# 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 fiel - Mathematics

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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 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 xx 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 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

Concept: Areas of Sector and Segment of a Circle
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 11 Area Related To Circles
Exercise 11.4 | Q 3 | Page 132
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