Pitagora 2 media

perimetro 2p = 432 = lo+lo+(lo-36)

468 = 3lo

lato obliquo lo = 468/3 = 156 cm

base b = 156-36 = 120 cm 

altezza h = √lo^2-(b/2)^2 = √156^2-60^2 = 144 cm

area A = b*h/2 = 144*60 = 8.640 cm^2

2l+b=432   l-b=36   l=b+36   2b+72+b=432   b=120   l=120+36=156

h=V 156^2-60^2=144   A=120*144/2=8640cm2

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Base $\small b= \dfrac{432-2×36}{3} = \dfrac{432-72}{3} = \dfrac{360}{3} = 120\,cm;$

ciascun lato obliquo $\small l= 120+36 = 156\,cm;$

altezza $\small h= \sqrt{l^2-\left(\dfrac{b}{2}\right)^2} = \sqrt{156^2-\left(\dfrac{120}{2}\right)^2} = \sqrt{156^2-60^2} = 144\,cm$ (teorema di Pitagora);

area $\small A= \dfrac{b×h}{2} = \dfrac{\cancel{120}^{60}×144}{\cancel2_1} = 60×144 = 8640\,cm^2.$