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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.

 May 2, 2022
edited by Guest  May 2, 2022
 #1
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The answer is 3 + sqrt(6).

 May 2, 2022
 #2
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The perimeter of the trapezoid is the sum of the sides + the sum of the line cutting the square. 

 

The length of the sides (not including the line) is \(4a\). One of the sides is 2a (labeled a in the diagram), and the other two sides (b and c in the diagram) will always add to 2a. 

 

We can then use the Pythagorean Theorem for the length of the bisecting line, and we find that its length is \(a \sqrt 5 \)

 

Thus, the perimeter of the shape is \(a(4 + \sqrt 5)\). Can you do the rest?

 

Diagram:

 

 

 May 3, 2022
edited by BuilderBoi  May 3, 2022
edited by BuilderBoi  May 3, 2022
edited by BuilderBoi  May 3, 2022
 #3
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Thank you, but I am a bit confused.

Guest May 3, 2022
 #4
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Would the answer then be 4+sqrt5?

Guest May 3, 2022
 #5
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Yes, that is correct!!!

BuilderBoi  May 3, 2022
 #6
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Thank you! I couldn't have done that without you. 

 

I have revised your solution and I finally understand.smiley

Guest May 3, 2022
 #7
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Aww... Thank you!!!

BuilderBoi  May 3, 2022

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