
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<math>\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 94 \qquad \textbf{(C)}\ 96 \qquad \textbf{(D)}\ 98 \qquad \textbf{(E)}\ 100</math>
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<math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math>
  
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=== Solution 2 ===
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==Solution 1==
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{{solution}}
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<center><asy>
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pointpen = black; pathpen = black + linewidth(0.62);  /* cse5 */
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pen sm = fontsize(10);            /* small font pen */
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pair D=(0,0),C=(6,0), K=(3.5,8/3); /* note that K.x is arbitrary, as generator for A,B */
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pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;
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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));
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MP("9",(A+B)/2,N,sm);MP("12",(C+D)/2,sm);MP("24",(A+D)/2+(1,0),E);
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</asy></center>
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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>.
  
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== See also ==
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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>.
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Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>.
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==Solution 2==
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We may consider that trapezoid to be right, as there is nothing specifying its angles.
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Consider D and A right. Let the length of DA be h. Now we let A be (0,0) and we compute the x-coordinate of K from lines AC and DB. <math>y=\frac{h}{9}x</math> for line DB, <math>y=-\frac{h}{12}x+h</math> for line AC. Solving for K, <math>\frac{h}{9}x=-\frac{h}{12}x+h</math> simplifying, <math>(\frac{1}{9}+\frac{1}{12})x=1</math>, <math>x=\frac{72}{14}=\frac{36}{7}</math>. Using the fact that <math>\frac{1}{2}*h*x=24</math>, we solve for h. <math>\frac{1}{2}*\frac{36}{7}*h=24, h=\frac{7}{18}*24=\frac{7*4}{3}</math>. Applying trapezoid area formula: <math>\frac{7*4}{3}*\frac{9+12}{2}=7*7*2=98</math>. Thus, the area is 98 and the answer is <math>\mathrm{(D)}</math>
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==See also==
 
{{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}
 
{{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}
  
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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[[Category:Triangle Area Ratio Problems]]
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{{MAA Notice}}

Latest revision as of 13:39, 13 October 2024

Problem

Trapezoid $ABCD$ has bases $\overline{AB}$ and $\overline{CD}$ and diagonals intersecting at $K.$ Suppose that $AB = 9$, $DC = 12$, and the area of $\triangle AKD$ is $24.$ What is the area of trapezoid $ABCD$?

$\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100$

Solution 1

[asy] pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */ pen sm = fontsize(10); /* small font pen */ pair D=(0,0),C=(6,0), K=(3.5,8/3); /* 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]

Since $\overline{AB} \parallel \overline{DC}$ it follows that $\triangle ABK \sim \triangle CDK$. Thus $\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}$.

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 $\triangle AKB, \triangle AKD$ share a common altitude to $\overline{BD}$, it follows that (we let $[\triangle \ldots]$ denote the area of the triangle) $\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}$, so $[\triangle AKB] = \frac{3}{4}(24) = 18$. Similarly, we find $[\triangle DKC] = \frac{4}{3}(24) = 32$ and $[\triangle BKC] = 24$.

Therefore, the area of $ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}$.

Solution 2

We may consider that trapezoid to be right, as there is nothing specifying its angles. Consider D and A right. Let the length of DA be h. Now we let A be (0,0) and we compute the x-coordinate of K from lines AC and DB. $y=\frac{h}{9}x$ for line DB, $y=-\frac{h}{12}x+h$ for line AC. Solving for K, $\frac{h}{9}x=-\frac{h}{12}x+h$ simplifying, $(\frac{1}{9}+\frac{1}{12})x=1$, $x=\frac{72}{14}=\frac{36}{7}$. Using the fact that $\frac{1}{2}*h*x=24$, we solve for h. $\frac{1}{2}*\frac{36}{7}*h=24, h=\frac{7}{18}*24=\frac{7*4}{3}$. Applying trapezoid area formula: $\frac{7*4}{3}*\frac{9+12}{2}=7*7*2=98$. Thus, the area is 98 and the answer is $\mathrm{(D)}$

See also

2008 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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All AMC 10 Problems and Solutions

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