Difference between revisions of "2018 AMC 8 Problems/Problem 22"
(→Solution 2) |
(→Solution 1.5) |
||
Line 25: | Line 25: | ||
==Solution 1.5== | ==Solution 1.5== | ||
− | We notice some similar triangles. | + | We notice some similar triangles. This is good, but you want to know the ratios in order to begin using them. Say that 45 is the area of half the triangle subtracted by <math>a</math>, so then the area of the whole square would be <math>90+2a</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 09:33, 23 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 1
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
Solution 1.5
We notice some similar triangles. This is good, but you want to know the ratios in order to begin using them. Say that 45 is the area of half the triangle subtracted by , so then the area of the whole square would be .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.