Sides of a triangular fiald are 15 m, 16m and 17m. With the three cormers of the field a cow, a buffalo and a horse are tied separately with ropes of length 7m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Explanation

Given that, a triangular field with the three corners of the field a cow, a buffalo and a horse are tied separatelp with ropes. So, each animal grazed the field in each corner of triangular field as a sectorial form.
Given, radius of each sectro ( r) = 7m
Now, area of sector with `angleC = (angleC)/(360^(@))xxpir^(2)=(angleC)/(360^(@))xxpixx(7)^(2) m^(2)`
Area of the sector with `angleB = (angleB)/(360^(@))xxpir^(2) = (angleB)/(360^(@))xxpixx(7)^(2)m^(2)`
and area of the sector with `angleH = (angleH)/(360^(@))xxpir^(2)=(angleH)/(360^(@))xxpixx(7)^(2)m^(2)`
Therfore, sum of the arese (in `cm^(2)` ) of the three sectors
`(angleC)/(360^(@))xxpixx(7)^(2)+(angleB)/(360^(@))xxpixx(7)^(2)+(angleH)/(360^(@))xxpixx(7)^(2)`
= `(angleC+angleB+angleH)/(360^(2))xxpixx49`.
= `(180^(@))/(360^(@))xx(22)/(7)xx49=11xx7 = 77 cm^(2)`
Given that, sides of triangle are a = 15,b = 16and c= 17
Now, spmi-perimeter of triangle, s = `(a+b+c)/(2)`
`rArr = (15 + 16 + 17)/(2) = (48)/(2) = 24`
`:.` Area of triangular field = `sqrt(s(s-a)(s-b)(s-c))` [by Heron's formula]
= `sqrt(24.9.8.7)`
= `sqrt(64. 9 . 21)`
= `8xx3 sqrt(21)=24sqrt(21)m^(2)`
So, area of the field which cannot be grazed by the three animals
= Area of triangular field - Area of each sectorial field
`24sqrt(21)-77m^(2)`
Hence, the required area of the field which can not be grazed by the three animals is `(24sqrt(21)-77)m^(2)`