Question

in parallelogram ABCD diagonals AC and BD intersect at Point E, BE = 2x^2 -3x + D = x^2 + 10 what is BD

Answer 1

Answer

Given the statement: In parallelogram ABCD, diagonals AC and BD intersects at point E.

Given that ABCD is a parallelogram.  

As, we know that in a parallelogram diagonals bisect each other.

Since AC and BD intersect at E, and we get E is the mid point of both diagonals AC and BD.

⇒ BE = DE                  .....[1]

Substitute the given values of $BE = 2x^2 -3x$ and $DE = x^2+10$ in [1] we have;

$2x^2-3x = x^2+10$

Subtract $x^2$ on both sides we get;

$2x^2-3x-x^2 = x^2+10-x^2$

Simplify:

$x^2-3x =10$

Subtract 10 on both sides we get;

$x^2-3x-10 =0$

or

$x^2-5x+2x-10 =0$

$x(x-5)+2(x-5)=0$

$(x-5)(x+2)=0$

equate these factors equal to zero we get;

(x-5) = 0  and (x+2) = 0

we have;

x = 5 and x = -2

Since, x cannot be negative.

So, x =5

$BE = DE = x^2+10 = (5)^2+10 = 25+10 = 35 units$

Diagonals BD = BE + DE = 35 + 35 =70 units.

Therefore, the value of BD = 70 units.