+1836 In quadrilateral ABCD, we have AB=BC=CD=DA, AC=14 and BD=48. Find the perimeter of ABCD.
+23252 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 + XB2 = 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
+23252 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 + XB2 = 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
