Question 1205024
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In a right angle triangle ABC side AC is 4cm shorter than the hypotenuse and side BC 
is also 4cm shorter than the hypotenuse. Find the dimensions of the triangle
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<pre>
x^2 + x^2 = (x+4)^2    <<<---=== Pythagorea equation;  x is the common length of any of the two its legs.

2x^2 = x^2 + 8x + 16

x^2 - 8x - 16 = 0


x = {{{(8 +- sqrt((-8)^2 - 4*1*(-16)))/2}}} = {{{(8 +- sqrt(128))/2}}} = {{{(8 +- 8*sqrt(2))/2}}} = {{{4 +- 4*sqrt(2)}}}.


Only positive root is the solution to the problem.


<U>ANSWER</U>.  The triangle is isosceles right angled triangle.

         Its legs are {{{4*sqrt(2)+4}}}  cm long.  Its hypotenuse is  {{{(4*sqrt(2)+4)*sqrt(2)}}} = {{{8+4*sqrt(2)}}} cm long.
</pre>

Solved.


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<H3>Warning to a reader</H3>

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solution by @mananth, giving two possible solutions for the hypotenuse, is NOT precisely correct.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Only one (greater) of the two values of the hypotenuse length is admittable; 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;the second (lesser) value is NOT admittable, since it leads to negative value of the leg length.