Question

The length of segment S V is 8 units.



A graph.Short description, A graph.,Long description,
A graph showing rectangle S T U V has point S at (negative 9, 2), point T at (3, negative 8), point U at (8, negative 2), and point V at (negative 4, 8).


Question
What is the perimeter of the rectangle?

Answer options with 4 options
A.
16 plus 2 StartRoot 72 EndRoot units

B.
16 plus 2 StartRoot 136 EndRoot units

C.
16 plus 2 StartRoot 180 EndRoot units

D.
16 plus 2 StartRoot 244 EndRoot units

Answer 1

To find the perimeter of rectangle STUV, we first need to determine the lengths of its sides. The points are given as follows:

- Point S: \((-9, 2)\)
- Point T: \((3, -8)\)
- Point U: \((8, -2)\)
- Point V: \((-4, 8)\)

### Step 1: Calculate the length of segment SV

Given that the length of segment SV is 8 units, we need to check if this distance is correct.

Using the distance formula:
\[
\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]
For points S \((-9, 2)\) and V \((-4, 8)\):
\[
\text{Distance} = \sqrt{((-4) - (-9))^2 + (8 - 2)^2}
= \sqrt{(5)^2 + (6)^2}
= \sqrt{25 + 36}
= \sqrt{61}
\]

This is not equal to 8 units, so we may need to determine the lengths of the actual sides of the rectangle \(ST\) and \(SU\).

### Step 2: Calculate the lengths of sides ST and SV

#### Length of ST:
For points S \((-9, 2)\) and T \((3, -8)\):
\[
\text{Distance ST} = \sqrt{(3 - (-9))^2 + (-8 - 2)^2}
= \sqrt{(12)^2 + (-10)^2}
= \sqrt{144 + 100}
= \sqrt{244}
\]

#### Length of SV:
As calculated, this is \( \sqrt{61} \) which doesn't lead to a rectangle’s length.

To find the vertical distance from \(S\) to some other point, irrespective of \(V\), we could compare it with segments directly.

### Step 3: Check the vertical segment US
For points U (8, -2) and S (-9, 2):
\[
\text{Distance SU} = \sqrt{(8 - (-9))^2 + (-2 - 2)^2}
= \sqrt{(17)^2 + (-4)^2}
= \sqrt{289 + 16}
= \sqrt{305}
\]

### Step 4: Total Perimeter
The perimeter \(P\) of a rectangle is given by:
\[
P = 2(\text{length} + \text{width})
\]
Taking length \(=\sqrt{244}\) and width (perpendicular distance) from S to T (let’s refer that as \(sqrt{305}\)).
From above, we could have lengths and it leads:

Assuming lengths are properly evaluated for rectangle:
Thus,
\[
P = 2(\sqrt{244} + 8)
\]
Bearing, from \(p= 16 + 2\sqrt{244}\).\

Thus, this suggests an answer will reflect based on rectangles’ interpretation, but largely we're close to size as \(16 + 2\sqrt{244}\).

### Conclusion

Hence **the closest option reflecting evaluations:**

**D. 16 plus 2 StartRoot 244 EndRoot units.**