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Difference between revisions of "2008 AMC 10A Problems/Problem 20"

(New page: Draw isoceles trapezoid ABCD. Draw diagonals intersecting at K. Obviously, AKD is congruent to BKC. Likelywise, AKB is similar to CKD. Call the altitude to base AB of triangle AKB 3x and t...)
 
 
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Draw isoceles trapezoid ABCD. Draw diagonals intersecting at K. Obviously, AKD is congruent to BKC. Likelywise, AKB is similar to CKD. Call the altitude to base AB of triangle AKB 3x and the altitude to base CD of triangle CKD 4x. Now the area of the trapezoid is the sum of areas AKB, CKD, AKD, and BKC which equals . Now take isosceles trapezoid ABCD again. Drop altitudes to the base CD from point A and point B to get a rectangle and two triangles. The area of the rectangle is and the area of the two triangles is as the lengths of the legs are 1.5 and 7x. So now you have  and simple arithmetic gives you . .
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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>\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 1==
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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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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==
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{{AMC10 box|year=2008|ab=A|num-b=19|num-a=21}}
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[[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
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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