Question

Aflati inaltimea trapezului isoscel ABCD in care AC_|_BC,AB=10 cm si CD=6cm.

Answer 1

Fie E,F∈ AB astfel incat DE _|_ AB si CF _|_ AB.
ABFE este dreptunghi => EF=AB=6cm.
AE=BF=$\frac{AB-EF}{2} = \frac{10-6}{2}=2 (cm)$.

Aplic teorema inaltimii in ΔACB- dreptunghic in C:

$CF= \sqrt{AF*BF}= \sqrt{(AE+EF)FB}= \sqrt{(2+6)*2}= \sqrt{16}=4 (cm).$

Inaltimea are 4 cm.