Question

a triangular field of side 20m each, has a cow tied at each of it's three corners. if the length of each rope is 7m, find the area in which the three cows can graze?​

Answer 1

Answer:

136.44m2

Step-by-step explanation:

Answer 2

The area in which the three cows can graze = 77 sq. m

Explanation:

If each side of the triangular field  is 20 meters , then the triangle must be an equilateral triangle.

The interior angle of any equilateral triangle is 60°.

The length of each rope = 7m

For each cow , Area for graze = Area of sector with radius 7 m and central angle 60°.

Area of sector = $\dfrac{x}{360^{\circ}}\pi r^2$ , where x= central angle and r is radius.

Put r= 7 and x=60°

Area of each sector = $\dfrac{60^{\circ}}{360^{\circ}}(\dfrac{22}{7}) (7)^2$

$=\dfrac{1}{6}\times22\times7=\dfrac{77}{3}\ m^2$

i.e. Area for each cow to graze = $\dfrac{77}{3}\ m^2$

Since each cow is tied at each end of triangle.

⇒The area in which the three cows can graze = 3 x Area for each cow to graze

= $3\times\dfrac{77}{3}=77\ m^2$

Hence, the area in which the three cows can graze = 77 sq. m

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