Question

Construct an isosceles triangle whose altitude is $5.6$ cm and angle at the vertex is ${90}^o$.

Answer 1

Step 1: Draw a line segment $BC$ of any measure.

Step 2: Construct a perpendicular $AE$ of line segment $BC$ which is also the median.

Step 3: Cut $5.6cm$ at $AE$ and extend the point $A$ to $X{X}^{\prime }$.

Step 4: Construct ${90}^o$ at $A$ and bisect the angle, i.e $\angle BAE=\angle CAE={45}^o$. Extend the line $AB$ and $AC$ until it cuts line $BC$.

$\therefore$ We get the $△ABC$ with altitude $AE=5.6cm$ and vertex $A={90}^o$