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I'm really struggling with this.  Can anyone help?

 

A square DEFG is inscribed in triangle ABC as shown. BC = 14, AB = 15, AC = 13.  Whats the length of the side of the square?

 

 Aug 8, 2020
 #1
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Let's start with the area of the triangle using Heron's formula. \(A = \sqrt{[s(s − a)(s − b)(s − c)]}\), where \(s = 1/2*(a+b+c)\) is the semi-perimeter of the triangle.

 

\(a = 14 \\b = 13 \\c = 15 \\s = 1/2*(14+13+15) = 21 \\A = \sqrt{[21(21-14)(21-13)(21-15)} \\A = 84\)

 

Now consider the height going from the top of the triangle, through the square and to the base of the traingle. Given that \({A}_{triangle}= 1/2*b*h\) and our base is 14, we get \(h = 2*A/b \\h = 2*84/14 \\h = 12\). Now we know that the upper triangle (above the square) is "similar" to the main triangle, because all of the angles are the same. We know this because the top angle is shared and because the bases of both triangles are parallel, we know that all of the angles and proportions of the triangle are the same. Then we can solve for the side length of the square through the following proportions.

 

\(h/{base}_{large} = (h-x)/{base}_{small} \\12/14 = (12-x)/x \\12x/14 = 12-x \\13x = 168 \\x = 168/13\)

 Aug 8, 2020
 #2
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Use the law of cosines to find angles B and C.

 

 ∠ B = 53.13°           ∠ C = 67.38°

     

x / tan(B) + x + x / tan(C) = 14   

 

x ≈ 6.461      

 Aug 8, 2020
edited by Dragan  Aug 8, 2020
edited by Dragan  Aug 10, 2020

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