Question

Un triunghi dreptunghic ABC are cateta AB = 8 cm  si ipotenuza BC = 20 cm. Calculati distanta  de la punctul A la BC

Answer 1

$AB^2+AC^2=BC^2 \Rightarrow AC= \sqrt{BC^2-AB^2}= \sqrt{20^2-8^2}= \\ \\ = \sqrt{400-64}= \sqrt{336}= 4 \sqrt{21}~(cm). \\ \\ d(A,BC)= \frac{AB \cdot AC}{BC}= \frac{8 \cdot 4 \sqrt{21}}{20}= \frac{8 \sqrt{21}}{5}~(cm).$