Question

what is the value of $x$ in the figure shown below?
$x =$

Explanation:

Step1: Determine the type of quadrilateral

Since \(AB = CD = 3.9\) and \(AC = BD = 4\), the quadrilateral \(ABCD\) is a parallelogram (by SSS - Side - Side - Side congruence of \(\triangle ABC\) and \(\triangle CDB\)). In a parallelogram, \(AB\parallel CD\).

Step2: Use the angle - sum property in \(\triangle ABC\)

In \(\triangle ABC\), by the angle - sum property of a triangle (\(\angle A+\angle ABC+\angle BCA = 180^{\circ}\)), \(\angle ABC=180^{\circ}-\angle A - \angle BCA=180^{\circ}-100^{\circ}-30^{\circ}=50^{\circ}\).

Step3: Use the property of parallel lines

Because \(AB\parallel CD\), \(\angle BCD+\angle ABC = 180^{\circ}\) (consecutive interior angles). Also, \(\angle BCD=x + 30^{\circ}\).
Since \(\triangle ABC\cong\triangle CDB\) (by SSS), \(\angle CBD=\angle BCA = 30^{\circ}\) and \(\angle CDB=\angle A=100^{\circ}\). Another way: Since \(AB\parallel CD\), \(\angle x=\angle ABC\) (alternate - interior angles for \(AB\parallel CD\) and transversal \(BC\)). But we can also use the congruence of triangles.
Since \(\triangle ABC\cong\triangle CDB\) (SSS: \(AB = CD = 3.9\), \(AC = BD = 4\), \(BC\) is common), \(\angle BAC=\angle CDB = 100^{\circ}\), \(\angle ABC=\angle BCD\).
In \(\triangle ABC\), \(\angle ABC=180^{\circ}-\angle A-\angle BCA=180 - 100-30 = 50^{\circ}\).
Since \(\angle BCD=x + 30^{\circ}\) and \(\angle ABC=\angle BCD\) (from congruent triangles \(\triangle ABC\cong\triangle CDB\)), \(x+30^{\circ}=50^{\circ}\).

Step4: Solve for \(x\)

Subtract \(30^{\circ}\) from both sides of the equation \(x + 30^{\circ}=50^{\circ}\).
\(x=50^{\circ}-30^{\circ}\)

Answer:

\(x = 20\)