We are given a trapezoid $ABCD$. We are given that $m\angle ABC = (4x + 11)^\circ$ and $m\angle DAB = (2x + 33)^\circ$. We need to find the value of $x$ such that the trapezoid $ABCD$ is isosceles.

1. Problem Description

We are given a trapezoid $ABCD$. We are given that $m\angle ABC = (4x + 11)^\circ$ and $m\angle DAB = (2x + 33)^\circ$. We need to find the value of $x$ such that the trapezoid $ABCD$ is isosceles.

2. Solution Steps

In an isosceles trapezoid, the base angles are equal. In trapezoid $ABCD$, if $AB$ and $CD$ are the parallel sides, then $\angle DAB$ and $\angle ABC$ are adjacent angles on the same side of the transversal $AB$. The sum of these two angles is $180^\circ$.

Therefore, we have the equation:

$m\angle DAB + m\angle ABC = 180^\circ$

$(2x + 33) + (4x + 11) = 180$

$6x + 44 = 180$

$6x = 180 - 44$

$6x = 136$

$x = \frac{136}{6} = \frac{68}{3}$

In an isosceles trapezoid, the base angles are equal. If $ABCD$ is an isosceles trapezoid, then the angles on the bases are equal.

So $\angle DAB = \angle ABC$

$2x + 33 = 4x + 11$

$22 = 2x$

$x = 11$

Also, if the non-parallel sides are the same, then the angles are equal. Therefore, $\angle DAB = \angle CBA$ or $\angle ADC = \angle BCD$.

$\angle DAB + \angle ADC = 180$

$\angle ABC + \angle BCD = 180$

In an isosceles trapezoid $ABCD$ with bases $AB$ and $CD$, we have $\angle DAB = \angle ABC$. Thus,

$2x + 33 = 4x + 11$

$22 = 2x$

$x = 11$

3. Final Answer

11