The problem asks for the perimeter of triangle $\triangle ABC$, which is circumscribed around circle $D$. Given:

• $AQ = 13 \, \text{cm}$
• $BR = 10.5 \, \text{cm}$
• $SC = 6 \, \text{cm}$

By a property of a triangle circumscribed around a circle, the tangents from a common external point to the circle are equal in length. Thus, we have the following pairs of equal segments:

• $AQ = AS$
• $BR = BQ$
• $CS = CR$

Let's denote these equal segments:

• $AS = AQ = 13 \, \text{cm}$
• $BQ = BR = 10.5 \, \text{cm}$
• $CR = CS = 6 \, \text{cm}$

Now, to find the perimeter of triangle $ABC$, we add the sides: $\text{Perimeter} = AB + BC + CA$ Where: $AB = AQ + BQ = 13 + 10.5 = 23.5 \, \text{cm}$ $BC = BR + CR = 10.5 + 6 = 16.5 \, \text{cm}$ $CA = CS + AS = 6 + 13 = 19 \, \text{cm}$

Thus, the perimeter of triangle $ABC$ is: $\text{Perimeter} = 23.5 + 16.5 + 19 = 59 \, \text{cm}$

The perimeter of triangle $ABC$ is 59 cm.

Would you like more details or further clarifications? Here are some related questions you might find interesting:

1. What is the formula for the area of a triangle circumscribed around a circle?
2. How do we derive the tangents' equal length property in such triangles?
3. How do we find the radius of the incircle given the sides of a triangle?
4. What are other properties of tangents drawn from an external point to a circle?
5. How can we find the area of triangle $ABC$ if we know its perimeter and inradius?

Tip: In any triangle circumscribed around a circle, the tangents drawn from a common external point to the circle are always equal.