Question

1. find the value of x for which abcd must be a parallelogram. (x + 31)° (2x - 1)°

Explanation:

Step1: Recall parallelogram property

In a parallelogram, opposite - angles are equal. So, $\angle A=\angle C$ and $\angle B=\angle D$. Also, adjacent - angles are supplementary ($\angle A+\angle B = 180^{\circ}$, $\angle B+\angle C = 180^{\circ}$, $\angle C+\angle D = 180^{\circ}$, $\angle D+\angle A = 180^{\circ}$). Here, we can use the property that alternate - interior angles formed by the diagonal and the parallel sides are equal. The given angles $(x + 31)^{\circ}$ and $(2x-1)^{\circ}$ are equal (alternate - interior angles for the parallel sides of the parallelogram).

Step2: Set up the equation

Set $(x + 31)=(2x - 1)$.

Step3: Solve the equation

Subtract $x$ from both sides: $x+31−x=2x - 1−x$, which gives $31=x - 1$.
Then add 1 to both sides: $31 + 1=x-1 + 1$. So, $x = 32$. But this is wrong. Let's use the property of opposite - angles.
If we assume the angles are opposite angles of the parallelogram, we set up the equation $x + 31=2x-1$.
Subtract $x$ from both sides: $31=2x - 1-x$.
Simplify to get $x=32$.

Answer:

$x = 32$