Odpowiedź:
Szczegółowe wyjaśnienie:
zad 4
a) twierdz. Pitagorasa
[tex]a^{2}[/tex] + [tex]b^{2}[/tex] = [tex]c^{2}[/tex]
[tex]2^{2}[/tex] + [tex]h^{2}[/tex] = [tex]3^{2}[/tex]
4 + [tex]h^{2}[/tex] = 9
[tex]h^{2}[/tex] =5
h = [tex]\sqrt{5}[/tex]
P = a*h
a = 2+5 = 7
P = 7[tex]\sqrt{5}[/tex]
b) [tex]3^{2}[/tex] + [tex]h^{2}[/tex] = [tex]4^{2}[/tex]
9 + [tex]h^{2}[/tex] = 16
[tex]h^{2}[/tex] = 7
h = [tex]\sqrt{7}[/tex]
a = 3+4 = 7
P = 7[tex]\sqrt{7}[/tex]
zad 5
a) a = 6
P = a*b
[tex]a^{2}[/tex] + [tex]b^{2}[/tex] = [tex]c^{2}[/tex]
[tex]6^{2}[/tex] + [tex]b^{2}[/tex] = [tex]10^{2}[/tex]
36 + [tex]b^{2}[/tex] = 100
[tex]b^{2}[/tex] = 64
b = 8
P = 6 * 8 = 48
b) obliczamy bok rombu a
[tex]5^{2} + 5^{2} = a^{2}[/tex]
[tex]a^{2}[/tex] = 25 + 25
a = [tex]\sqrt{50}[/tex] = [tex]\sqrt{2*25}[/tex] = 5[tex]\sqrt{2}[/tex]
P = a*h
P = 5[tex]\sqrt{2}[/tex] * 5 = 25[tex]\sqrt{2}[/tex]
zad 6
a) [tex]a^{2} = 8^{2} + 3^{2}[/tex]
[tex]a^{2}[/tex] = 64 + 9
[tex]a^{2}[/tex] = 73
a = [tex]\sqrt{73}[/tex]
O = 4a = 4[tex]\sqrt{73}[/tex]
b) [tex]3x^{2} + x^{2} = 20^{2}[/tex]
[tex]4x^{2}[/tex] = 400
[tex]x^{2}[/tex] = 100
x = [tex]\sqrt{100}[/tex] = 10
O = 2x + 2*3x = 2x + 6x = 8x
O = 8 * 10 = 80
