Oblicz długość odcinków oznaczonych literami
z^2+8^2=10^2
z^2+64=100
z^2=16
z=6
3^2+(10-4)^2=c^2
9+6^2=c^2
9+36=c^2
45=c^2
c=\sqrt{45}
c=3\sqrt{5}
4^2+( 2\sqrt{5} )^2=x^2
16+4*5=x^2
16+20=x^2
36=x^2
x=6
4^2+y^2=6^2
16+y^2=36
y^2=20
y=2\sqrt{5}
mam nadzieje ze pomoglam
Zadanie 1
Trójkąt w kuli:
a = z
b = 16 : 2 = 8
c = 10
a² + b² = c²
z² + 8² = 10²
z² + 64 = 100
z² = 100 - 64
z² = 36
z = √36
z = 6
Trójkąt w trapezie:
a = 3
b = 10 -4 = 6
c = c
3² + 6² = c²
9 + 36 = c²
45 = c²
c = √ 45 = √ 9 * 5 = 3√ 5
Trójkąt:
a = 4
b = 2√ 5
c = x
4² + 2√ 5 ² = c²
16 + 20 = c²
36 = c²
c = √36
c = 6
b = x
4² + b² = 6²
16 + b² = 36
b² = 36 - 16
b = √ 20 = √ 4 * 5 = 2√5
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z^2+8^2=10^2
z^2+64=100
z^2=16
z=6
3^2+(10-4)^2=c^2
9+6^2=c^2
9+36=c^2
45=c^2
c=\sqrt{45}
c=3\sqrt{5}
4^2+( 2\sqrt{5} )^2=x^2
16+4*5=x^2
16+20=x^2
36=x^2
x=6
4^2+y^2=6^2
16+y^2=36
y^2=20
y=2\sqrt{5}
mam nadzieje ze pomoglam
Zadanie 1
Trójkąt w kuli:
a = z
b = 16 : 2 = 8
c = 10
a² + b² = c²
z² + 8² = 10²
z² + 64 = 100
z² = 100 - 64
z² = 36
z = √36
z = 6
Trójkąt w trapezie:
a = 3
b = 10 -4 = 6
c = c
a² + b² = c²
3² + 6² = c²
9 + 36 = c²
45 = c²
c = √ 45 = √ 9 * 5 = 3√ 5
Trójkąt:
a = 4
b = 2√ 5
c = x
a² + b² = c²
4² + 2√ 5 ² = c²
16 + 20 = c²
36 = c²
c = √36
c = 6
Trójkąt:
a = 4
b = x
c = 6
a² + b² = c²
4² + b² = 6²
16 + b² = 36
b² = 36 - 16
b = √ 20 = √ 4 * 5 = 2√5