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In a pentagon ABCDE, EA = AB = BC = CD = 10cm, (angle)EAB = (angle)ECD = 90(degrees)...

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theeking | eNoter

Posted June 16, 2012 at 11:18 PM via web

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In a pentagon ABCDE, EA = AB = BC = CD = 10cm, (angle)EAB = (angle)ECD = 90(degrees) and (angle)ABC = 135(degrees). Find the length of ED.

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thilina-g | College Teacher | (Level 1) Educator

Posted June 17, 2012 at 12:07 AM (Answer #1)

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If you draw a diagram you would see that if we find EC length, we can find ED, since ECD is a right angled triangle.

To find EC we can draw a perpendicular line to EA from C. The point it cuts AC would be F. Now if you observe properly you would see that,

`EF + BC xx cos45 = EA`

`EF + 10 xx 1/sqrt(2) = 10`

`EF = 10 - 10/sqrt(2)`

`EF = 10(1-1/sqrt(2))`

 

Also CF is given by,

`CF = AB + BC xx sin45`

`CF = 10 + 10/sqrt(2)`

`CF = 10(1+1/sqrt(2))`

 

Now the triangle EFC is a right angled triangle,

`EC^2 = EF^2 + CF^2`

We also know from ECD triangle,

`ED^2 = CD^2 + EC^2`

`ED^2 = CD^2 + EF^2 + CF^2`

`ED^2 = 10^2 + (10(1-1/sqrt(2)))^2 + (10(1+1/sqrt(2)))^2`

`ED^2 = 10^2(1+1-2/sqrt(2) + 1/2+1+2/sqrt(2)+1/2)`

`ED^2 = 10^2 (1+1+1/2+1+1/2)`

`ED^2 = 10^2(4)`

`ED^2 = 400`

Therefore, `ED = sqrt(400)`

`ED = 20`

 

Therefore the length of ED is 20.

 

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