The altitude of a triangle is 2 cm shorter that its base. The area is 15 cm squared. Find the base of the triangle.
+130511 Let b = the base ....then the height = b-2 ....so....
15 = (1/2(b)(b-2) multiply both sides by 2
30 = b^2 - 2b rearranging, we have
b^2 - 2b - 30 = 0 using the on-site solver, we have....
$${{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,-\,}}{\mathtt{30}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{b}} = {\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{31}}}}\\
{\mathtt{b}} = {\sqrt{{\mathtt{31}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{b}} = -{\mathtt{4.567\: \!764\: \!362\: \!830\: \!021\: \!9}}\\
{\mathtt{b}} = {\mathtt{6.567\: \!764\: \!362\: \!830\: \!021\: \!9}}\\
\end{array} \right\}$$
So the base is (√31 + 1)cm [and the height = (√31 - 1)cm]
+130511 Let b = the base ....then the height = b-2 ....so....
15 = (1/2(b)(b-2) multiply both sides by 2
30 = b^2 - 2b rearranging, we have
b^2 - 2b - 30 = 0 using the on-site solver, we have....
$${{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,-\,}}{\mathtt{30}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{b}} = {\mathtt{1}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{31}}}}\\
{\mathtt{b}} = {\sqrt{{\mathtt{31}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{b}} = -{\mathtt{4.567\: \!764\: \!362\: \!830\: \!021\: \!9}}\\
{\mathtt{b}} = {\mathtt{6.567\: \!764\: \!362\: \!830\: \!021\: \!9}}\\
\end{array} \right\}$$
So the base is (√31 + 1)cm [and the height = (√31 - 1)cm]
