A circle is inscribed in `Delta` PQR having sides 8 cm, 10 cm and 12 cm. A circle touches the sides PQ, QR and RP at D, E and F. Find PD

Options

A. 5 cm
B. 6 cm
C. 4 cm
D. 10 cm

Answer

To find the length of PD in triangle PQR, where a circle is inscribed, we can follow these steps:

### Step 1: Identify the sides of the triangle
The sides of triangle PQR are given as:
- PQ = 8 cm
- QR = 10 cm
- PR = 12 cm

### Step 2: Understand the properties of tangents
When a circle is inscribed in a triangle, the points where the circle touches the sides are called the points of tangency. The lengths of the tangents drawn from an external point to a circle are equal.

### Step 3: Assign variables to the tangent segments
Let:
- PD = x (the length we want to find)
- DQ = 8 - x (since PQ = PD + DQ = 8 cm)
- QE = y (the tangent from Q to the circle)
- ER = 10 - y (since QR = QE + ER = 10 cm)
- RF = z (the tangent from R to the circle)
- FP = 12 - z (since PR = RF + FP = 12 cm)

### Step 4: Set up equations based on the tangent properties
From the properties of tangents:
1. PD = PF (since they are tangents from point P)
- Therefore, PD = PF = x
2. DQ = QE (since they are tangents from point Q)
- Therefore, DQ = QE = 8 - x
3. ER = RF (since they are tangents from point R)
- Therefore, ER = RF = 10 - y

### Step 5: Relate the segments to the sides of the triangle
From the triangle's side lengths, we can write:
- For side PQ: PD + DQ = 8
- Thus, x + (8 - x) = 8
- For side QR: QE + ER = 10
- Thus, (8 - x) + (10 - y) = 10
- For side PR: PF + FP = 12
- Thus, x + (12 - z) = 12

### Step 6: Solve for x
From the equation for side PR:
- x + (12 - z) = 12
- This simplifies to z = 12 - x

From the equation for side QR:
- (8 - x) + (10 - (12 - x)) = 10
- Simplifying gives: 8 - x + 10 - 12 + x = 10
- Which simplifies to: 6 = 10 (not useful)

Instead, we can directly use the relationship:
- Since PD = PF, we can say:
- PD + PF = PR
- x + x = 12
- 2x = 12
- x = 6

### Step 7: Find PD
Thus, PD = x = 6 cm.

### Final Answer
PD = 6 cm.

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