Question 986115
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We are given a right-angled triangle &nbsp;<B>ABC</B>&nbsp; (see the <B>Figure</B>)&nbsp; with the legs &nbsp;|<B>AB</B>| = 8 &nbsp;and &nbsp;|<B>AC</B>| = 16.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

A square &nbsp;<B>ADEF</B>&nbsp; is inscribed to the triangle &nbsp;<B>ABC</B>&nbsp; in such a way that the triangle and 

the square have the common right angle &nbsp;<I>L</I><B>A</B>.

We need to find the length of the square side. 


Let <B>x</B> be the square side length. 

Then the length of the segment &nbsp;<B>DB</B>&nbsp; is &nbsp;8 - x&nbsp; and the length of the segment &nbsp;<B>CF</B>&nbsp; is &nbsp;16 - x. 


The triangles &nbsp;{{{DELTA}}}<B>DBE</B>&nbsp; and  &nbsp;{{{DELTA}}}<B>FEC</B>&nbsp; are right angled triangles. 

They have congruent acute angles &nbsp;<I>L</I><B>FCE</B>&nbsp; and &nbsp;<I>L</I><B>DEB</B>&nbsp; as these angles are the corresponding 

angles at the parallel lines &nbsp;<B>AC</B>&nbsp; and &nbsp;<B>DE</B>&nbsp; and the transverse &nbsp;<B>BC</B>. 


Hence, the triangles &nbsp;{{{DELTA}}}<B>DBE</B>&nbsp; and  &nbsp;{{{DELTA}}}<B>FEC</B>&nbsp; are similar right-angled triangles. 

Therefore, &nbsp;their legs are proportional: 

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{{{drawing( 200, 250, 0.5, 4.5, 0.5, 5.5, 
            line( 1, 1, 4, 1), 
            line( 1, 1, 1, 5),
            line( 4, 1, 1, 5), 
       blue(line( 1, 2.714, 2.714, 2.714)), 
       blue(line( 2.714, 1, 2.714, 2.714)), 

            locate ( 0.9,  1,    A),
            locate ( 4.0,  1,    B),
            locate ( 0.9,  5.3,  C),
            locate ( 0.78, 2.95, F),
            locate ( 2.8,  2.95, E),
            locate ( 2.65, 1.0,  D)
)}}}


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure</B> 

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{{{abs(CF)/abs(FE)}}} = {{{abs(ED)/abs(DB)}}}, &nbsp;&nbsp;&nbsp;&nbsp;or&nbsp;&nbsp;&nbsp;&nbsp; {{{(16-x)/x}}} = {{{x/(8-x)}}}.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3) 


Now we need to solve the last equation for the unknown &nbsp;<B>x</B>. &nbsp;Simplify the equation &nbsp;(3)&nbsp; step by step: 


{{{(16-x)*(8-x)}}} = {{{x^2}}},


{{{16*8 - 16x - 8x + x^2}}} = {{{x^2}}}, 


{{{128 - 24x + x^2}}} = {{{x^2}}}, 


{{{24x}}} = {{{128}}}, &nbsp;&nbsp;{{{x}}} = {{{128/24}}} = {{{16/3}}} = {{{5}}}{{{1/3}}}. 


<B>Answer</B>. &nbsp;The side measure of the square is of &nbsp;{{{5}}}{{{1/3}}}&nbsp; m long. 


For a similar problem see the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Triangles/Miscellaneous-problems-on-similar-triangles.lesson>Miscellaneous problems on similar triangles</A>&nbsp; in this site.