SOLUTION: In the diagram, AC is the median of triangle ABD. Find the area of the triangle ABD. Diagram link: https://imgur.com/GkL6y3v

Algebra ->  Triangles -> SOLUTION: In the diagram, AC is the median of triangle ABD. Find the area of the triangle ABD. Diagram link: https://imgur.com/GkL6y3v      Log On


   

Question 1149735: In the diagram, AC is the median of triangle ABD. Find the area of the triangle ABD.
Diagram link: https://imgur.com/GkL6y3v
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!


Use the law of cosines on triangles ABC and ABD

cos%28B%29=%28a%5E2%2Bc%5E2-b%5E2%29%2F%282ac%29

Using it on ABC

cos%28B%29=%2817%5E2%2Bx%5E2-17%5E2%29%2F%282%2A17%2Ax%29

which simplifies to

cos%28B%29=x%2F34

Using it on ABD

cos%28B%29=%2817%5E2%2B%282x%29%5E2-27%5E2%29%2F%282%2A17%2A2x%29

which simplifies to

%28x%5E2-110%29%2F17x

Equating the two expressions for cos(B),


x%2F34=%28x%5E2-110%29%2F17x

Solve that and get

x=2sqrt%2855%29

The height of both triangles is the green line h = AE:



AE%5E2=AB%5E2-BE%5E2
AE%5E2=17%5E2-%28sqrt%2855%29%2F4%29%5E2
AE%5E2=234
AE=3sqrt%2826%29

Area=expr%281%2F2%29base%2Aheight

Area=expr%281%2F2%294sqrt%2855%29%2A3sqrt%2826%29

Area=6sqrt%281430%29

approx. 226.8920448

Edwin

Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.

I will borrow the picture from the Edwin' solution to produce my own, much more simple. 




The triangle ABC is isosceles, therefore, its altitude AE is the median, at the same time.


So, I introduce the variable y = x/2.


Then BE = EC = y;  CD = x = 2y;  ED = 3y;  BD = 4y.


Therefore,

    h^2 = 17^2 - y^2 = 27^2 - (3y)^2,   or

    17 - y^2 = 27^2 - 9y^2

    9y^2 - y^2 = 27^2 - 17^2

    8y^2       = (27-17)*(27+17) = 10*44 = 440

     y^2                                 = 440/8 = 55

     y                                   = sqrt%2855%29.


Then   h^2 = 17^2 - 55 = 234  and  h = sqrt%28234%29 = sqrt%289%2A26%29 = 3%2Asqrt%2826%29.


Now the area of the triangle ABD is 


     area = %281%2F2%29%2ABD%2AAE%29 = %281%2F2%29%2A4y%2Ah = 2y*h = 2%2Asqrt%2855%29%2A3%2Asqrt%2826%29 = 6%2Asqrt%281430%29 = 226.89  (approx.)

Solved.