+77 In quadrilateral $ABCD$, we have $AB = BC = CD = DA$, $AC = 14$, and $BD = 48$. Find the perimeter of $ABCD$.
+489 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)
\(s=√(7²+24²) =25 \)
+489 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)
\(s=√(7²+24²) =25 \)
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