in quadrilateral ABCD AC is a diagonol, m<ACD=2x+4 and m<ACB=5x-11 what proves ABCD a rhombus
^^^ is correct
indeed
how do you work out x ?
A decade later we still looking for the right answer lol
To prove that quadrilateral ABCD is a rhombus, we need to show that all four sides are congruent.
In this case, AC is a diagonal, and we are given two angles, ∠ACD and ∠ACB, in terms of x.
To start, we should recall the properties of a rhombus:
1. All sides of a rhombus are congruent.
2. Diagonals of a rhombus bisect each other at right angles.
To find the values of x, we can use the fact that opposite angles in a quadrilateral sum up to 180 degrees. Therefore, we have:
∠ACD + ∠ACB = 180 degrees.
Substituting the given angle values, we get:
(2x + 4) + (5x - 11) = 180.
Now, we can solve this equation for x:
2x + 4 + 5x - 11 = 180
7x - 7 = 180
7x = 187
x = 187/7
x = 26.71
Now that we have found the value of x, we can substitute it back into the given angles to find their measures:
∠ACD = 2x + 4 = 2(26.71) + 4 = 57.42 degrees
∠ACB = 5x - 11 = 5(26.71) - 11 = 133.55 degrees
Next, we need to determine if all four sides are congruent. We can do this by comparing opposite sides of the rhombus.
Comparing side AB and side CD:
Since AC is a diagonal, it divides the rhombus into two congruent triangles (ACD and ABC). Therefore, the corresponding sides AB and CD are congruent.
Comparing side BC and side AD:
Since AC is a diagonal, it bisects the angles at B and D, making them right angles. The opposite sides BC and AD are also congruent.
Therefore, we have shown that all four sides in quadrilateral ABCD are congruent. Hence, quadrilateral ABCD is a rhombus.
It's a rhombus if the diagonal bisects the angles. So, if
2x+4 = 5x-11
the ABCD is a rhombus