Difference between revisions of "2018 AMC 8 Problems/Problem 22"
Virjoy2001 (talk | contribs) m (→Solution 1) |
Virjoy2001 (talk | contribs) m (→=Solution 1) |
||
Line 16: | Line 16: | ||
<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> | ||
− | + | =Solution 1= | |
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>. | ||
Line 32: | Line 32: | ||
The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}</math>. | The area of the square is then <math>\frac{45}{\frac{5}{12}}=9\cdot12=\boxed{\textbf{(B)}108}</math>. | ||
+ | |||
=See Also= | =See Also= | ||
{{AMC8 box|year=2018|num-b=21|num-a=23}} | {{AMC8 box|year=2018|num-b=21|num-a=23}} | ||
Set s to be the bottom left triangle. | Set s to be the bottom left triangle. | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:27, 20 August 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.