Difference between revisions of "1991 AHSME Problems/Problem 23"
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\text{(E) } \frac{3}{5}</math> | \text{(E) } \frac{3}{5}</math> | ||
− | == Solution == | + | == Solution 1: Coordinate Geometry== |
− | <math>\ | + | Solution by e_power_pi_times_i |
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+ | First, we find out the coordinates of the vertices of quadrilateral <math>BEIH</math>, then use the Shoelace Theorem to solve for the area. Denote <math>B</math> as <math>(0,0)</math>. Then <math>E (0,1)</math>. Since I is the intersection between lines <math>DE</math> and <math>AF</math>, and since the equations of those lines are <math>y = \dfrac{1}{2}x + 1</math> and <math>y = -2x + 2</math>, <math>I (\dfrac{2}{5}, \dfrac{6}{5})</math>. Using the same method, the equation of line <math>BD</math> is <math>y = x</math>, so <math>H (\dfrac{2}{3}, \dfrac{2}{3})</math>. Using the Shoelace Theorem, the area of <math>BEIH</math> is <math>\dfrac{1}{2}\cdot\dfrac{14}{15} = \boxed{\textbf{(C) } \dfrac{7}{15}}</math>. | ||
== See also == | == See also == |
Revision as of 13:25, 14 December 2016
Problem
If is a square, is the midpoint of , is the midpoint of , and intersect at , and and intersect at , then the area of quadrilateral is
Solution 1: Coordinate Geometry
Solution by e_power_pi_times_i
First, we find out the coordinates of the vertices of quadrilateral , then use the Shoelace Theorem to solve for the area. Denote as . Then . Since I is the intersection between lines and , and since the equations of those lines are and , . Using the same method, the equation of line is , so . Using the Shoelace Theorem, the area of is .
See also
1991 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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