Difference between revisions of "2008 AMC 10A Problems/Problem 20"
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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> | <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== | ==Solution 1== | ||
<center><asy> | <center><asy> | ||
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 */ | ||
− | pair D=(0,0),C=( | + | 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; | 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)); | 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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Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>. | Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>. | ||
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==Solution 2== | ==Solution 2== | ||
− | We | + | 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. <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> | ||
==See also== | ==See also== |
Latest revision as of 09:25, 9 August 2024
Contents
Problem
Trapezoid has bases
and
and diagonals intersecting at
Suppose that
,
, and the area of
is
What is the area of trapezoid
?
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]](http://latex.artofproblemsolving.com/a/e/9/ae9e6780ac0e2f899814f783c0dadde9dfd16a40.png)
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 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. for line DB,
for line AC. Solving for K,
simplifying,
,
. Using the fact that
, we solve for h.
. Applying trapezoid area formula:
. Thus, the area is 98 and the answer is
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.