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In quadrilateral $ABCD$, we have $AB = BC = CD = DA$, $AC = 14$, and $BD = 48$. Find the perimeter of $ABCD$.

 
eileenthecoolbean  Aug 9, 2017
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 #1
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ANSWER: \(\boxed{100}\)

Thus if s=length of each side of the rhombus, then
\(s²=(AC/2)²+(BD/2)²\)
using Pythagoras theorem. 
Calculate (Multiply S by 4 to get final answer because all four sides are equal)
\(s=√(7²+24²)  =25 \)

 
RektTheNoob  Aug 9, 2017
edited by RektTheNoob  Aug 9, 2017
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 #1
avatar+74 
+3
Best Answer

ANSWER: \(\boxed{100}\)

Thus if s=length of each side of the rhombus, then
\(s²=(AC/2)²+(BD/2)²\)
using Pythagoras theorem. 
Calculate (Multiply S by 4 to get final answer because all four sides are equal)
\(s=√(7²+24²)  =25 \)

 
RektTheNoob  Aug 9, 2017
edited by RektTheNoob  Aug 9, 2017

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