Question

In quadrilateral ABCD, m< ACD= 2x + 4 and m< ACB= 5x - 8. For what value of x is ABCD a rhombus?

Answer 1

Opposite angles in a rhombus are equal, so by setting m

Step-by-step explanation:

To determine the value of x for which quadrilateral ABCD is a rhombus, we need to recognize that in a rhombus, opposite angles are equal. This means that m<ACD should be equal to m<ACB since they are opposite angles. Given that m<ACD = 2x + 4 and m<ACB = 5x - 8, setting them equal to each other and solving for x will give us the value we are looking for:

2x + 4 = 5x - 8
Adding 8 to both sides: 2x + 12 = 5x
Subtracting 2x from both sides: 12 = 3x
Dividing both sides by 3: x = 4

So, for x = 4, the measurements of the angles indicate that ABCD is indeed a rhombus.

Answer 2

x = 4

Step-by-step explanation:

If a quadrilateral is a rhombus, then the diagonals of it will bisect each angle of the quadrilateral.

Now, the diagonal AC bisects angle C into ∠ ACD and ∠ ACB.

Hence, ∠ ACD = ∠ ACB

⇒ 2x + 4 = 5x - 8 {Given that ∠ ACD= 2x + 4 and ∠ ACB= 5x - 8 }

⇒ 3x = 12

⇒ x = 4 (Answer)