Question

calculati aria paralelogramului ABCD , Ac intersectat cu BD={O}

 1 AB=6 radical din 6 cm, BC=2 radical din 2 cm si m(<ABC)=60 grade
 2.AB=13 cm BC= 20 cm , AC=21 cm

Answer 1

1.AB=6 √6 cm
 BC=2 √2 cm
m(<ABC)=60 grade
Aria ABCD= AB*BC*sin (<ABC)=6√6*2√2*sin 60=24*√3/2=12√3

2.AB=13 cm BC= 20 cm , AC=21 cm
Ducem AM_|_BC si avem:
in ΔAMC
AM²=AC²-MC²=21²-(20-BM)²=441-400-BM²+40*BM=41-BM²+40*BM
in ΔABM avem
AM²=AB²-BM²=13²-BM²=169-BM²

=> 41-BM²+40*BM=169-BM²
40*BM=169-41=128
BM=128/40=16/5

AM²=AB²-BM²=13²-(16/5)²=169-256/25==(4225-256)/25=3969/25
AM=63/5

Aria ABCD=BC*AM=20*63/5=252