in trapez ducem inaltimile notate cu M si N
MN=AB=16
DM=NC=(28-16)/2=6
in Δ dreptunghic format de inaltime (ΔBNC) aplicam T Pitagora
BC²=NC²+BN² , unde NC=6, BN=4/5BC
6²+(4/5BC)²=BC²
aducem la acelasi numitor comun 5²
6²*5²+4²BC²=5²BC²
6²*5²+16BC²=25BC²
6²*5²=9BC²
6²*5²=3²BC²⇒ BC²=6²*5²/3²⇒BC=6*5/3⇒BC=30/3⇒BC=10