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The square with vertices $(-a, -a), (a, -a), (-a, a), (a, a)$ is cut by the line $y = x/2$ into congruent quadrilaterals. The perimeter of one of these congruent quadrilaterals divided by $a$ equals what? Express your answer in simplified radical form.

michaelcai  Oct 31, 2017
 #1
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y  =  x/2

 

This line will  intersect the square at x values  -a  and a

 

So....the associated y values are

 

y =  (-a)/2  =  -a/2     and   (a)/2  

 

So....the perimeter of one of the quadrilaterals =

 

2a  +  [(a) -  (-a/2)] + [(a) - a/2 ]  +  sqrt  [ ( a - (-a)) ^2 + ( a/2 - (-a/2)^2]  =

 

2a  + 3a/2  + a/2  +  sqrt  [ 4a^2  + a^2 ]  =

 

4a  +  a√ 5    dividing this by a  produces  =

 

4  + √ 5

 

 

cool cool cool

CPhill  Oct 31, 2017

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