[Risolto] GEOMETRIA

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Base maggiore (B) $AB= 25\,cm;$

base minore (b) $CD= 15\,cm;$

altezza $DH= \dfrac{4}{3}·CD = \dfrac{4}{\cancel3_1}×\cancel{15}^5 = 4×5 =20\,cm;$

area del trapezio isoscele:

$A= \dfrac{(B+b)·h}{2} = \dfrac{(AB+CD)·DH}{2}=\dfrac{(25+15)×20}{2}=\dfrac{40×20}{2}=400\,cm^2.$