Let's analyze the problem step by step.

Step 1: Understand the Given Information

• PQRS is a rhombus, which means all sides are equal, and opposite angles are equal.
• PQT is an isosceles triangle with $PT = QT$.
• The measure of angle $\angle PSR$ is given as $110^\circ$.

Step 2: Find the Other Angles of the Rhombus

In a rhombus, opposite angles are equal, and adjacent angles are supplementary (they add up to $180^\circ$). Given that $\angle PSR = 110^\circ$, the angle $\angle PQR$ opposite to $\angle PSR$ is also $110^\circ$.

Since adjacent angles in a rhombus are supplementary: $\angle PQS = 180^\circ - 110^\circ = 70^\circ$ Therefore, both $\angle PQS$ and $\angle QRP$ are $70^\circ$.

Step 3: Analyze Triangle PQT

Given that $PT = QT$ in the isosceles triangle $PQT$, the base angles $\angle PQT$ and $\angle QPT$ are equal.

Let's denote the base angles as $x$. Since $\angle PQS = 70^\circ$, $\angle PQT$ is the exterior angle for triangle $PQT$ at vertex $Q$. Therefore, the exterior angle theorem gives: $\angle PQT = \angle P + \angle Q = x + x = 2x$ However, $\angle PQS = 70^\circ$ is not the exterior angle directly affecting $\angle SQT$, since $SQT$ doesn't include $\angle PQT$.

Step 4: Identify $\angle SQT$

The angle $\angle SQT$ is external to triangle $PQT$. Knowing that the entire $\angle PSR$ covers $\angle PQS + \angle SQT$, we conclude: $\angle PSR = \angle PQS + \angle SQT = 110^\circ$ Since $\angle PQS$ is $70^\circ$: $\angle SQT = \angle PSR - \angle PQS = 110^\circ - 70^\circ = 40^\circ$

Conclusion

The size of angle $\angle SQT$ is $\boxed{40^\circ}$.

Would you like further details or have any questions?

Here are some related questions:

1. What is the sum of the angles in any quadrilateral?
2. How do you determine the area of a rhombus given its side lengths?
3. How does the interior angle of an equilateral triangle compare with that of a rhombus?
4. How would this problem change if the triangle $PQT$ were equilateral?
5. What is the significance of isosceles triangles in different geometric proofs?

Tip: When dealing with rhombuses, remember that adjacent angles add up to $180^\circ$, which can help solve various angle-related problems.