Question 1151146
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<pre>

Triangle ABC is a right-angled triangle.


Its leg AB is  {{{sqrt(25^2-7^2)}}} = 24 meters long.


Its area is  {{{(1/2)*7*24}}} = 7*12 = 84 square meters.


Triangles ABC and EDC are similar (since they both are right-angled and have common acute angle C).


Hence, their corresponding sides are proportional.


In particular,  EC = k*AC = 24k,  ED = k*AB = 7k,  where "k" is the proportionality coefficient.


Then the area of the triangle EDC is  {{{(1/2)*abs(EC)*abs(ED)}}} = {{{(1/2)*24k*7k}}} = {{{84k^2}}}.


It is half the area of the triangle ABC, i.e.

     {{{84k^2}}} = {{{84/2}}} = 42,   or


     {{{k^2}}} = {{{42/84}}} = {{{1/2}}}.


Therefore, the proportionality coefficient (the similarity coefficient)

     k = {{{sqrt(1/2)}}} = {{{sqrt(2)/2}}}.


It implies  EC = k*AC = {{{(sqrt(2)/2)*24}}} = {{{12*sqrt(2)}}} = 16.971 (approximately).    <U>ANSWER</U>
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Solved.