Question

12)Sa se afle aria Trapezului isoscel ABCD cu AB||CD ,AB=4cm ,DC=10cm si m(<C)=45°

Answer 1

raspunsul final :
Aria =
$= \frac{(10 + 4) \times 3}{2} = \frac{14 \times 3}{2} = 21cm$

Answer 2

Aria lui ABCD= (AB+CD)xBE/2= (4+10)x3/2=14x3/2= 21 cm²