Question

In parallelogram ABCD , BE=7x−2 and DE=x^2−10.Segment AC intersects segment BD at point E. What is BD ? Enter your answer in the box.

Answer 1

Answer:

Length of BD = 108 units

Step-by-step explanation:

Given: In parallelogram ABCD , BE = 7x-2 and DE = $x^2-10$.

Segment AC intersects segment BD at point E.

Properties of parallelograms:-

• Opposite sides are congruent (AB=CD)
• Opposite angels are congruent (D=B).
• Also, consecutive angles are supplementary (A + D = 180°).

• Diagonals of a parallelogram bisect each other.

• Every diagonal of a parallelogram separates it into two congruent.

The diagonal BD;

BD = BE + ED

since, by properties of parallelogram;

BE = ED                     ......[1]

Substitute the given values of BE and DE in [1] to solve for x;

$7x-2=x^2-10$

or

$x^2-10-7x+2=0$

$x^2-7x-8=0$

$x^2-8x+x-8=0$

$x(x-8)+1(x-8)=0$

(x-8)(x+1) = 0

equate each factor equals to 0 we get;

x = 8 and x = -1

Since sides are always in positive,

⇒x =8

Then;

BE = 7x -2 = 7(8) -2 = 56-2 = 54 units.

To find the Length of BD:

As we know;

BD = BE + ED = BE + BE = 2BE              [Since, BE = ED]

BD = $2 \times 54 =108$ units

Therefore, the length of BD is 108 units.