Question

three coins are tossed within 10 M long rope at the three corners of a triangular field having side 42 m, 20m,34 m find the area of plot which can be grown by the cows also find the area of remaining field​

Answer 1

Answer:

157 m²

179 m²

Step-by-step explanation:

Three sides of triangle are 42 m , 20 m & 34 m

Length of each rope = 10m

shortest side = 20 m and half of this = 10 m

so there will be no overlapping of circular field grazed by cows.

Let say angle of triangle are a , b & c deg

Area grazed by cows will be

$= (\frac{a}{360} ) \times \pi \times R^{2} + (\frac{b}{360} ) \times \pi \times R^{2} + (\frac{c}{360} ) \times \pi \times R^{2}$

Where R = 10 m

$= (\frac{a+b+c}{360} ) \times \pi \times 10^{2}$

a + b + c = 180 deg  ( sum of angles of triangle)

$= (\frac{180}{360} ) \times \pi \times 10^{2}\\\\=(\frac{1}{2} ) \times 3.14 \times 10^{2}\\\\= 1.57 \times 100\\\\= 157 \:m^2$

Area of field grazed by cows = 157 m²

To find remaining field

Area of triangle need to be find

s = (42 + 20 + 34)/2 = 48

Area of triangle = √((48)(48-42)(48-20)(48-34))

=√48×6×28×14 = √2×2×2×6×6×2×14 = 2×2×6×14 = 336 m²

Remaining field = 336 - 157  = 179 m²