Difference between revisions of "2018 AMC 8 Problems/Problem 22"
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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>. | ||
==Solution 3== | ==Solution 3== | ||
− | Note that triangle <math>ABC</math> has half the area of the square and triangle <math>FEC</math> has <math>\dfrac1{12}</math>th. | + | Note that triangle <math>ABC</math> has half the area of the square and triangle <math>FEC</math> has <math>\dfrac1{12}</math>th. Thus the area of the quadrilateral is <math>1-1/2-1/12=5/12</math>th the area of the square. The area of the square is then <math>45\cdot\dfrac{12}{5}=\boxed{\textbf{(B)}108}</math>. |
=See Also= | =See Also= |
Revision as of 12:28, 21 November 2019
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 .
Solution 3
Note that triangle has half the area of the square and triangle has th. Thus the area of the quadrilateral is th the area 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.