In quadrilateral abcd m acd =2 x+4 and m acb=5x-11. for what value of x is abcd a rhombus?
a.4
b.5
c.6
d.7

The value of x is 5

Explanation

Given <ACD=2x+4

<ACB=5x-11

Given that ABCD is a rhombus. So the opposite angles will be same and all the sides will be equal.

diagonal AC divides the rhombus into two triangles ACD and ACB

according to angle sum property, the sum of all angles of a triangle is 180.

consider triangle ACD

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

96 39 cause 135 -96=39 im smart (i used a calculater)