Question

NO LINKS!!

Please help me with this graphs

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Answer 1

• triangle: 18.81 units
• parallelogram: 17.21 units

Explanation:

You want the perimeter of each of the figures defined by the coordinates of their vertices. You are told to use the distance formula as necessary.

Triangle

The first attachment shows the graph of the triangle. The distance formula is needed only for the length of the diagonal segment:

d = √((x2 -x1)² +(y2 -y1)²)

d = √((3 -(-2))² +(-3 -3)²) = √(5² +(-6)²) = √61 ≈ 7.81

The lengths of the horizontal and vertical legs of the triangle are the difference of their x- and y-coordinates, respectively.

CB = 3 -(-2) = 5

CA = 3 -(-3) = 6

The perimeter is the sum of the side lengths:

P = CA +CB +AB = 6 +5 +7.81 = 18.81

The perimeter of the triangle is 18.81 units.

Parallelogram

The second attachment shows the graph of the parallelogram. As with the triangle, we only need to use the distance formula for the length of the diagonal side. Here is the length of ML.

d = √((4 -2)² +(1 -(-2))²) = √(2² +3²) = √13 ≈ 3.606

The length of the horizontal legs is the difference of their x-coordinates.

KL = 4 -(-1) = 5

Opposite sides are congruent, so the perimeter is double the length of two adjacent sides.

P = 2(3.606 +5) ≈ 17.21

The perimeter of the parallelogram is about 17.21 units.

Answer 2

1. 18.81 units

2. 17.21 units

Explanation:

`\boxed{\begin{minipage}{7.3 cm}\underline{Distance between two points}\\\\$d=√((x_2-x_1)^2+(y_2-y_1)^2)$\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}`

Question 1

Given vertices of ΔABC:

• A = (-2, 3)
• B = (3, -3)
• C = (-2, -3)

Plot the vertices on the given graph paper and join with line segments to create the triangle.

As points B and C share the same y-coordinate:

`\implies BC = |x_B-x_C|=|3-(-2)|=5\:\: \sf units`

As points A and C share the same x-coordinate:

`\implies AC = |y_A-y_C|=|3-(-3)|=6\:\: \sf units`

Use the distance formula to find the length AB:

`\begin{aligned}AB & =√((x_B-x_A)^2+(y_B-y_A)^2)\\& =√((3-(-2))^2+(-3-3)^2)\\& =√((5)^2+(-6)^2)\\& =√(25+36)\\& =√(61)\\\end{aligned}`

The perimeter of a two-dimensional shape is the distance all the way around the outside.

`\begin{aligned}\textsf{Perimeter of $ABC$} & = AB + BC + AC\\& = √(61)+5+6\\& = 11+√(61)\\& = 18.81\:\: \sf units\:(nearest\:hundredth)\end{aligned}`

Question 2

Given vertices of polygon KLMN:

• K = (-1, 1)
• L = (4, 1)
• M = (2, -2)
• N = (-3, -2)

Plot the vertices on the given graph paper and join with line segments to create the polygon.

As the y-coordinate of points K and L, and M and N are the same, KL and MN are parallel line segments.

As the difference between the x-coordinates of K and N, and L and M is 2 units, KN and LM are parallel line segments.

Therefore, the polygon is a parallelogram.

A parallelogram has two pairs of opposite sides that are equal in length.

Therefore, KL = NM and KN = LM.

As points K and L share the same y-coordinate:

`\implies KL = |x_K-x_L|=|-1-4|=5\:\: \sf units`

Use the distance formula to find the length KN:

`\begin{aligned}KN & =√((x_N-x_K)^2+(y_N-y_K)^2)\\& =√((-3-(-1))^2+(-2-1)^2)\\& =√((-2)^2+(-3)^2)\\& =√(4+9)\\& =√(13)\\\end{aligned}`

The perimeter of a two-dimensional shape is the distance all the way around the outside.

`\begin{aligned}\textsf{Perimeter of $KLMN$} & = 2\:KL + 2 \:KN\\& = 2 \cdot 5 + 2\cdot √(13)\\& =10 + 2√(13)\\& = 17.21\:\: \sf units\:(nearest\:hundredth)\end{aligned}`