SOLUTION: 1. The triangle has an area of 2500 m2. The sides are in the ratio 5:10:11. Find the following a) Length of the longest side. b) Length of the median to the shortest side. c)

Algebra ->  Triangles -> SOLUTION: 1. The triangle has an area of 2500 m2. The sides are in the ratio 5:10:11. Find the following a) Length of the longest side. b) Length of the median to the shortest side. c)       Log On


   



Question 1185136: 1. The triangle has an area of 2500 m2. The sides are in the ratio 5:10:11. Find the following
a) Length of the longest side.
b) Length of the median to the shortest side.
c) Length of altitude to the longest side
d) Radius of the incircle)
e) Radius of the circumcircle

Answer by ikleyn(52786) About Me  (Show Source):
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The triangle has an area of 2500 m2. The sides are in the ratio 5:10:11. Find the following
a) Length of the longest side.
b) Length of the median to the shortest side.
c) Length of altitude to the longest side
d) Radius of the incircle)
e) Radius of the circumcircle
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From the condition, the sides are a = 5x, b = 10x and c = 11x units long.


Apply the Heron's formula for the area of a triangle.


Semi-perimeter is  s = %285x+%2B+10x+%2B+11x%29%2F2 = 13x.

                       s - a = 13x -  5x = 8x

                       s - b = 13x - 10x = 3x

                       s - c = 13x - 11x = 2x


According to the Heron's formula, the area of the triangle is


    area = sqrt%28s%2A%28s-a%29%2A%28s-b%29%2A%28s-c%29%29 = x%5E2%2Asqrt%2813%2A8%2A3%2A2%29 = 4x%5E2%2Asqrt%2839%29.


So, the "area" equation to find x is


    4x%5E2%2Asqrt%2839%29 = 2500.


From the equation,  x^2 = 625%2Fsqrt%2839%29;    x = 25%2Froot%284%2C39%29.


Hence, the sides are  a =  5x = 125%2Froot%284%2C39%29;  

                      b = 10x = 250%2Froot%284%2C39%29;

                      c = 11x = 275%2Froot%284%2C39%29.


The longest side is c = 11x = 275%2Froot%284%2C39%29.      <<<---===  It is the ANSWER to question (a).


Find the altitude  h%5Bc%5D drawn to this side from the area equation


     %281%2F2%29%2Ac%2Ah%5Bc%5D = 2500,  


which gives  


     h%5Bc%5D = 5000%2F%28%28275%2Froot%284%2C39%29%29%29 = %28200%2F11%29%2Aroot%284%2C39%29.      <<<---===  It is the ANSWER to question (c).



The radius of the incircle (inscribed circle) is

    r = area%2Fsemiperimeter = 2500%2F%2813x%29 = 2500%2F%2813%2A%2825%2Froot%284%2C39%29%29%29 = %28100%2F13%29%2Aroot%284%2C39%29.      <<<---===  It is the ANSWER to question (d).


The radius of the circumscribed circle is

    R = %28a%2Ab%2Ac%29%2F%284%2Aarea%29,


and you just have EVERYTHING from me to complete this calculation ON YOUR OWN.

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PLEASE  next time do not place so many assignments/questions into one post.

It  ALWAYS  works against your own interests.