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

 Dec 30, 2015

Best Answer 

 #1
avatar+23245 
+11

Since all 4 sides are equal, this is a rhombus.

The diagonals of a rhombus intersect at their midpoint.

If the midpoint is labeled "X", then AX = 7  and  BX = 24.

Also, the diagonals of a rhombus are perpendicular to each other; this means that triangle AXB is a right triangle with a right angle at corner X.

Using the Pythagorean Theorem:  AX2 + XB = AB2     --->     (7)2 + (24)2  =  (AB)2      --->     49 + 576  =  (AB)2

     --->     625  =  (AB)2     --->     AB = 25

Since all four sides are equal, the perimeter  =  4 x 25  =  100

 Dec 30, 2015
 #1
avatar+23245 
+11
Best Answer

Since all 4 sides are equal, this is a rhombus.

The diagonals of a rhombus intersect at their midpoint.

If the midpoint is labeled "X", then AX = 7  and  BX = 24.

Also, the diagonals of a rhombus are perpendicular to each other; this means that triangle AXB is a right triangle with a right angle at corner X.

Using the Pythagorean Theorem:  AX2 + XB = AB2     --->     (7)2 + (24)2  =  (AB)2      --->     49 + 576  =  (AB)2

     --->     625  =  (AB)2     --->     AB = 25

Since all four sides are equal, the perimeter  =  4 x 25  =  100

geno3141 Dec 30, 2015

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