A building with base in the form of equillateral triangle of side 14 m is built in a huge grass field . In one corner of a building a cow is tethered with a rope of length 21 m . Find the area grazed by the cow

Dear Student,

Please find below the solution to the asked query:

From given information we form our diagram , As :



Here , Sum of area of Sector OCD , Sector BDE and Sector ACE represent the area cow can graze , We have base of building is equilateral . ( As we know all three internal angles of equilateral triangle is at 60$°$ )

So, Cow can graze total area =  Area of Sector OCD + Area of Sector BDE + Area of Sector ACE

And we get radius of each sector from above diagram .

We know area of sector of circle  = $\frac{\mathrm{Center} \mathrm{Angle}}{360°}\times \mathrm{\pi } r^2$ , So

Area of Sector OCD  $\frac{300°}{360°}\times \frac{22}{7}\times {\left(21\right)}^2 = \frac{5}{6}\times \frac{22}{7}\times 21 \times 21 =\frac{5}{3}\times 11\times 3 \times 21 =5\times 11\times 21=\mathbf{ }\mathbf{1155}\mathbf{ }{\mathbf{m}}^{\mathbf{2}}$

And

Area of Sector BDE  $\frac{120°}{360°}\times \frac{22}{7}\times {\left(7\right)}^2 = \frac{1}{3}\times \frac{22}{7}\times 7\times 7 =\frac{1}{3}\times 22 \times 7 =\frac{154}{3}=\mathbf{ }\mathbf{51}\mathbf{.}\mathbf{33}\mathbf{ }{\mathbf{m}}^{\mathbf{2}}$

And

Area of Sector ACE  $\frac{120°}{360°}\times \frac{22}{7}\times {\left(7\right)}^2 = \frac{1}{3}\times \frac{22}{7}\times 7\times 7 =\frac{1}{3}\times 22 \times 7 =\frac{154}{3}=\mathbf{ }\mathbf{51}\mathbf{.}\mathbf{33}\mathbf{ }{\mathbf{m}}^{\mathbf{2}}$


So,

Cow can graze total area  = 1155 + 51.33 + 51.33 = 1257.66 m²                                   ( Ans )


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