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(C) For a triangular field,the side lengths are $a = 51 \,m$,$b = 52 \,m$,and $c = 53 \,m$.First,calculate the semi-perimeter $(s)$:$s = \frac{a + b +

Vedclass · Accepted Answer

$1170$

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(C) For a triangular field,the side lengths are $a = 51 \,m$,$b = 52 \,m$,and $c = 53 \,m$.First,calculate the semi-perimeter $(s)$:$s = \frac{a + b + c}{2} = \frac{51 + 52 + 53}{2} = \frac{156}{2} = 78 \,m$.Now,calculate the differences:$s - a = 78 - 51 = 27 \,m$$s - b = 78 - 52 = 26 \,m$$s - c = 78 - 53 = 25 \,m$Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$= \sqrt{78 	imes 27 	imes 26 	imes 25}$$= \sqrt{(13 	imes 6) 	imes (9 	imes 3) 	imes (13 	imes 2) 	imes 25}$$= \sqrt{13^2 	imes 6 	imes 27 	imes 25}$$= \sqrt{13^2 	imes (2 	imes 3) 	imes (3^3) 	imes 5^2}$$= \sqrt{13^2 	imes 2 	imes 3^4 	imes 5^2}$$= 13 	imes 3^2 	imes 5 	imes \sqrt{2}$Wait,let's re-calculate the prime factorization:$78 = 2 	imes 3 	imes 13$$27 = 3^3$$26 = 2 	imes 13$$25 = 5^2$$	ext{Area} = \sqrt{(2 	imes 3 	imes 13) 	imes (3^3) 	imes (2 	imes 13) 	imes 5^2}$$= \sqrt{2^2 	imes 3^4 	imes 13^2 	imes 5^2}$$= 2 	imes 3^2 	imes 13 	imes 5$$= 2 	imes 9 	imes 13 	imes 5 = 1170 \,m^2$.

The shape of a field is a triangle. The measurements of the sides of the field are $51 \,m$,$52 \,m$ and $53 \,m$. Find its area. (in $,m^2$)

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