2005 AMC 10A Problems/Problem 25
Problem
In we have , , and . Points and are on and respectively, with and . What is the ratio of the area of triangle to the area of the quadrilateral ?
Solution 1(no trig)
We have that
But , so
Solution 2(no trig)
We can let [ADE]=x. Since EC=$2\cdotEC$ (Error compiling LaTeX. ! Undefined control sequence.), [DEC]=2x. So, [ADC]=3x. This means that [BDC]=$\frac{6}{19}\cdot3x=\frac{18x}{19}. Thus,
Solution (trig)
Using this formula:
Since the area of is equal to the area of minus the area of ,
.
Therefore, the desired ratio is
Note: was not used in this problem
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
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