The question asks to determine the size of angle $x$ in a quadrilateral. The diagram provided shows that:

• $AB = AC = AD$ (all sides are equal),
• $\angle BAD = 90^\circ$,
• $\angle ABC = 72^\circ$.

Analyzing the Problem:

1. The triangle $ABD$ is isosceles since $AB = AD$, and one of the angles ($\angle BAD$) is $90^\circ$.

2. The angles in triangle $ABD$ must sum to $180^\circ$. Thus, the sum of the other two angles, $\angle ABD$ and $\angle ADB$, must equal $90^\circ$. Since the triangle is isosceles, $\angle ABD = \angle ADB$. Therefore, $\angle ABD = \angle ADB = \frac{90^\circ}{2} = 45^\circ$.

3. Now look at quadrilateral $ABCD$. The total interior angles of a quadrilateral sum to $360^\circ$. Using the known angles:

   • $\angle BAD = 90^\circ$,
   • $\angle ABC = 72^\circ$,
   • $\angle ABD = 45^\circ$.

4. To find $\angle x$ (or $\angle ACD$): $\angle x = 360^\circ - (90^\circ + 72^\circ + 45^\circ) = 360^\circ - 207^\circ = 153^\circ.$

Thus, $\angle x$ should be $153^\circ$, not 72° as stated in the answer provided in the worksheet.

Let me know if you'd like further details on this or any other questions.

Here are five related questions:

1. How do you calculate the sum of angles in any polygon?
2. How do you determine angles in isosceles triangles?
3. What is the significance of the property $AB = AC = AD$ in this problem?
4. How would the problem change if one of the sides was not equal?
5. What are the key properties of cyclic quadrilaterals?

Tip:

For any quadrilateral, always remember that the interior angles sum to $360^\circ$.