Question

three horses are tethered with 7 meter long ropes at three corners of triangular field having sides 20 meter, 34 meter and 42 meter. find the area of the plot which can be grazed by horses. also find the area of the plot which remain ungrazed.

Answer 1

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

Answer 2

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