A small town in Louisiana is over 300 years old. Over the years as the town grew, the property owners established the property lines informally using landmarks to determine property division. However, due to a recent property dispute, the town council has drafted a new law that all property markers and boundaries need to be well defined by modern standards. The law also states property divisions must be fenced. Along with the boundaries and fences, the town council is also instituting a yearly property tax to help build a new school. Property tax will be assessed at the rate of $1.00 per 100 square feet. Landowners can offset their property tax burden this year by the amount it will cost them to build a fence around their property. To establish the taxes and build fences, people need to know the perimeter and area of their property. Mr. Augustin (A), Mrs. Bertrand (B), Mrs. Chaisson (C), and Old Farmer Daigle (D) have hired a land surveyor to mark and measure their property lines so they can install fences and pay their taxes. Use the property shapes and dimensions in the following image to answer the questions below (all dimensions in the image are in feet. Drawing is not to scale; all of the sub-shapes are triangles, rectangles, parallelograms, rhombi, or trapezoids.):

PICTURES ARE ATTACHED!!!

a. Determine each owner’s tax rate. Explain how you derived your answers.
b. Determine how many feet of fencing each owner will need to enclose their property. Explain how you derived your answers.

A small town in Louisiana is over 300 years old Over the years as the town grew the property owners established the property lines informally using landmarks to class=

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SHAPE  A

Shape A can be split into two shapes as shown in the diagram below. One is a rectangle and the other is an isosceles triangle. The dimension of the rectangle is 80×180 but we do not know the base of the triangle.

Area of rectangle = 80×180 = 14400

Base of rectangle can be worked out using the Pythagoras theorem
[tex]x= \sqrt{65^{2}- 60^{2} } =25[/tex]
the base of the triangle = 2×25 = 50

Area of rectangle = 1/2(base)(height) = 0.5×50×60 = 1950

Total area of shape A = 1950 + 14400 = 16350
Tax = 16350÷100 = $163.50
Total perimeter of shape A = 80+80+180+65+65 = 470

SHAPE B

We split shape B into a parallelogram and a trapezium
Area of parallelogram = base × height = 250 × 30 = 7500
Area of trapezium = [tex] \frac{a+b}{2} [/tex]×[tex]h[/tex] =[tex] \frac{100+40}{2} [/tex]×[tex]80=5600[/tex]

Total area = 5600 + 7500 = 13100
Tax = 13100÷100 = $131

Perimeter  = 250+250+35+35+100+40+80 = 790

SHAPE C

Shape C is an isosceles shape which allows us to split side bc into 100÷2=50. The diagram is shown in the third diagram.

Quadrilateral PODY = Quadrilaterla AXPO
Side CY = [tex] \sqrt{107^{2}- 100^{2} } =38.1[/tex]
Side PY = 50+38.1 = 88.1
Side DS = [tex] \sqrt{117^{2}- 88.1^{2} }= 77[/tex] (rounded to whole number)
Side OP = 100-77=23

Area of PODY = [tex] \frac{100+23}{2}*88.1=5418.15 [/tex]
Area of DCY = [tex] \frac{1}{2}*38.1*100=1905 [/tex]
Area of DCPO = 5418.15-1905 = 3513.15
Area of ABCD = 3513.15×2 = 7026.3

Tax = 7026.3÷100 = $70.263

Perimeter = 117+117+107+107+100 = 548


SHAPE D

Shape D can be split into a triangle and a trapezium as shown in the fourth diagram below

Area of triangle = [tex] \frac{1}{2}*150*300=22500 [/tex]
Area of trapezium = [tex] \frac{500+300}{2}*400=160000 [/tex]
Total area = 22500 + 160000 = 182500

Tax = 182500÷100 = $1825

Perimeter = 280+165+406+500+425 = 1776

Ver imagen merlynthewhizz
Ver imagen merlynthewhizz
Ver imagen merlynthewhizz
Ver imagen merlynthewhizz