SOLUTION: In an acute triangle ABC, the altitudes AD and CE are drawn. Find the length of the line segment DE if AB = 15 cm, BC = 18 cm, and AD = 10 cm.

Algebra ->  Triangles -> SOLUTION: In an acute triangle ABC, the altitudes AD and CE are drawn. Find the length of the line segment DE if AB = 15 cm, BC = 18 cm, and AD = 10 cm.      Log On


   

Question 1080661: In an acute triangle ABC, the altitudes AD and CE are drawn. Find the length of the line segment DE if AB = 15 cm, BC = 18 cm, and AD = 10 cm.
Answer by ikleyn(52794) About Me  (Show Source):
You can put this solution on YOUR website!
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In an acute triangle ABC, the altitudes AD and CE are drawn.
Find the length of the line segment DE if AB = 15 cm, BC = 18 cm, and AD = 10 cm.
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0.  Make a sketch to follow my arguments.


1.  The area of the triangle ABC is

    S = %281%2F2%29%2Aabs%28AB%29%2Aabs%28CE%29 = %281%2F2%29%2Aabs%28BC%29%2Aabs%28AD%29, 


    which gives you an equation 

    15*|CE| = 18*10.

    It implies |CE| = %2818%2A10%29%2F15 = 12 cm.


2.  Then from the right-angled triangle BCE  |BE| = sqrt%28abs%28BC%29%5E2-abs%28CE%29%5E2%29 = sqrt%2818%5E2-12%5E2%29 = sqrt%28180%29.


3.  From the right-angled triangle ABD  |BD| = sqrt%28abs%28AB%29%5E2-abs%28AD%29%5E2%29 = sqrt%2815%5E2-12%5E2%29 = sqrt%2881%29 = 9.


4.  From the right-angled triangle ABD  cos(B) = abs%28BD%29%2Fabs%28AB%29 = 9%2F15 = 3%2F5.


5.  Now, to answer the problem's question, apply the cosine law:


    abs%28DE%29%5E2 = abs%28BE%29%5E2+%2B+abs%28BD%29%5E2+-+2%2Aabs%28BE%29%2Aabs%28BD%29%2Acos%28B%29
 = %28sqrt%28180%29%29%5E2+%2B+9%5E2+-+2%2Asqrt%28180%29%2A9%2A%283%2F5%29 = 180+%2B+81+-+2%2Asqrt%28180%29%2A9%2A%283%2F5%29 = 261-%282%2A2%2A3%2A9%2A3%2Asqrt%285%29%29%2F5 = 261-%28324%2Asqrt%285%29%29%2F5.


    Hense,  |DE| = sqrt%28261-%28324%2Asqrt%285%29%29%2F5%29 = 10.78 (approximately).

Solved.