Fie E,F∈ AB astfel incat DE _|_ AB si CF _|_ AB.
ABFE este dreptunghi => EF=AB=6cm.
AE=BF=[tex] \frac{AB-EF}{2} = \frac{10-6}{2}=2 (cm) [/tex].
Aplic teorema inaltimii in ΔACB- dreptunghic in C:
[tex] CF= \sqrt{AF*BF}= \sqrt{(AE+EF)FB}= \sqrt{(2+6)*2}= \sqrt{16}=4 (cm). [/tex]
Inaltimea are 4 cm.
