2008 AMC 10A Problems/Problem 20
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 ![$[\triangle \ldots]$](http://latex.artofproblemsolving.com/1/8/4/18412a965332198f2b2d5c0bb1c491330047d92a.png)
denote the area of the triangle) ![$\frac{[\triangle AKB]}{[\triangle AKD]} = \frac{KB}{KD} = \frac{3}{4}$](http://latex.artofproblemsolving.com/e/e/d/eed58cd07515b1733fcc6a5b95ea36fb385b7c10.png)
, so ![$[\triangle AKB] = \frac{3}{4}(24) = 18$](http://latex.artofproblemsolving.com/0/1/c/01c8718ee2a2aa3bd4eb6e803892547b8d61fa1b.png)
. Similarly, we find ![$[\triangle DKC] = \frac{4}{3}(24) = 32$](http://latex.artofproblemsolving.com/e/d/4/ed41cd2ecadd00ff045b470f1e65bf28da3d6dfe.png)
and ![$[\triangle BKC] = 24$](http://latex.artofproblemsolving.com/e/7/c/e7ce9a8758dfdea64da26bdff894339c1379bfe7.png)
.
Therefore, the area of ![$ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}$](http://latex.artofproblemsolving.com/e/2/8/e2891d9d69739764a5a6e2042414a91fa836d880.png)
.
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 | ||
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