Odpowiedź:
a ) e = 10 f = 24
a² = 5² + 12² = 25 + 144 = 169
a = [tex]\sqrt{169} = 13[/tex]
Pole rombu
P = 0,5 e*f = 0,5*10*24 = 120
oraz
P = a*h = 13*h = 120 / : 13
h = [tex]\frac{120}{13} = 9 \frac{3}{13}[/tex]
=================
b ) d = 0,5 h = [tex]\frac{60}{13}[/tex]
a = y + x
więc
x² + d² = 12²
x² =144 - ( [tex]\frac{60}{13}[/tex] )² = [tex]\frac{24 336}{169} - \frac{3600}{169} = \frac{20 736}{169} =\frac{144^2}{13^2}[/tex]
x = [tex]\frac{144}{13} =[/tex] 11 [tex]\frac{1}{13}[/tex]
y = a - x = 13 - 11 [tex]\frac{1}{13} =[/tex] 1 [tex]\frac{12}{13}[/tex]
Odp. 1 [tex]\frac{12}{13}[/tex] i 11 [tex]\frac{1}{13}[/tex]
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Odpowiedź:
a ) e = 10 f = 24
a² = 5² + 12² = 25 + 144 = 169
a = [tex]\sqrt{169} = 13[/tex]
Pole rombu
P = 0,5 e*f = 0,5*10*24 = 120
oraz
P = a*h = 13*h = 120 / : 13
h = [tex]\frac{120}{13} = 9 \frac{3}{13}[/tex]
=================
b ) d = 0,5 h = [tex]\frac{60}{13}[/tex]
a = y + x
więc
x² + d² = 12²
x² =144 - ( [tex]\frac{60}{13}[/tex] )² = [tex]\frac{24 336}{169} - \frac{3600}{169} = \frac{20 736}{169} =\frac{144^2}{13^2}[/tex]
x = [tex]\frac{144}{13} =[/tex] 11 [tex]\frac{1}{13}[/tex]
y = a - x = 13 - 11 [tex]\frac{1}{13} =[/tex] 1 [tex]\frac{12}{13}[/tex]
Odp. 1 [tex]\frac{12}{13}[/tex] i 11 [tex]\frac{1}{13}[/tex]
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Szczegółowe wyjaśnienie: