The perimeter of a triangle is 16 text{ cm},o | vedclass

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(B) Given that the perimeter $2s = 16 	ext{ cm}$,so the semi-perimeter $s = 8 	ext{ cm}$.Let the sides be $a, b, c$. Given $a = 6 	ext{ cm}$ and ar

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isosceles

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(B) Given that the perimeter $2s = 16 	ext{ cm}$,so the semi-perimeter $s = 8 	ext{ cm}$.Let the sides be $a, b, c$. Given $a = 6 	ext{ cm}$ and area $\Delta = 12 	ext{ cm}^2$.Using Heron's formula: $\Delta^2 = s(s-a)(s-b)(s-c)$.$144 = 8(8-6)(8-b)(8-c)$.$144 = 8(2)(8-b)(8-c) \Rightarrow 144 = 16(8-b)(8-c)$.$9 = (8-b)(8-c)$.Since $a+b+c = 16$ and $a=6$,we have $b+c = 10$,so $c = 10-b$.Substituting $c$: $9 = (8-b)(8-(10-b)) = (8-b)(b-2)$.$9 = 8b - 16 - b^2 + 2b \Rightarrow b^2 - 10b + 25 = 0$.$(b-5)^2 = 0 \Rightarrow b = 5 	ext{ cm}$.Then $c = 10 - 5 = 5 	ext{ cm}$.Since two sides are equal $(b=c=5 	ext{ cm})$,the triangle is isosceles.

The perimeter of a triangle is $16 \text{ cm}$,one of the sides is of length $6 \text{ cm}$. If the area of the triangle is $12 \text{ cm}^2$,then the triangle is:

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