Question 1210470
This problem is solvable by using the property of **tangents drawn from an external point to a circle**.

The perimeter of $\triangle \text{XYZ}$ is **$28$**.

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## 📐 Tangent Segment Theorem

The key geometric principle here is the **Tangent Segment Theorem**: If two tangent segments are drawn to a circle from the same external point, then the segments are equal in length.

In the provided figure, the inscribed circle (incircle) is tangent to the sides of $\triangle \text{XYZ}$ at points $\text{P}$, $\text{Q}$, and a third point, let's call it $\text{R}$, on side $\text{XZ}$.

* From external point $\mathbf{X}$: The tangent segments are $\mathbf{XQ}$ and $\mathbf{XR}$ (assuming $\text{R}$ is the tangent point on $\text{XZ}$).
* From external point $\mathbf{Y}$: The tangent segments are $\mathbf{YP}$ and $\mathbf{YQ}$.
* From external point $\mathbf{Z}$: The tangent segments are $\mathbf{ZP}$ and $\mathbf{ZR}$.



## 📏 Calculating Side Lengths

Using the Tangent Segment Theorem and the given lengths:

1.  **Find the length of the side $\text{YZ}$**:
    * $\text{YZ}$ is given as $10$.
    * $\text{YZ} = \text{YP} + \text{PZ}$.
    * Given $\text{YP} = 6$.
    * Therefore, $\text{PZ} = \text{YZ} - \text{YP} = 10 - 6 = 4$.

2.  **Find the lengths of $\text{XQ}$ and $\text{YQ}$**:
    * Since $\text{YP}$ and $\text{YQ}$ are tangents from $\text{Y}$, we have $\mathbf{\text{YQ} = \text{YP}}$.
        $$\text{YQ} = 6$$
    * Since $\text{XQ}$ and $\text{XR}$ (let's use $\text{XR}$ for the tangent to $\text{XZ}$) are tangents from $\text{X}$, we have $\mathbf{\text{XR} = \text{XQ}}$.
        $$\text{XR} = 4$$

3.  **Find the length of $\text{ZP}$ and $\text{ZR}$**:
    * Since $\text{ZP}$ and $\text{ZR}$ are tangents from $\text{Z}$, we have $\mathbf{\text{ZR} = \text{ZP}}$.
    * We calculated $\text{ZP} = 4$ in step 1.
        $$\text{ZR} = 4$$

**Note:** The length $\text{XP} = 8$ and $\text{PQ} = 7$ are extraneous/conflicting if the tangency points are $\text{P}$, $\text{Q}$, and $\text{R}$ on the sides of $\triangle \text{XYZ}$. $\text{X}$, $\text{P}$, and $\text{Q}$ would not be vertices of a known figure, and $\text{XP}$ and $\text{XQ}$ are two tangent lengths from $\text{X}$, which must be equal.

The only way the problem is solvable is if the tangency points are $\text{Q}$ on $\text{XY}$ and $\text{P}$ on $\text{YZ}$, and the third point is $\text{R}$ on $\text{XZ}$. Given that $\mathbf{\text{XQ} = 4}$ and $\mathbf{\text{XP} = 8}$, the points **$\text{P}$ and $\text{Q}$ cannot both be tangency points** to the same circle from the vertices $\text{X}$ and $\text{Y}$.

Let's assume the labels in the diagram mean:
* $\mathbf{XQ}$ and $\mathbf{XR}$ (on $\text{XZ}$) are tangents from $\text{X}$. $\implies \text{XQ} = \text{XR} = 4$.
* $\mathbf{YP}$ and $\mathbf{YQ}$ are tangents from $\text{Y}$. $\implies \text{YP} = \text{YQ} = 6$.
* $\mathbf{ZP}$ and $\mathbf{ZR}$ are tangents from $\text{Z}$. $\implies \text{ZP} = \text{ZR}$.

The length $\mathbf{\text{XP} = 8}$ and $\mathbf{\text{PQ} = 7}$ must be disregarded as they directly contradict the Tangent Segment Theorem for the perimeter calculation, which only requires the tangent lengths from the vertices.

## ➕ Perimeter Calculation

The **Perimeter** $(\mathbf{Per})$ of $\triangle \text{XYZ}$ is the sum of its three side lengths:
$$\text{Per} = \text{XY} + \text{YZ} + \text{XZ}$$

1.  **Side $\text{XY}$**:
    $$\text{XY} = \text{XQ} + \text{QY} = 4 + 6 = 10$$
2.  **Side $\text{YZ}$**:
    $$\text{YZ} = 10 \text{ (Given)}$$
3.  **Side $\text{XZ}$**:
    $$\text{XZ} = \text{XR} + \text{RZ}$$
    * $\text{XR} = \text{XQ} = 4$
    * $\text{RZ} = \text{ZP}$
    * $\text{ZP} = \text{YZ} - \text{YP} = 10 - 6 = 4$
    * So, $\text{XZ} = 4 + 4 = 8$

**Total Perimeter:**
$$\text{Per} = \text{XY} + \text{YZ} + \text{XZ} = 10 + 10 + 8 = 28$$