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