A worksheet has a area of 54 square inches and a perimeter of 30 inches what are the dimensions of the figure
+118613 A worksheet has a area of 54 square inches and a perimeter of 30 inches what are the dimensions of the figure
xy=54 y=54/x
2(x+y)=30 x+y=15
$$\\x+\frac{54}{x}=15\\
x^2+54=15x\\
x^2-15x+54=0\\$$
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{54}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{6}}\\
{\mathtt{x}} = {\mathtt{9}}\\
\end{array} \right\}$$
+118613 A worksheet has a area of 54 square inches and a perimeter of 30 inches what are the dimensions of the figure
xy=54 y=54/x
2(x+y)=30 x+y=15
$$\\x+\frac{54}{x}=15\\
x^2+54=15x\\
x^2-15x+54=0\\$$
$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{15}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{54}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{6}}\\
{\mathtt{x}} = {\mathtt{9}}\\
\end{array} \right\}$$
