Question

Inaltimea trapezului dreptunghic ABCD cu AB paralel cu CD , AB<CD,  dreptunghic in A , este de 8 cm , iar tgC=1.(3). Se stie ca AB=BC.
a) Calculati BC
b) Calculati aria trapezului ABCD.

Answer 1

construiesti inaltimea BM
In ΔBMC, tg C=BM/MC=1,(3)
BM/MC=4/3
8/MC=4/3⇒MC=6
In ΔBMC, aplicam Pitagora ⇒BC²=BM²+MC²⇒BC²=64+36=100
BC=10
B)
A=(AB+DC)*BM/2
In ΔDBC, aplicam th catetei⇒BC²=MC*DC⇒100=6*DC⇒DC=100/6=16,6
DM//AB⇒DM≡AB
DM=DC-MC
DM=16,(6)-6=10,6
A=(10,6+16,6)*8/2=108,8