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# help

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In the diagram, AB = 13 cm, DC = 20 cm, and AD = 5 cm. What is the length of AC, to the nearest tenth of a centimeter? Jan 2, 2019

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By the Pythagorean Theorem, BD = √ [ AB^2 - AD^2]  =√  [ 13^2 - 5^2] = √[ 169 - 25] = √144 = 12

And, again, by the Pythagorean Theorem,

BC = √ [ DC^2 - BD^2 ] = √ [ 20^2 - 12^2 ] = √ [ 400 - 144 ] = √256  =  16

Call the intersection of BD and AC , E

Angle AED  = Angle CEB   [ vertical angles ]

Angle ADE = Angle CBE   [ right angles ]

Therefore ....triangle AED is similar to triangle CEB

And

AD/BC = DE /BE

5/16 =  DE/BE

Therefore....there are 21 equal parts of BD...and BE is 16 of these...and DE is 5 of these....

So BE =  (16/21) *BD = (16/21)* 12 =  64/7

And DE = (5/21)* BD = (5/21)  *12 = 60/21 = 20/ 7

And by the Pythagorean Theorem

EC = √ [BE^2 + BC^2]  = √ [ (64/7)^2 + 16^2 ]

And

AE = √[ AD^2 + DE^2 ] = √ [ 5^2 + (20/7)^2 ]

So

AC  = AE  + EC  =  √ [ (64/7)^2 + 16^2 ] + √ [ 5^2 + (20/7)^2 ]   ≈  24.2 cm   Jan 2, 2019
edited by CPhill  Jan 2, 2019