Question

In a triangle,$\angle BAC={90}^{\circ }$ and $AD\perp BC.$

If AC=13.5 cm and CD =4.5cm; calculate the length of 2BC

Answer 1

In $△ADC$ and $△BAC$,
$\angle ADC=\angle BAC$ (Each ${90}^{\circ }$)
$\angle DCA=\angle BCA$ (Common)
$\angle DAC=\angle ABC$ (Third angle)
Thus, $△ADC\sim △BAC$ (AAA rule)
$\frac{AC}{BC}=\frac{CD}{AC}$ (Corresponding sides)
$A{C}^2=CD\times BC$
$(13.5{)}^2=4.5\times BC$
$BC=40.5cm$

$2BC=81cm$