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A pentagon is drawn by placing an isosceles right triangle on top of a square as pictured. What percent of the area of the pentagon is the area of the right triangle?

 May 7, 2015

Best Answer 

 #2
avatar+129852 
+11

Area of square = s^2

Area of isoceles right triangle  = (1/2)(s/√2)^2  = (1/2) [(s^2) / 2]  = (1/4)s^2

So...the area of the pentagon = s^2 + (1/4) s^2   = (5/4)s^2

So....the triangle is   [  (1/4)s^2 ] / [ (5/4)s^2 ]  = (1/4) / (5/4) = (1/4) (4/5) = 4/20 =   1/5 of the pentagon's area  = 20%

 

  

 May 7, 2015
 #1
avatar+1314 
+6

no need for explanation

 May 7, 2015
 #2
avatar+129852 
+11
Best Answer

Area of square = s^2

Area of isoceles right triangle  = (1/2)(s/√2)^2  = (1/2) [(s^2) / 2]  = (1/4)s^2

So...the area of the pentagon = s^2 + (1/4) s^2   = (5/4)s^2

So....the triangle is   [  (1/4)s^2 ] / [ (5/4)s^2 ]  = (1/4) / (5/4) = (1/4) (4/5) = 4/20 =   1/5 of the pentagon's area  = 20%

 

  

CPhill May 7, 2015

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