Question

In parallelogram ABCD , diagonals AC¯¯¯¯¯ and BD¯¯¯¯¯ intersect at point E, AE=2x2−3x , and CE=x2+4 .

Answer 1

Answer:

The length of AC is either 10 units or 40 units.

Step-by-step explanation:

it is given that parallelogram ABCD and diagonals AC and BD intersect at point E.

According to the property of parallelogram, the diagonals are intersecting each other at their midpoint.

$AE=CE$

$2x^2-3x=x^2+4$

$2x^2-3x-x^2-4=0$

$x^2-3x-4=0$

$x^2-4x+x-4=0$

$x(x-4)+1(x-4)=0$

$(x-4)(x+1)=0$

By zero product property, equate each factor equal to 0. So the value of x is 4 and -1.

$AC=2CE$

$AC=2(x^2+4)$

Let he value of x=4.

$AC=2(4^2+4)$

$AC=2(20)$

$AC=40$

Therefore the length of AC is 40.

Let he value of x=-1.

$AC=2((-1)^2+4)$

$AC=2(5)$

$AC=10$

Therefore the length of AC is 10.

Answer 2

If the diagonals of the parallelogram intersect with each other, this means that the diagonals bisect each other. Thus,

    AE = CE

Substituting the equations,

  2x² - 3x = x² + 4

The values of x from the equation are equal to 4 and -1. 

  AE = 2(4)² - 3(4) = 20

Hence, AC is equal to 40. 

Answer: 40 units