In `Delta ABC,` perpendiculars AP and CQ are droped from vertices A and C on opposite sides BC and AB, respectively. If `AB = 10 cm , BC = 16 cm, AP = 8.8 cm, ` find PQ .

Answer

To solve the problem, we need to find the length of segment PQ in triangle ABC, where perpendiculars AP and CQ are dropped from vertices A and C onto sides BC and AB, respectively. We are given the lengths of AB, BC, and AP.

### Step-by-Step Solution:

1. **Identify Given Values:**
- AB = 10 cm
- BC = 16 cm
- AP = 8.8 cm

2. **Understanding the Area of Triangle ABC:**
The area of triangle ABC can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
We can use either pair of perpendiculars (AP and CQ) to find the area.

3. **Using the First Pair (AP and BC):**
We can calculate the area of triangle ABC using AP as the height and BC as the base:
\[ \text{Area} = \frac{1}{2} \times BC \times AP = \frac{1}{2} \times 16 \times 8.8 \]
\[ \text{Area} = \frac{1}{2} \times 16 \times 8.8 = 8 \times 8.8 = 70.4 \text{ cm}^2 \]

4. **Using the Second Pair (CQ and AB):**
Now, we can express the area using CQ as the height and AB as the base:
\[ \text{Area} = \frac{1}{2} \times AB \times CQ = \frac{1}{2} \times 10 \times CQ \]

5. **Setting the Areas Equal:**
Since both expressions represent the area of the same triangle, we can set them equal to each other:
\[ 70.4 = \frac{1}{2} \times 10 \times CQ \]

6. **Solving for CQ:**
We can simplify the equation:
\[ 70.4 = 5 \times CQ \]
Dividing both sides by 5:
\[ CQ = \frac{70.4}{5} = 14.08 \text{ cm} \]

7. **Conclusion:**
The length of segment PQ, which is equal to CQ, is:
\[ PQ = CQ = 14.08 \text{ cm} \]