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?

Guest Jan 2, 2019

#1**+1 **

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

CPhill Jan 2, 2019