Difference between revisions of "2008 AMC 10A Problems/Problem 20"
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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>. | ||
==Solution 2== | ==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</math>, as desired. | + | 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== | ==See also== |
Revision as of 01:53, 27 January 2022
Contents
[hide]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
![[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]](http://latex.artofproblemsolving.com/a/8/5/a8512170c7bb9c8c44a4195790890290d8352b64.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 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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