Question

The city council is planning a new city park on a triangular plot of land
formed by three intersecting streets.
The designer placed the triangle on a grid to determine the side lengths and
angle measures needed for the purchase and placement of a fence
surrounding the park. Each unit on the grid is one inch, which represents
20 feet of actual length.
The measure of angle CAB in the designer’s diagram is (5x + 1)°, and the
measure of angle ACB is (3x + 12.5)°.


Create a report that provides a detailed
description of the park. Include the following in your report:
• the actual side lengths of the triangular plot of land
• the type of triangle formed by the intersection of the three streets
• the amount of fencing needed to surround the city park
• the actual measure of each angle of the triangle

Answer 1

Answer:

• AB = AC = 172.2 feet , CB = 200 feet
• Isosceles triangle
• Perimeter of the park = 544.4 feet
• ∠CAB = 71° , ∠ACB = ∠ABC = 54.4°

Step-by-step explanation:

The actual side lengths of the triangular plot of land:

The length of AC = $\sqrt{(6 -1)^2 + (8 - 1)^2}  = \sqrt{25 + 49}  = \sqrt{74}$ inches

The length of AB = $\sqrt{(11 -6)^2 + (1 - 8)^2}  = \sqrt{25 + 49}  = \sqrt{74}$ inches

The length of CB = 11 inches - 1 inch = 10 inches

The type of triangle formed by the intersection of the three streets:

Since AC = AB , the triangle ACB is an isosceles triangle.

Actual measure of each angle of the triangle:

∠ACB = ∠ABC = 3x + 12.5 = 3x + 12.5 (isosceles triangle has two equal angles)

(5x + 1)° + 2(3x  + 12.5)° = 180° (angles in a triangle adds up to 180°)

5x + 6x + 1 + 25 = 180°

11x = 154

x = 14

∠CAB = (5(14) + 1)° = 71°

∠ACB = ∠ABC = 3(14) + 12.5 = 54.5°

The actual side lengths of the triangular plot of land:

Given; 1 inch = 20 feet

CB = 10 inches = 200 feet

$\frac{CB}{SINangle CAB}=\frac{AB}{SINangle ACB}$

i.e $\frac{200}{sin71} = \frac{AB}{sin 54.5}$

AB = AC = ( 200 × sin 54.5°) ÷ sin 71° = 172.2 feet (correct to one decimal place)

The amount of fencing needed to surround the city park

:

The perimeter of the city pack = CB + AB + AC = 200 feet + 2(172.2) feet = 544.4 feet