Question

A field in the shape of equilateral triangle of side 10.5m. A cow is tethered to the corner of the field on the outside with the rope of length 14 m. Find the area that the cow can graze.
​

Answer 1

In the diagram, the cow will only graze 'outside the 60° sector' and 'in a circle of a radius of 14 m'.

Now we should calculate area by:-

$\boxed{\sf{(Area)=(Whole\;Circle)-(60^{\circ}\;Sector)}}$

But.

Though this seems perfect, we didn't consider that the cow will go graze around the triangle. Here, we add more areas.

This gives us the correct method:-

$\boxed{\sf{(Area)=(Whole\;Circle)-(60^{\circ}\;Sector)+(Missed\;Area)}}$

• $\sf{(Whole\;Circle)-(Sector\;Area)=(300^{\circ}\;sector)}$ ...(1)

$\to \sf{(Area)=(300^{\circ}\;Sector)+(Missed\;Area)}$ From (1)

Now,

we calculate each area.

• $\boxed{\sf{(300^{\circ}\;Sector)=\dfrac{300^{\circ}}{360^{\circ}} \times\pi \times(14)^2=\dfrac{980}{6} \pi }}$
• $\boxed{\sf{(Missed\;Area)=2\times \frac{120^{\circ}}{360^{\circ}} \times\pi \times(3.5)^2=\frac{49}{6} \pi }}$

So the area is:-

$\sf{(Area)=\dfrac{1029}{6}\pi }$

$\sf{(Area)\approx \dfrac{1029}{6}} \times\dfrac{22}{7}}$ (Taking π≈22/7)

$\sf{(Area)\approx\dfrac{1617}{3}}$

$\sf{(Area)\approx539\;m^2}$

Used in Solving

The missed areas are two circular sectors. [Central angle 120°]

For your information

(-Edit: This problem doesn't require the triangle area.)

(+Edit: This is proof of an equilateral triangle area.)

The triangle area can be proven by the Pythagorean Theorem.

The perpendicular line drawn to one side will make two right triangles.

By letting one side of the triangle hypotenuse 'a', half of the side becomes 'a/2'.

$\boxed{\sf{(Hypotenuse)^2=(Base)^2+(Height)^2\;[Pythagorean Theorem]}}$

$\sf{\to a^2=(\dfrac{a}{2} )^2+(Height)^2}$

$\sf{\to(Height)^2=\dfrac{3}{4} a^2}$

$\sf{\to (Height)=\dfrac{\sqrt{3} }{2} a}$

$\boxed{\sf{(Triangle\;Area)=\dfrac{1}{2} \times(Base)\times(Height)}}$

$\sf{\to (Triangle\;Area)=\dfrac{1}{2}\times a \times \dfrac{\sqrt{3} }{2} a}$

$\sf{\to (Triangle\;Area)=\dfrac{\sqrt{3} }{4} a^2}$