Question 1132035
.


            Another solution is possible based on totally different idea.



<pre>
Let the triangle be ABC with the base AB and the lateral sides AC and BC.


Let AD be the altitude of the triangle drawn from A to the lateral side BC.


Then DC is the leg in the right angle triangle ADB with the hypotenuse AB= 130 m and the other leg AD = 120 m - hence


    |DB| = {{{sqrt(130^2-120^2)}}} = 50 m.


Draw the altitude CE from vertex C to the base AB.


The triangles ADB and CEB are similar (since they are right-angled triangles with the common acute angle B). Hence,


    {{{abs(AD)/abs(BD)}}} = {{{abs(CE)/abs(BE)}}},  or  {{{120/50}}} = {{{h/65}}},


where  h is the altitude CE:  h = |CE|.  It gives for h


    h = {{{(120*65)/50}}} = 12*13  meters.


Now the area of the triangle  ABC  is   


    Area = {{{(1/2)*130*(12*13))}}} = 130*6*13 = 10140 square meters.    <U>ANSWER</U>
</pre>