Odpowiedź :
[tex]zad.1\\P_{1} =98~cm^{2} ~~\land~~P_{1}=a\cdot b~~\land~~a=14~cm\\\\14~cm\cdot b = 98~cm^{2} ~~\mid \div 14~cm\\\\b=7~cm\\\\Obw_{1} =2\cdot a + 2\cdot b~~\land ~~a=14~cm~~\land ~~b=7~cm~~\Rightarrow ~~Obw_{1} =42~cm\\\\P_{2} =98~cm^{2} ~~\land~~P_{1}=c\cdot d~~\land~~c=2~cm\\\\2~cm\cdot d = 98~cm^{2} ~~\mid \div 2~cm\\\\d=49~cm\\\\Obw_{2} =2\cdot c + 2\cdot d~~\land ~~c=2~cm~~\land ~~d=49~cm~~\Rightarrow ~~Obw_{1} =102~cm[/tex]
[tex]zad.2\\\\P_{trapezu} =\dfrac{(a+b)}{2} \cdot h \\\\a,b-podstwy~~trapezu\\h-wysokosc~~trapezu\\\\a+b=40\\\\h=(a+b)\cdot 40\% =40\cdot 0,4=16\\\\P_{trapezu} =\dfrac{40}{2} \cdot 16 =320~[j^{2} ]\\\\Odp:~~B.~~320[/tex]
[tex]zad.3\\\\P_{szesciokata~~foremnego} =6\cdot P\Delta_{rownobocznego} \\\\P\Delta_{rownobocznego} =\dfrac{a^{2} \sqrt{3} }{4} ~~\land~~a=3\sqrt{2} \\\\P\Delta_{rownobocznego} =\dfrac{(3\sqrt{2}) ^{2} \sqrt{3} }{4} =\dfrac{9\cdot 2\cdot \sqrt{3} }{4} =\dfrac{9\sqrt{3} }{2}~[j^{2} ] \\\\P_{szesciokata~~foremnego} =6\cdot \dfrac{9\sqrt{3} }{2}=27\sqrt{3}~ [j^{2} ]\\\\Odp:~~A.~~27\sqrt{3}[/tex]