+912 In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, BE=2x2−3x , and DE=x2+10 .
What is BD ?
+129852 The diagonals of a parallelogram bisect each other....therefore...
2x^2 - 3x = x^2 + 10 simplify as
x^2 - 3x - 10 = 0 factor
(x - 5) ( x + 2) = 0
Setting each factor to 0 and solving for x, we have that x = 5 or x = -2
Therefore we have two possible values for BD
If x = 5 ...BD = 2 [ (5)^2 + 10] = 2 [ 35 ] = 70 units
If x = - 2, .....BD = 2 [ (-2)^2 + 10 ] = 2 [ 14 ] = 28 units
+633 BD = \(\sqrt{3x^{2}-3x+10}\) . This is because, with the Pythagorean theorem, you add the 2 legs of a right triangle and take the square root to get the hypotenuse of it. This only applies if ABCD is a rhombus.
+129852 The diagonals of a parallelogram bisect each other....therefore...
2x^2 - 3x = x^2 + 10 simplify as
x^2 - 3x - 10 = 0 factor
(x - 5) ( x + 2) = 0
Setting each factor to 0 and solving for x, we have that x = 5 or x = -2
Therefore we have two possible values for BD
If x = 5 ...BD = 2 [ (5)^2 + 10] = 2 [ 35 ] = 70 units
If x = - 2, .....BD = 2 [ (-2)^2 + 10 ] = 2 [ 14 ] = 28 units
