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(B) Let the sides of the given triangle be $a = 12 \, cm$,$b = 17 \, cm$,and $c = 25 \, cm$.First,calculate the semi-perimeter $(s)$:$s = \frac{a + b

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(B) Let the sides of the given triangle be $a = 12 \, cm$,$b = 17 \, cm$,and $c = 25 \, cm$.First,calculate the semi-perimeter $(s)$:$s = \frac{a + b + c}{2} = \frac{12 + 17 + 25}{2} = \frac{54}{2} = 27 \, cm$.Now,calculate the differences:$s - a = 27 - 12 = 15 \, cm$$s - b = 27 - 17 = 10 \, cm$$s - c = 27 - 25 = 2 \, cm$Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s - a)(s - b)(s - c)}$$	ext{Area} = \sqrt{27 	imes 15 	imes 10 	imes 2} \, cm^2$$	ext{Area} = \sqrt{(9 	imes 3) 	imes (3 	imes 5) 	imes (5 	imes 2) 	imes 2} \, cm^2$$	ext{Area} = \sqrt{9 	imes 9 	imes 25 	imes 4} \, cm^2$$	ext{Area} = 3 	imes 3 	imes 5 	imes 2 = 90 \, cm^2$.

The measurements of the sides of a triangle are $12 \, cm$,$17 \, cm$,and $25 \, cm$. Find the area of this triangle. (in $, cm^2$)

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