Question

three cows are tethred with 10m long rope at the corner of a triangular side 42m 20m and 34m find the area of the plot which can be grassed by the cow​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

For a triangle with the sides a, b, c, and of half perimeter s,

$\implies\boxed{ Area =\sqrt{s(s-a)(s-b)(s-c)}}\\ where, s=\frac{a+b+c}{2}$

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According to the Question three cows are tethered at the corner of a triangle of side 42 m,20 m and 34m,

therefore, the area of the plot which can be grasped by the cow will be equal to the area of the triangle formed by their position.

$\therefore half of perimeter of the plot=s=\frac{42+20+34}{2}\:m\\ \implies s=\frac{\cancel{94}}{\cancel2}\ \implies \boxed{s=48\: m}$

$\therefore Area \:of\: the\: plot=\sqrt{48(48-42)(48-20)(48-34)}\\ \bf\red{\rightarrow}Area \:of\: the\: plot =\sqrt{48\times6\times28\times14}\\ \bf\red{\rightarrow} Area \:of\: the\: plot=\sqrt{3\times4\times4\times3\times2\times3\times3\times2\times7}\\ \bf\red{\rightarrow} Area \:of\: the\: plot=4\times3\times3\times2\sqrt{7}\\ \bf\red{\rightarrow} Area \:of\: the\: plot=72\sqrt{7}\\ \bf\red{\rightarrow}\boxed{ \red{Area \:of\: the\: plot}=190.49 \:m{}^{2}\:(approx)}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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