2011 UNCO Math Contest II Problems/Problem 3
Problem
The two congruent rectangles shown have dimensions in. by in. What is the area of the shaded overlap region?
Solution
[asy] filldraw((0,0)--(13,0)--(25,5)--(12,5)--cycle,blue); pair A1=(0,0),B1=(25,0),C1=(25,5),D1=(0,5); draw(A1--B1--C1--D1--cycle,black); pair P,R; P=unit(A1-D1)+D1; R=unit(C1-D1)+D1; draw(P--(P+R-D1)--R,black); pair A2=(0,0),B2=(25/13,-60/13),C2=(25,5),D2=(25-25/13,5+60/13); draw(A2--B2--C2--D2--cycle,black); P=unit(A2-D2)+D2; R=unit(C2-D2)+D2; draw(P--(P+R-D2)--R,black); label("A",(0,0),SW); label("B",(0,5),NW); label("M",(12,5),NW); label("C",(25,5),NE); label("D",(25,0),SE); label("N",(13,0),SE); label("O",(25/13,-60/13),S); label("P",(25-25/13,5+60/13); [/asy] We notice that triangles and are congruent, since they share two angles and a side (a right angle, and the opposite angle, and the side 5) This means that if we label then and and so since and are congruent, then and so solving for we get 12, meaning that the area of the two triangles and (which are also congruent since the is parallel to ) is 30, so the total area of them is 60, and so taking this away from rectangle we get the blue, so so the answer is 65.
See Also
2011 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |