2009 UNCO Math Contest II Problems/Problem 9
Problem
A square is divided into three pieces of equal area by two parallel lines as shown. If the distance between the two parallel lines is what is the area of the square?
Solution
Let x be the length of a side. Then the square has area and each portion has area If x is the base of one of the triangles, then the height will be . By the pythaogrean theorem, longer side of the parallelogram has length Thus sqrt(13)*x/3*8 = x^2/3. Solving this gives x = 8*sqrt(13). Thus, the area of the square is 64*13 = 832.
Solution 2 (Better Explained)
In ADC, = because = (According to question)
= (AECF is a parallelogram and = )
= (same reason as before)
= * 8 * EC
= * EC * (Perpendicular from D to EC)
2 * * 8 * EC = * EC * (Perpendicular from D to EC)
(Perpendicular from D to EC) = 16
Let AE = , then ED = . So the side length of square is .
(Perpendicular from D to EC) =
EC = (By Pythagoras Theorem)
16 =
16 =
=
=
Area of square ABCD = = = 832
See also
2009 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |