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# pls help

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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.

Apr 21, 2020

#1
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If you draw the square with vertices:  (a, a)     (-a, a)     (a, -a)     (-a, a)

and the line  y  =  x/2  this line will pass through the points  (a, a/2)   and  (-a, -a/2).

Consider the quadrilateral whose enpoints are (a, a/2)    (a, a)     (-a, a)     (-a, -a/2).

The distance from  (a, a/2)  to  (a, a)  is  a/2.

The distance from  (a, a)  to  (-a, a)  is  2a.

The distance from  (-a, a)  to (-a, -a/2)  is  3/2·a

To find the distance from  (-a, -a/2)  to  (a, a/2)  we can use the distance formula:

distance  =  sqrt[ (a - a)2 + (a/2 - -a/2)2 ]  =  sqrt[ (2a)2 + (a)2 ]  =  sqrt[ 4a2 + a2 ]  =  sqrt( 5a2 )  =  a·sqrt(5).

Adding these togerther:  a/2 + 2a + 3/2·a + a·sqrt(5)  =  4a + a·sqrt(5)

Dividing this by  a  gives  4 + sqrt(5).

Apr 21, 2020
#2
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thank u so much!

Guest Apr 21, 2020