College

We appreciate your visit to A polygon has vertices tex W 4 28 X 8 23 Y 13 16 tex and tex Z 4 11 tex Part A What is. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

A polygon has vertices [tex] W(-4, 28), X(8, 23), Y(-13, 16) [/tex], and [tex] Z(-4, 11) [/tex].

Part A: What is the perimeter of [tex] WXYZ [/tex] to the nearest tenth of a unit?

Part B: What is the area of [tex] WXYZ [/tex] to the nearest tenth of a square unit?

Enter the correct answers in the boxes.

Answer :

Sure, let's break down the solution to find the perimeter and area of the polygon with vertices [tex]\( W(-4,28) \)[/tex], [tex]\( X(8,23) \)[/tex], [tex]\( Y(-13,16) \)[/tex], and [tex]\( Z(-4,11) \)[/tex].

Part A. Perimeter:

To find the perimeter of the polygon, we need to calculate the distances between each pair of consecutive vertices and then sum them up.

1. Calculate distance [tex]\( W \)[/tex] to [tex]\( X \)[/tex]:
[tex]\[
\text{Distance} = \sqrt{(X_x - W_x)^2 + (X_y - W_y)^2}
\][/tex]
[tex]\( W(-4, 28) \)[/tex] and [tex]\( X(8, 23) \)[/tex]:
[tex]\[
= \sqrt{(8 - (-4))^2 + (23 - 28)^2}
= \sqrt{(12)^2 + (-5)^2}
= \sqrt{144 + 25}
= \sqrt{169}
= 13
\][/tex]

2. Calculate distance [tex]\( X \)[/tex] to [tex]\( Y \)[/tex]:
[tex]\[
\text{Distance} = \sqrt{(Y_x - X_x)^2 + (Y_y - X_y)^2}
\][/tex]
[tex]\( X(8, 23) \)[/tex] and [tex]\( Y(-13, 16) \)[/tex]:
[tex]\[
= \sqrt{(-13 - 8)^2 + (16 - 23)^2}
= \sqrt{(-21)^2 + (-7)^2}
= \sqrt{441 + 49}
= \sqrt{490}
\approx 22.1
\][/tex]

3. Calculate distance [tex]\( Y \)[/tex] to [tex]\( Z \)[/tex]:
[tex]\[
\text{Distance} = \sqrt{(Z_x - Y_x)^2 + (Z_y - Y_y)^2}
\][/tex]
[tex]\( Y(-13, 16) \)[/tex] and [tex]\( Z(-4, 11) \)[/tex]:
[tex]\[
= \sqrt{(-4 - (-13))^2 + (11 - 16)^2}
= \sqrt{(9)^2 + (-5)^2}
= \sqrt{81 + 25}
= \sqrt{106}
\approx 10.3
\][/tex]

4. Calculate distance [tex]\( Z \)[/tex] to [tex]\( W \)[/tex]:
[tex]\[
\text{Distance} = \sqrt{(W_x - Z_x)^2 + (W_y - Z_y)^2}
\][/tex]
[tex]\( Z(-4, 11) \)[/tex] and [tex]\( W(-4, 28) \)[/tex]:
[tex]\[
= \sqrt{(-4 - (-4))^2 + (28 - 11)^2}
= \sqrt{(0)^2 + (17)^2}
= \sqrt{17^2}
= 17
\][/tex]

Now, sum up the distances to get the perimeter:
[tex]\[
\text{Perimeter} = 13 + 22.1 + 10.3 + 17 \approx 62.4 \text{ units}
\][/tex]

Part B. Area:

To find the area of the polygon, we can use the Shoelace formula (or Gauss's area formula), which is given by:
[tex]\[
\text{Area} = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i y_{i+1} - y_i x_{i+1}) + (x_n y_1 - y_n x_1) \right|
\][/tex]

For vertices [tex]\( W(-4,28) \)[/tex], [tex]\( X(8,23) \)[/tex], [tex]\( Y(-13,16) \)[/tex], and [tex]\( Z(-4,11) \)[/tex]:
[tex]\[
\begin{aligned}
\text{Area} &= \frac{1}{2} \left| (-4 \cdot 23 + 8 \cdot 16 + (-13) \cdot 11 + (-4) \cdot 28) - (28 \cdot 8 + 23 \cdot (-13) + 16 \cdot (-4) + 11 \cdot (-4)) \right| \\
&= \frac{1}{2} \left| ( -92 + 128 - 143 - 112) - ( 224 - 299 - 64 - 44) \right| \\
&= \frac{1}{2} \left| -219 -115 \right| \\
&= \frac{1}{2} \left| -104 \right|\\
&= 52
\end{aligned}
\][/tex]

Thus, the area is:
[tex]\[
\text{Area} = 52 \text{ square units}
\][/tex]

So, the answers are:
[tex]\[
\begin{aligned}
\text{Part A: Perimeter} &= 62.4 \text{ units} \\
\text{Part B: Area} &= 52 \text{ square units}
\end{aligned}
\][/tex]

Thanks for taking the time to read A polygon has vertices tex W 4 28 X 8 23 Y 13 16 tex and tex Z 4 11 tex Part A What is. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada