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In $\triangle ABC,$ $AB=AC=25$ and $BC=23.$ Points $D,E,$ and $F$ are on sides $\overline{AB},$ $\overline{BC},$ and $\overline{AC},$ respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB},$ respectively. What is the perimeter of parallelogram $ADEF$?

 Aug 7, 2022

Best Answer 

 #1
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Let \(\overline{AD} = x\) and \(\overline{BD} = y\). We know that \(\overline{AB} = \overline{AD} + \overline{BD}\) so \( x + y = 25 \)

 

Because \(\triangle BDE \) is similar to \(\triangle ABC\), we know that \(\triangle BDE\) is isoceles, meaning \(\overline{DE} = y\)

 

Also, because \(ADEF\) is a parallelogram, we know that \( \overline{EF} = x\) and \(\overline{AF} = y\)

 

So, the perimeter of the rectangle is just \(\overline{AD} + \overline{DE} + \overline{EF} + \overline{AF} = x + y + x + y = 2(x+y) = 2 \times 25 = \color{brown}\boxed{50}\)

 Aug 7, 2022
edited by BuilderBoi  Aug 7, 2022
 #1
avatar+2446 
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Best Answer

Let \(\overline{AD} = x\) and \(\overline{BD} = y\). We know that \(\overline{AB} = \overline{AD} + \overline{BD}\) so \( x + y = 25 \)

 

Because \(\triangle BDE \) is similar to \(\triangle ABC\), we know that \(\triangle BDE\) is isoceles, meaning \(\overline{DE} = y\)

 

Also, because \(ADEF\) is a parallelogram, we know that \( \overline{EF} = x\) and \(\overline{AF} = y\)

 

So, the perimeter of the rectangle is just \(\overline{AD} + \overline{DE} + \overline{EF} + \overline{AF} = x + y + x + y = 2(x+y) = 2 \times 25 = \color{brown}\boxed{50}\)

BuilderBoi Aug 7, 2022
edited by BuilderBoi  Aug 7, 2022

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