HELPPP PLEASE!!!
Luke purchased a warehouse on a plot of land for his business. The gure
represents a plan of the land showing the location of the warehouse and parking
area. The coordinates represent points on a rectangular grid with units in feet.
(0, 60)
(70, 60)
(450)
(39,50)
Warehouse
(4, 20)
(39, 20)
Parking Area
(0, 0).
(50, 0)
(4,0)
Enter your answers to each part in the boxes
Part A
What is the perimeter of the plot of land rounded to the nearest tenth of a foot?
Part B
What is the area of the plot of land that does not include the warehouse and the parking area?
Part C
Luke is planning to put a fence along two interior sides of the parking area. The sides are
represented in the plan by the legs of the trapezoid What is the total length of fence needed to the
nearest tenth of a foot?

Answer:

A: 243.2 B: 1740 C: 42.8

Step-by-step explanation:

The  perimeter of the plot of land is 180 + 20√10. THe are aof the plot of land that doesn't include the warehouse and the parking area is 85400 sq. units. The total length of fence needed is 20 + √520 units

The shortest distance(length of the straight line segment's length connecting both given points) between points ( p,q) and (x,y) is:

[tex]D = \sqrt{(x-p)^2 + (y-q)^2} \: \rm units.[/tex]

Its slope would be:

[tex]m = \dfrac{y-q}{x-p}[/tex]

Slope of parallel lines are same. Slopes of perpendicular lines are negative reciprocal of each other.

For this case, the renamed image is given below.

The slope of AB is: [tex]m_{AB} = \dfrac{50-50}{39-4} = 0[/tex]

The slope of CD is: [tex]m_{CD} = \dfrac{20-20}{39-4} = 0[/tex]

That shows AB and CD are parallel. Similarly, we can show that AC and BD are parallel. Also, we can show that AB and AC are perpendicular.

Thus, ABCD is a rectangle.

|AB| = length of AB = [tex]\sqrt{(39-4)^2 + (50-50)^2} = 35 \: \rm units[/tex]

|AC| = length of AC = [tex]\sqrt{(4-4)^2 + (50-20)^2} = 30 \: \rm units[/tex]

Thus, |AB| and |CD| are of same length because ABCD is a rectangle, and so as |AC| and |BD|.

The area of ABCD = multiplication of length of its two adjacent sides= [tex]35 \times 30 = 1350 \: \rm unit^2[/tex]Area of Warehouse

Similarly, we can show that the sides EI and CD are parallel, thus making CEID a trapezoid. Also, CE is perpendicular to CD and EI, thus, its length can serve as height of the trapezoid CEID.

The length of CD = length of AB = |AB| = 35 units.

The length of EI = [tex]\sqrt{(50-4)^2 + (0-0)^2} = 46 \: \rm units[/tex]

The length of CE = [tex]\sqrt{(4-4)^2 + (20-0)^2} = 20 \: \rm units[/tex]

Thus, area of CEID = [tex]\dfrac{1}{2}(\text{Sum of parallel sides})\times \text{height}[/tex] = [tex]\dfrac{35 \times 46 \times20}{2} = 16100\: \rm unit^2[/tex] = Area of Parking Area

Similarly, IFGH is a trapezoid, and its height can be taken as the length of GF as GF is perpendicular to its parallel sides.

The length of GF = [tex]\sqrt{(0-0)^2 + (0-60)^2} = 60 \: \rm units[/tex]

The length of IF = [tex]\sqrt{(50-0)^2 + (0-0)^2} = 50 \: \rm units[/tex]

The length of GH = [tex]\sqrt{(0-70)^2 + (60-60)^2} = 70 \: \rm units[/tex]

Thus, area of IFGH = [tex]\dfrac{1}{2}(\text{Sum of parallel sides})\times \text{height}[/tex] = [tex]\dfrac{50 \times 70 \times 60}{2} = 105000\: \rm unit^2[/tex] = Area of Plot

Length of HI = [tex]\sqrt{(70-50)^2 + (60-0)^2} = 20\sqrt{10} \: \rm units[/tex]
Thus, perimeter of plot = |GF| + |IF| + |GH| + |HI| = 180 + 20√10 units

Area of the plot of land that does not include the warehouse and the parking area = 105000 - 16100 - 3500 = 85400 sq. units

The fencing is done on the legs of the trapezoid CEID which are CE and DI

Length of CE = 20 units,

Length of DI = [tex]\sqrt{(39-50)^2 + (20-0)^2} = \sqrt{521} \: \rm units[/tex]

Thus, the length of the fencing is [tex]20 + \sqrt{521}[/tex] units.

Thus, the  perimeter of the plot of land is 180 + 20√10. THe are aof the plot of land that doesn't include the warehouse and the parking area is 85400 sq. units. The total length of fence needed is 20 + √520 units

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