Difference between revisions of "2018 AMC 8 Problems/Problem 22"
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<math>\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144</math> | <math>\textbf{(A) } 100 \qquad \textbf{(B) } 108 \qquad \textbf{(C) } 120 \qquad \textbf{(D) } 135 \qquad \textbf{(E) } 144</math> | ||
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+ | ==Solution== | ||
+ | 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>. | ||
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+ | By AAA 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>: | ||
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+ | <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>. -scrabbler94 | ||
{{AMC8 box|year=2018|num-b=21|num-a=23}} | {{AMC8 box|year=2018|num-b=21|num-a=23}} |
Revision as of 17:22, 21 November 2018
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
Let the area of be . Thus, the area of triangle is and the area of the square is .
By AAA 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 . -scrabbler94
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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All AJHSME/AMC 8 Problems and Solutions |