Question

question
find the value of ( x ), ( y ), and ( z ) in the rhombus below.
answer
attempt 1 out of 2
( x=quad y=quad z= )

Explanation:

Step1: Use the property of rhombus (opposite angles are equal)

In a rhombus, opposite angles are equal.
We know that one angle is $90^{\circ}$.
The angle $(-2x + 6)^{\circ}$ is opposite to the angle $(-6y+6)^{\circ}$. Also, adjacent angles in a rhombus are supplementary (sum to $180^{\circ}$).
Since one angle is $90^{\circ}$, the adjacent angles:
For the angle $(-2x + 6)^{\circ}$:
$-2x+6+90 = 180$
$-2x=180 - 96$
$-2x = 84$
$x=\frac{84}{- 2}=-42$
For the angle $(-6y + 6)^{\circ}$:
$-6y+6+90=180$
$-6y=180 - 96$
$-6y = 84$
$y=\frac{84}{-6}=-14$

Step2: Use the property of rhombus (opposite angles are equal)

The angle $(-5z + 10)^{\circ}$ is equal to $90^{\circ}$ (opposite angles in a rhombus)
$-5z+10 = 90$
$-5z=90 - 10$
$-5z = 80$
$z=\frac{80}{-5}=- 16$ (Wait, no! Wait, adjacent angles in a rhombus: actually, in a rhombus, if one angle is $90^{\circ}$, it is a square. All angles are $90^{\circ}$.
So $-2x + 6=90$, $-2x=90 - 6=84$, $x=-42$; $-6y + 6=90$, $-6y=90 - 6 = 84$, $y=-14$; $-5z+10 = 90$, $-5z=90 - 10=80$, $z = 16$

Answer:

$x = -42$, $y = -14$, $z = 16$