lato obliquo=(150 - (57 + 25))/2 = 34 dm
altezza trapezio isoscele= √(34^2 - 16^2) = 30 dm
Area=1/2·(57 + 25)·30 = 1230 dm^2
158)
Perimetro $2p= 150\,dm;$
base maggiore $(B)$ $AB= 57\,dm;$
base minore $(b)$ $CD= 25\,dm;$
proiezione lato obliquo $(pl)$ $AH=CD=\dfrac{AB-CD}{2} = \dfrac{57-25}{2} = 16\,dm;$
lato obliquo $(l)$:
$BC=DA=\dfrac{2p-(AB+CD)}{2} = \dfrac{150-(57+25)}{2} = \dfrac{150-82}{2}=34\,dm;$
altezza $(h)$ $DH=CK= \sqrt{(BC)^2-(KB)^2}=\sqrt{34^2-16^2} = 30\,dm$ (teorema di Pitagora);
area $A= \dfrac{(B+b)×h}{2} = \dfrac{(AB+CD)×CK}{2} = \dfrac{(57+25)×30}{2} = 1230\,dm^2.$