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
What is the distance between two points ( p,q) and (x,y)?
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]
What is the slope of a line which passes through points ( p,q) and (x,y)?
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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