[Risolto] Geometria grazie

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Ipotenusa $\overline{AC}= 52\,cm;$

cateto minore $\overline{BC}= \dfrac{1}{4}×\overline{AC}+7 = \dfrac{1}{4}×52+7 = 13+7 = 20\,cm;$

cateto maggiore $AB= \sqrt{(\overline{AC})^2-(\overline{BC})^2} = \sqrt{52^2-20^2} = 48\,cm$ $(teorema\,di\,Pitagora);$

perimetro $2p= \overline{AC}+\overline{BC}+\overline{AB} = 52+20+48 = 120\,cm;$

area $A= \dfrac{\overline{AB}×\overline{BC}}{2} = \dfrac{48×20}{2} = 480\,cm^2.$

BC= 1/4*52= 13 +7= 20 cm 

AB ( con Pitagora) = radice quadrata 52^2 - 20^2= 2304= 48cm

Perimetro = 52+20+48= 120 cm

Area=  48*20/2=  480 cm quadrati