Answers

2015-03-09T10:57:32+05:30
3 areas of 3 horses = 3 sectors of a circle
with radius= 7m
let the 3 angles be Ф₁, Ф₂ & Ф₃
now, 
area = (pi * r² * Ф₁/360) + (pi * r² * Ф₂/360) + (pi * r² * Ф₃/360)
on taking common
22/7* 49 * 1/360 ( Ф₁ + Ф₂ +  Ф₃)
22/7* 49 * 1/360 * 180 (angle sum prop of triangle)
on solving, we get 
area = 77 m²
0
2015-03-09T12:04:35+05:30

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See diagram.

The area that can be grazed by the horse at each vertex is the area of the sector of radius 7 m at each vertex. 
To find that area we need to know the angle at each vertex.
We use the cosine rule in a triangle as we know the lengths of the sides.

AC² = AB² + BC² - 2 AB * BC * Cos B
20² = 34² + 42² - 2 * 34 * 42 * Cos B
Cos B = 0.88235      => B = 28.07⁰

AB² = AC² + BC² - 2 AC * BC * Cos C
34² = 42² + 20² - 2 * 42 * 20 * Cos C
Cos C = 0.6          => C = 53.13°
A = 180° - B - C = 98.80°

Area grazed by the horse at the vertex A = (π * 7²) * (98.80°/360°) m²
           = 42.247 m²
Area grazed by the horse at the vertex B =  (π * 7² * (28.07°/360°) m²
         = 12 m²
Area grazed by the horse at the vertex C = (π 7² * (53.13°/360°) m²
       = 22.718 m²

Total area of the triangle ABC can be found by Heron's formula as:
  s = semi perimeter = (AB+BC+CA)/2 =  48 m
Area\ \Delta ABC=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{48*6*28*14}=336\ m^2

The area left ungrazed is = 336 - 22.718 - 12 - 42.247 = 259.035 m²

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