Problem 6
Jim is planning to fence off this triangular plot of land and divide it into 3 sections. Triangle ACE and Triangle BCE are 30°-60°-90° triangles. How much fencing does Jim need to fence the area off and divide it into 3 parts.
ECD is a 45°-45°-90° triangle.
That means ED is 10(Sqrt 2) m, EC is 10 m and CD is 10 m
Triangle BCE is 30°-60°-90°. That means that BE is 5 m, EC is 10 m and BC is 5(Sqrt 3)
Triangle ACE is 30°-60°-90°. That means that AE is 20 m, EC is 10 m and AC is 10(Sqrt 3)
In order to find out the total amount of fencing we need to add the lengths of AE, ED, AC, CD, BC and CE
AE is 20 m, ED is 10(Sqrt 2) m, AC is 10(Sqrt 3), CD is 10 m, BC is 5(Sqrt 3), EC is 10 m
20+10+10=40m
10(Sqrt 3)+5(Sqrt 3)=15(Sqrt 3)
10(Sqrt 2)
The amount of fencing Jim needs=40+10(Sqrt 2)+15(Sqrt 3)
That means ED is 10(Sqrt 2) m, EC is 10 m and CD is 10 m
Triangle BCE is 30°-60°-90°. That means that BE is 5 m, EC is 10 m and BC is 5(Sqrt 3)
Triangle ACE is 30°-60°-90°. That means that AE is 20 m, EC is 10 m and AC is 10(Sqrt 3)
In order to find out the total amount of fencing we need to add the lengths of AE, ED, AC, CD, BC and CE
AE is 20 m, ED is 10(Sqrt 2) m, AC is 10(Sqrt 3), CD is 10 m, BC is 5(Sqrt 3), EC is 10 m
20+10+10=40m
10(Sqrt 3)+5(Sqrt 3)=15(Sqrt 3)
10(Sqrt 2)
The amount of fencing Jim needs=40+10(Sqrt 2)+15(Sqrt 3)