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

#### Solution

Given that, a triangular field with the three corners of the field a cow, a buffalo and a horse are tied separately with ropes.

So, each animal grazed the field in each corner of triangular field as a sectorial form.

Given, radius of each sector (r) = 7 m

Now, area of sector with ∠C = `(∠C)/360^circ xx pir^2 = (∠C)/360^circ xx pi xx (7)^2m^2`

Area of the sector with ∠B = `(∠B)/360^circ xx pir^2 = (∠B)/360^circ xx pi xx (7)^2 m^2`

And area of the sector with ∠H = `(∠H)/360^circ xx pir^2 = (∠H)/360^circ xx pi xx (7)^2m^2`

Therefore, sum of the area (in cm^{2}) of the three sectors

= `(∠C)/360^circ xx pi xx (7)^2 + (∠B)/360^circ xx pi xx (7)^2 + (∠H)/360^circ xx pi xx (7)^2`

= `((∠C + ∠B + ∠H))/360^circ xx pi xx 49`

= `180^circ/360^circ xx 22/7 xx 49`

= `11 xx 7`

= 77 cm^{2}

GIven that, sides of triangle are a = 15, b = 16 and c = 17

Now, semi-perimeter of triangle, s = `(a + b + c)/2`

⇒ `(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)`

= `8 xx 3sqrt(21)`

= `24sqrt(21) m^2`

So, area o the field which cannot be grazed by the three animals

= Area of triangular field - Area of each sectorial field

= `24sqrt(21) - 77 m^2`

Hence, the required area of the field which cannot be grazed by the three animals is `(24sqrt(21) - 77)m^2`