We are given a pentagon ABCDE with angles $A = 120^\circ$, $C = 110^\circ$, and $D = 135^\circ$. We are also given that $\angle AED = 4 \times \angle ABC$. We need to find the size of $\angle AED$.

1. Problem Description

We are given a pentagon ABCDE with angles $A = 120^\circ$, $C = 110^\circ$, and $D = 135^\circ$. We are also given that $\angle AED = 4 \times \angle ABC$. We need to find the size of $\angle AED$.

2. Solution Steps

First, recall that the sum of the interior angles of a pentagon is given by the formula:

$$(n-2) \times 180^\circ$$

where $n$ is the number of sides. In this case, $n = 5$, so the sum of the interior angles is:

$$(5-2) \times 180^\circ = 3 \times 180^\circ = 540^\circ$$

We know that the sum of the angles in pentagon ABCDE is:

$$\angle A + \angle B + \angle C + \angle D + \angle E = 540^\circ$$

Substituting the given values, we have:

$$120^\circ + \angle ABC + 110^\circ + 135^\circ + \angle AED = 540^\circ$$

$$365^\circ + \angle ABC + \angle AED = 540^\circ$$

We also know that $\angle AED = 4 \times \angle ABC$. Let $x = \angle ABC$. Then $\angle AED = 4x$. Substituting this into the equation above, we get:

$$365^\circ + x + 4x = 540^\circ$$

$$365^\circ + 5x = 540^\circ$$

$$5x = 540^\circ - 365^\circ$$

$$5x = 175^\circ$$

$$x = \frac{175^\circ}{5}$$

$$x = 35^\circ$$

Therefore, $\angle ABC = 35^\circ$. Now we can find $\angle AED$:

$$\angle AED = 4 \times \angle ABC = 4 \times 35^\circ = 140^\circ$$

3. Final Answer

$\angle AED = 140^\circ$