zad. 20
|AG| = [tex]\frac{1}{4}[/tex] |AB|
|AG| = [tex]\frac{1}{4}[/tex] x 16 dm = 4 dm
|GH| = [tex]\frac{2}{4}[/tex]|AB|
|GH| = [tex]\frac{2}{4}[/tex] x 16 dm = 8 dm
Ośmiokąt EFGHIJKL dzielimy na 3 figury: trapez FGHI, prostokąt EFIJ, trapez EJKL
Pole trapezu FGHI
P = [tex]\frac{(a+b)h}{2}[/tex] = [tex]\frac{(8dm + 16dm)x 4dm}{2}[/tex] = 24 dm x 2 dm = 48 [tex]dm^{2}[/tex]
Trapez EJKL ma takie samo pole
Pole prostokąta EFIJ:
P = ab = 16 dm x 8 dm = 128 [tex]dm^{2}[/tex]
Pole ośmiokąta EFGHIJKL:
P = 2 x 48[tex]dm^{2}[/tex] + 128 [tex]dm^{2}[/tex] = 96 [tex]dm^{2}[/tex] + 128 [tex]dm^{2}[/tex] = 224 [tex]dm^{2}[/tex]
Odp.: Pole ośmiokąta wynosi 224 [tex]dm^{2}[/tex]
zad. 21
Przekątna graniastosłupa, wysokość graniastosłupa oraz przekątna podstawy (kwadratu) tworzą trójkąt prostokątny
d - przekątna kwadratu
[tex]7^{2}[/tex] + [tex]d^{2}[/tex] = [tex]11^{2}[/tex]
49 + [tex]d^{2}[/tex] = 121
[tex]d^{2}[/tex] = 121 - 49 = 72
d = [tex]\sqrt{72}[/tex] = 6[tex]\sqrt{2}[/tex] dm
d = a[tex]\sqrt{2}[/tex]
a = [tex]\frac{d}{\sqrt{2} }[/tex]
a = [tex]\frac{6\sqrt{2} }{\sqrt{2} }[/tex] = 6 dm
Pole podstawy: [tex]a^{2}[/tex]
P = [tex](6 dm)^{2}[/tex] = 36 [tex]dm^{2}[/tex]
Pole ściany: aH
P = 6 dm x 7 dm = 42 [tex]dm^{2}[/tex]
Powierzchnia całkowita:
Pc = 2 x Pp + 4 x S
Pc = 2 x 36 [tex]dm^{2}[/tex] + 4 x 42 [tex]dm^{2}[/tex] = 72 [tex]dm^{2}[/tex] + 168 [tex]dm^{2}[/tex] = 240 [tex]dm^{2}[/tex]
