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# Geometry

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In triangle ABC, AB =10 and AC =17. Let D be the foot of the perpendicular from A to BC. If BD:CD = 2:5, then find AD.

Jun 4, 2021

#1
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Let  BC  = x

Then   BD =  (2/7)x     and   DC =   (5/7)x

And

AB^2   -  BD^2   =    AD^2      ⇒   10^2  - [ (2/7)x]^2   =  AD^2       (1)

AC^2  -   CD^2  =    AD^2      ⇒    17^2  -  [(5/7)x^2]  = AD^2        (2)

Equate (1)   and (2)

10^2   -  [ (2/7)x]^2     =   17^2 - [ (5/7)x]^2

100   -  (4/49)x^2  =   289  -  (25/49)x^2

(25/49)x^2 - (4/49)x^2   =   289  -  100

(21/49)x^2  =  189

(3/7)x^2  = 189

x^2  =   189  ( 7/3)

x^2   =   441

x =  21

So

BD = (2/7) (21)  =   6

And   Jun 4, 2021
#2
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In triangle ABC, AB =10 and AC =17. Let D be the foot of the perpendicular from A to BC. If BD:CD = 2:5, then find AD.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

BC = x

BD = 2/7x

CD = 5/7x

sqrt[102 - (2/7x)2] = sqrt[172 - (5/7x)2]

x = 21

BD = 2/7 * 21 = 6

CD = 5/7 * 21 = 15

AD = sqrt(AB2 - BD2) = 8

Jun 4, 2021