Question

Sides of a triangular field are 15 16 and 17 at the corner of a field 1 hour and a horse

Answer 1

Answer:

Let ABC be the triangular field with sides Ac = 15 m, AB = 16 m and BC = 17 m

And,

Let the place where the cow, the buffalo and the horse are tied, are three sectors i.e. sector ADE, sector BFG and sector CHI 

The area of triangular field by Heron's formula =

√s(s-a)(s-b)(s-c)

s = (a+b+c)/2

s = (15+16+17)/2

s = 48/2

s = 24 m

√24(24-15)(24-16)(24-17)

√24*9*8*7

√12096

Area of triangular field = 109.98 sq m

Area of the grazed part = Area of the triangular field = Area of the sector ADE + Area of sector BFG + Area of sector CHI

= π*7²*∠A/360 + π*7²*∠B/360 + π*7²*∠C/360

= π*7²(∠ A + ∠ B + ∠ C)/360

= 22/7*7*7*180/360

= 154/2

Area of the grazed part = 77 sq m

Now, the area of the field which cannot be grazed by these animals 

= 109.98 - 77

= 32.98 sq m

Answer.