\(y=-\dfrac{1}{4}x^4+2x^2\Rightarrow y'=-x^3+4x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\Leftrightarrow y=0\left(M\right)\\x=2\Leftrightarrow y=4\left(N\right)\\x=-2\Leftrightarrow y=4\left(P\right)\end{matrix}\right.\)
Tính được: \(c=MN=\sqrt{\left(x_M-x_N\right)^2+\left(y_M-y_N\right)^2}\)
\(=\sqrt{\left(0-2\right)^2+\left(0-4\right)^2}=2\sqrt{5}\)
Tương tự, cũng tính được \(b=MP=\sqrt{\left(x_M-x_P\right)^2+\left(y_M-y_P\right)^2}=2\sqrt{5}\)
\(a=NP=\sqrt{\left(x_N-x_P\right)^2+\left(y_N-y_P\right)^2}=4\)
\(\Rightarrow p=\dfrac{a+b+c}{2}=2+2\sqrt{5}\)
Công thức Heron: \(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)
Mà: \(S=pr\Rightarrow r=\dfrac{\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}}{p}=\sqrt{5}-1\approx1,24\)
