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| == Solution == | | == Solution == |
| <math>65</math> | | <math>65</math> |
− | [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);
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− | 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);
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− | 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);
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− | P=unit(A2-D2)+D2; R=unit(C2-D2)+D2;
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− | draw(P--(P+R-D2)--R,black);
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− | label("A",(0,0),SW);
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− | label("B",(0,5),NW);
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− | label("M",(12,5),NW);
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− | label("C",(25,5),NE);
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− | label("D",(25,0),SE);
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− | label("N",(13,0),SE);
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− | label("O",(25/13,-60/13),S);
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− | label("P",(25-25/13,5+60/13);
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− | [/asy]
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− | We notice that triangles <math>AON</math> and <math>NDC</math> 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 <math>ON</math> <math>x</math> then <math>NC=25-x</math> and <math>AN^2=5^2+x^2</math> and so since <math>AON</math> and <math>NDC</math> are congruent, then <math>AN=NC</math> and so <math>(25-x)^2=5^2+x^2</math> solving for <math>x</math> we get 12, meaning that the area of the two triangles <math>AON</math> and <math>MPC</math> (which are also congruent since the <math>AD</math> is parallel to <math>BC</math>) is 30, so the total area of them is 60, and so taking this away from rectangle <math>AOCP</math> we get the blue, so <math>125-60=65</math> so the answer is 65.
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| == See Also == | | == See Also == |