Question

Q (1, 6) R (-4, 2) P (1, 2)
Use the given coordinates to compute the perimeter of the triangle. Round your answer to the nearest tenth.

Answer 1

To find the perimeter of triangle QRP with vertices Q(1, 6), R(-4, 2), and P(1, 2), we first need to compute the lengths of the sides QR, RP, and PQ using the distance formula:

The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

1. **Calculate the length of QR:**
- \(Q(1, 6)\) and \(R(-4, 2)\)
\[
QR = \sqrt{((-4) - 1)^2 + (2 - 6)^2} = \sqrt{(-5)^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41}
\]

2. **Calculate the length of RP:**
- \(R(-4, 2)\) and \(P(1, 2)\)
\[
RP = \sqrt{(1 - (-4))^2 + (2 - 2)^2} = \sqrt{(1 + 4)^2 + 0^2} = \sqrt{5^2} = 5
\]

3. **Calculate the length of PQ:**
- \(P(1, 2)\) and \(Q(1, 6)\)
\[
PQ = \sqrt{(1 - 1)^2 + (6 - 2)^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4
\]

Now we can calculate the perimeter \(P\) of triangle QRP:

\[
\text{Perimeter} = QR + RP + PQ = \sqrt{41} + 5 + 4
\]

We need to compute \(\sqrt{41}\) to the nearest tenth:

\[
\sqrt{41} \approx 6.4
\]

So:

\[
\text{Perimeter} \approx 6.4 + 5 + 4 = 15.4
\]

Rounded to the nearest tenth, the perimeter of triangle QRP is:

\[
\boxed{15.4}
\]