r%7Bmagenta%7D%7B%20%5Cmaltese%7B%20%5Cunderline%7B%20%5Ccolor%7Bpink%7D%7B%20%5Ctextsf%7B%20%5Ctextbf%7BPRACTICAL%20GEOMETRY%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D%7D" id="TexFormula1" title="{ \fcolorbox{cyan}{black}{ \huge{ \boxed{ \mathbb{ \color{magenta}{ \maltese{ \underline{ \color{pink}{ \textsf{ \textbf{PRACTICAL GEOMETRY}}}}}}}}}}}" alt="{ \fcolorbox{cyan}{black}{ \huge{ \boxed{ \mathbb{ \color{magenta}{ \maltese{ \underline{ \color{pink}{ \textsf{ \textbf{PRACTICAL GEOMETRY}}}}}}}}}}}" align="absmiddle" class="latex-formula"> construct a triangle ABC in which AB = 6cm media CD to AB = 4cm and altitude CE to AB = 3cm
➢ 1. Draw base AB of 6cm then construct 90° on A then take a compass and open it 4cm then from point A cut the arc on that 90° and name that point C. Join B to C triangle ABC will be the required triangle.