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Two of the altitudes of an acute triangle divide the sides into segments of lengths 5,3,2 and x units, as shown. What is the value of x?

 

Maplesnowy  Apr 26, 2018
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We can equate the area of the the triangle on both sides of an equation

 

(1/2)*  (3 + 5) * sqrt [ (x + 2)^2 - 3^2 ]   =  (1/2) (x + 2) * sqrt  [ ( 8^2 - 2^2 ]

 

8 * sqrt [ x^2 + 4x + 4 - 9]  = (x +2 ) sqrt (60) 

 

8 * sqrt [ x^2 + 4x - 5]  = (x+ 2) sqrt (60)          square both sides

 

64 (x^2 + 4x  - 5 ]  =  (x^2 + 4x + 4 ) * 60

 

16(x^2 + 4x - 5)  = 15(x^2 + 4x + 4)

 

16x^2 + 64x - 80  =  15x^2 + 60x + 60

 

x^2 + 4x - 140  = 0       factor

 

(x - 10) ( x + 14)  = 0

 

Setting both factors to 0 and solving for x gives that x  = -14 (reject)  or  

 

x = 10

 

 

 

cool cool cool

CPhill  Apr 26, 2018

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