Step-by-step solution-:

The area of a triangle is given by:

A=\frac{1}{2}(b h)

Where **b** is the base of a triangle and **h** is the height of the triangle.

Differentiate on both sides with respect to t:

\textcolor{primary}{\frac{d}{dt}}(A)=\textcolor{primary}{\frac{d}{dt}}(\frac{1}{2}b h)

By The **Product Rule for Derivatives**:

{\frac{d}{dt}}(A)=\frac{1}{2}\left({(\textcolor{secondary}{h})\frac{d}{dt}}(\textcolor{primary}{b})+(\textcolor{primary}{b}){\frac{d}{dt}}(\textcolor{secondary}{h)}\right)

The base of the triangle is decreasing:

It means

\frac{d}{dt}{({b})}={-13}

\frac{d}{dt}{(\textcolor{primary}{b})}=\textcolor{primary}{-13}, {\frac{d}{dt}}(\textcolor{secondary}{h)}=\textcolor{secondary}{6},\textcolor{primary}{b=5} ,\textcolor{secondary}{h=1}

After substituting the above expression we get:

{\frac{d}{dt}}(A)=\frac{1}{2}\left({(\textcolor{secondary}{1})}(\textcolor{primary}{-13})+(\textcolor{primary}{5}){}(\textcolor{secondary}{6)}\right)\implies {\frac{d}{dt}}(A)=\frac{17}{2}=8.5

The required rate of change of the area of a triangle is:

{\frac{d}{dt}}(A)=8.5~square~millimeters ~per~minute

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