How do you balance this? Al + CuCl2 = AlCl3 + Cu
1 answer:
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1.
Draw a circle with center C And radius CA, as shown in the attached picture.
Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|=
.
these are all shown in the picture.
2.
Area of triangle ABC is 1/2 * 2R * R= R^2
3.
Let the area between arc BXA and chord AB be Y. (the yellow region).
and let G be the shaded region between arcs AB and AXB.
G=1/2(Area circle with center O)-Y
=
To find Y:
Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.
So Area of sector ACB =
Y =area of sector ABC-Area(triangle ABC)=
4.
Finally,
This proves that the 2 shaded regions have equal area.
Draw a circle with center C And radius CA, as shown in the attached picture.
Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|=
these are all shown in the picture.
2.
Area of triangle ABC is 1/2 * 2R * R= R^2
3.
Let the area between arc BXA and chord AB be Y. (the yellow region).
and let G be the shaded region between arcs AB and AXB.
G=1/2(Area circle with center O)-Y
=
To find Y:
Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.
So Area of sector ACB =
Y =area of sector ABC-Area(triangle ABC)=
4.
Finally,
This proves that the 2 shaded regions have equal area.