An equilateral

An equilateral triangle with a side 10 m represents a wooden platform standing in a lawn. A goat is tied to a corner with a 15 m rope. What is the maximum amount of grazing area available to the goat?

Result

S =  663.557 m2

Solution:

a=10 m r=15 m  S1=3/4⋅ a2=3/4⋅ 102≐25 3 m2≐43.3013 m2  S2=π⋅ r2=3.1416⋅ 152≐706.8583 m2  S=S2−S1=706.8583−43.3013≐663.5571≐663.557 m2a=10 \ \text{m} \ \\ r=15 \ \text{m} \ \\ \ \\ S_{ 1 }=\sqrt{ 3 }/4 \cdot \ a^2=\sqrt{ 3 }/4 \cdot \ 10^2 \doteq 25 \ \sqrt{ 3 } \ \text{m}^2 \doteq 43.3013 \ \text{m}^2 \ \\ \ \\ S_{ 2 }=\pi \cdot \ r^2=3.1416 \cdot \ 15^2 \doteq 706.8583 \ \text{m}^2 \ \\ \ \\ S=S_{ 2 }-S_{ 1 }=706.8583-43.3013 \doteq 663.5571 \doteq 663.557 \ \text{m}^2

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