[Risolto] GEOMETRIA

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Base maggiore (B) $AB=14\,cm;$

base minore (b) $DC= AB-8 = 14-8 = 6\,cm;$

altezza (h) $DH= 9\,cm;$ 

area del trapezio isoscele:

$A= \dfrac{(B+b)·h}{2} = \dfrac{(AB+CD)·DH}{2}=\dfrac{(14+6)×9}{2}=\dfrac{20×9}{2}=90\,cm^2.$