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abcd is a square .from the diagonal bd, a length bx is cut off equal toba.from x, a straight line xy is drawn perpendicular to bd to meet ad at y.then ab+ay=

 Aug 23, 2015

Best Answer 

 #1
avatar+129852 
+15

Call the side of the square S.....

 

Using the Law of Cosines, we have

 

AX^2 = 2S^2 -2S^2cos(45)   =  2S^2  - 2S^2(1/√2)  =  S^2 [ 2 - √2]

 

So........AX =  S*√[ 2 - √2]

 

And using some basic geometry <YXD = 90  and <XDA = 45......so <XYD = 45....so <AYX = [180- <XYD]= 135

 

And since AB = BX and <ABD = 45, then <AXB  = (180 - 45]/2 = 67.5

 

Then <AXY = [180 - 90 - 67.5] = 22.5

 

And using the  Law of Sines, again, we have

 

AY/sin(22.5) = AX/sin(135)

 

AY = sin(22.5)/sin(135)* S*√[ 2 - √2] = [√[1-1/√2] / √2] * √2* √[2 - √2]S  = [√[1-1/√2]*√[2 - √2]*S  = [ √[√2 -1] * √[2 - √2] / √2]*S =[√ [2√2 - √2 - 2 +√2] / √2]*S  = [ √[2√2 -2]/ √2]*S= [√[(2)(√2 -1 )] / √2] *S   = [√2 - 1]*S

 

So.... AB + AY = S + [√2 - 1]S  = S [ 1 + [√2  - 1] ] S  = √2S

 

Her's an (aproximate) picture......

 

 

 

 Aug 23, 2015
 #1
avatar+129852 
+15
Best Answer

Call the side of the square S.....

 

Using the Law of Cosines, we have

 

AX^2 = 2S^2 -2S^2cos(45)   =  2S^2  - 2S^2(1/√2)  =  S^2 [ 2 - √2]

 

So........AX =  S*√[ 2 - √2]

 

And using some basic geometry <YXD = 90  and <XDA = 45......so <XYD = 45....so <AYX = [180- <XYD]= 135

 

And since AB = BX and <ABD = 45, then <AXB  = (180 - 45]/2 = 67.5

 

Then <AXY = [180 - 90 - 67.5] = 22.5

 

And using the  Law of Sines, again, we have

 

AY/sin(22.5) = AX/sin(135)

 

AY = sin(22.5)/sin(135)* S*√[ 2 - √2] = [√[1-1/√2] / √2] * √2* √[2 - √2]S  = [√[1-1/√2]*√[2 - √2]*S  = [ √[√2 -1] * √[2 - √2] / √2]*S =[√ [2√2 - √2 - 2 +√2] / √2]*S  = [ √[2√2 -2]/ √2]*S= [√[(2)(√2 -1 )] / √2] *S   = [√2 - 1]*S

 

So.... AB + AY = S + [√2 - 1]S  = S [ 1 + [√2  - 1] ] S  = √2S

 

Her's an (aproximate) picture......

 

 

 

CPhill Aug 23, 2015
 #2
avatar+118673 
+5

Nice work chris,  Your diagram looks good :)

 Aug 24, 2015

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