Each side of `DeltaPQR` is 16 units. S is the foot of the perpendicular dropped from Q on PR, T is the mid point QS. The length of PT is

Options

A. `6sqrt(3)` units
B. `6sqrt(7)` units
C. `3sqrt(3)` units
D. `4sqrt(7)` units

Answer

To solve the problem step by step, we need to find the length of PT in the equilateral triangle PQR, where each side is 16 units.

### Step-by-Step Solution:

1. **Understand the Triangle**:
- We have an equilateral triangle PQR where each side is 16 units.
- This means that \( PQ = QR = RP = 16 \) units.

2. **Draw the Perpendicular**:
- Drop a perpendicular from point Q to side PR. Let the foot of the perpendicular be point S.
- Since it is an equilateral triangle, the perpendicular from a vertex bisects the opposite side. Therefore, \( PS = SR \).

3. **Calculate PS and SR**:
- Since PS and SR are equal and PR = 16 units, we have:
\[ PS = SR = \frac{PR}{2} = \frac{16}{2} = 8 \text{ units} \]

4. **Determine the Height of the Triangle**:
- The height (QS) of an equilateral triangle can be calculated using the formula:
\[ \text{Height} = \frac{\sqrt{3}}{2} \times \text{side} \]
- Substituting the side length:
\[ QS = \frac{\sqrt{3}}{2} \times 16 = 8\sqrt{3} \text{ units} \]

5. **Find the Midpoint T**:
- T is the midpoint of QS. Therefore, we can find ST:
\[ ST = \frac{QS}{2} = \frac{8\sqrt{3}}{2} = 4\sqrt{3} \text{ units} \]

6. **Use the Pythagorean Theorem**:
- In triangle PTS, we can apply the Pythagorean theorem:
\[ PT^2 = PS^2 + ST^2 \]
- We already know:
- \( PS = 8 \) units
- \( ST = 4\sqrt{3} \) units
- Now calculate:
\[ PT^2 = 8^2 + (4\sqrt{3})^2 \]
\[ PT^2 = 64 + 16 \times 3 = 64 + 48 = 112 \]

7. **Calculate PT**:
- Now take the square root to find PT:
\[ PT = \sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7} \text{ units} \]

### Final Answer:
The length of PT is \( 4\sqrt{7} \) units.

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