The perimeter of a triangle is $16 \mathrm{~cm}$, one of the sides is of length $6 \mathrm{~cm}$. If the area of the triangle is $12 \mathrm{sq} \mathrm{cm}$. Then the triangle is

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Aicharya · Accepted Answer

Given that $ \begin{aligned} & 2 s=16 \Rightarrow s=8 \mathrm{~cm} \ & 2 s=a+b+c, a=6 \mathrm{~cm}, \Delta=12 \ & \Delta^2=s(s-a)(s-b)(s-c) \ & 144=8(8-6)(8-b)(8-c) \ & {\left[\begin{array}{r} \because a+b+c=16 \ b+c=10 \ b=10-c \end{array}ight]} \ & 18=2(8-b)(8-c) \ & 18=2(8-10+c)(8-c) \ & 9=(-2+c)(8-c) \ & 9=-16+10 c-c^2 \Rightarrow c^2-10 c+25=0 \ & \Rightarrow \quad(c-5)^2=0 \ & \Rightarrow \quad c=5 \ & \end{aligned} $ Then, $b=10-5=5 \Rightarrow b=c$ Therefore, triangle is isosceles

The perimeter of a triangle is $16 \mathrm{~cm}$, one of the sides is of length $6 \mathrm{~cm}$. If the…

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