Difference between revisions of "2018 AMC 8 Problems/Problem 22"
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Let the area of <math>\triangle CEF</math> be <math>x</math>. Thus, the area of triangle <math>\triangle ACD</math> is <math>45+x</math> and the area of the square is <math>2(45+x) = 90+2x</math>. | Let the area of <math>\triangle CEF</math> be <math>x</math>. Thus, the area of triangle <math>\triangle ACD</math> is <math>45+x</math> and the area of the square is <math>2(45+x) = 90+2x</math>. | ||
− | By | + | By AA similarity, <math>\triangle CEF \sim \triangle ABF</math> with a 1:2 ratio, so the area of triangle <math>\triangle ABF</math> is <math>4x</math>. Now consider trapezoid <math>ABED</math>. Its area is <math>45+4x</math>, which is three-fourths the area of the square. We set up an equation in <math>x</math>: |
<cmath> 45+4x = \frac{3}{4}\left(90+2x\right) </cmath> | <cmath> 45+4x = \frac{3}{4}\left(90+2x\right) </cmath> | ||
Solving, we get <math>x = 9</math>. The area of square <math>ABCD</math> is <math>90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}</math>. | Solving, we get <math>x = 9</math>. The area of square <math>ABCD</math> is <math>90+2x = 90 + 2 \cdot 9 = \boxed{\textbf{(B)} 108}</math>. | ||
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==Solution 2== | ==Solution 2== |
Revision as of 14:51, 15 July 2019
Contents
Problem 22
Point is the midpoint of side in square and meets diagonal at The area of quadrilateral is What is the area of
Solution 1
Let the area of be . Thus, the area of triangle is and the area of the square is .
By AA similarity, with a 1:2 ratio, so the area of triangle is . Now consider trapezoid . Its area is , which is three-fourths the area of the square. We set up an equation in :
Solving, we get . The area of square is .
Solution 2
We can use analytic geometry for this problem.
Let us start by giving the coordinate , the coordinate , and so forth. and can be represented by the equations and , respectively. Solving for their intersection gives point coordinates .
Now, ’s area is simply or . This means that pentagon ’s area is of the entire square, and it follows that quadrilateral ’s area is of the square.
The area of the square is then .
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AJHSME/AMC 8 Problems and Solutions |
Set s to be the bottom left triangle. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.