Question

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.

Answer 1

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 pi"r"^2`

= `(∠"C")/360^circ xx pi xx (7)^2"m"^2`

Area of the sector with ∠B

= `(∠"B")/360^circ xx pi"r"^2`

= `(∠"B")/360^circ xx pi xx (7)^2"m"^2`

And area of the sector with ∠H

= `(∠"H")/360^circ xx pi"r"^2`

= `(∠"H")/360^circ xx pi xx (7)^2"m"^2`

Therefore, sum of the area (in cm²) 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 × 7

= 77 cm²

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 of 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`.