Odpowiedź:
c = [tex]\frac{15}{7} + \frac{20}{7} = \frac{35}{7} = 5[/tex]
a, b - dł. przyprostokątnych , a < b
Z tw. o dwusiecznej mamy
[tex]\frac{a}{\frac{15}{7} } = \frac{b}{\frac{20}{7} }[/tex]
[tex]\frac{7}{15} *a = \frac{7}{20} *b[/tex] / I [tex]\frac{15}{7}[/tex]
a = [tex]\frac{15}{20}[/tex] b = [tex]\frac{3}{4}[/tex] b
--------------------------
Z tw. Pitagorasa
a² + b² = c²
( [tex]\frac{3}{4} b )^2 + b^2 = 5^2[/tex]
[tex]\frac{9}{16} b^2 + b^2 = 25 / * 16[/tex]
9 b² + 16 b² = 25*16
25 b² = 25*16 / : 25
b² = 16
b = 4
=====
a² + 4² = 25
a² = 25 - 16 = 9
a = 3
=======
Obwód Δ
L = a + b + c = 3 + 4 + 5 = 12
==============================
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Odpowiedź:
c = [tex]\frac{15}{7} + \frac{20}{7} = \frac{35}{7} = 5[/tex]
a, b - dł. przyprostokątnych , a < b
Z tw. o dwusiecznej mamy
[tex]\frac{a}{\frac{15}{7} } = \frac{b}{\frac{20}{7} }[/tex]
[tex]\frac{7}{15} *a = \frac{7}{20} *b[/tex] / I [tex]\frac{15}{7}[/tex]
a = [tex]\frac{15}{20}[/tex] b = [tex]\frac{3}{4}[/tex] b
--------------------------
Z tw. Pitagorasa
a² + b² = c²
( [tex]\frac{3}{4} b )^2 + b^2 = 5^2[/tex]
[tex]\frac{9}{16} b^2 + b^2 = 25 / * 16[/tex]
9 b² + 16 b² = 25*16
25 b² = 25*16 / : 25
b² = 16
b = 4
=====
a² + 4² = 25
a² = 25 - 16 = 9
a = 3
=======
Obwód Δ
L = a + b + c = 3 + 4 + 5 = 12
==============================
Szczegółowe wyjaśnienie: