+0

0
320
2
+238

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=

matsunnymat  Aug 23, 2015

#1
+80785
+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......

CPhill  Aug 23, 2015
Sort:

#1
+80785
+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......

CPhill  Aug 23, 2015
#2
+91409
+5

Nice work chris,  Your diagram looks good :)

Melody  Aug 24, 2015

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