Cone is an important geometrical shape due to conics. Conics are the intersection of a cone and a plane. Look at ry1.
Have first a look at ry1 and try to calculate yourself.
Every cylinder has an height of 2 cm. The top cylinder's radius is 1 cm. and the radius increases by 1 cm. for each cylinder down the other.
So the volume is π x 2 x (1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10) cm.^{3}
The sum squares of first ten natural numbers is 1+4+9+16+25+36+49+64+81+100 = 385 .
The base circle may at anywhere be tangent to the upper sector .
The perimeter of the circular section of the the sector left should be as long as the circumference of the circle right.
Remember the relation of the previous question .
π/2 x 400 + π x 100 = π x 300 .
The measure of ∠A is 180° .
The orange circle is too big.
Since π is taken to be 3 , π/3 = 1 . We only need to multiply square of the radius by height .
The conic curves (circle, ellipse, parabola, and hyperbola) are pretty important .
As said, they should be learned .
It can be perceived as a fraction problem or as the difference of two cones.
As a fraction problem:
1- The upper cone's volume is 1/4 ths of the whole cone's if only the radiuses are considered (the volume is proportional with the square of the radius).
2- The upper cone's volume is one half of the whole cone's if only the heights are considered (the volume is proportional with height).
3- If both radiuses and heights are considered, the upper cone's volume is 1/8 ths of the whole cone's.
It is again one thirds of base area multiplied by height .
The sheet of paper covering the plastic sphere is supposed to be a sphere net .
This figure is called gores .
Easy qustion .
The radius of the sphere is given .
6 times the arc length given will be the circumference from which the radius should be derived .
The radius will also be 30 cm. bacause π is given as 3 and 2 x π = 6 .
Look at the triangle carefully; it is the well known 3-4-5 triangle we have mentioned before (2.5 ^{3} = 15.625) .
Diameter is height .
Compare 4/3 x π x (D/2)^{3} vs. π x (D/2)^{2} x D . Take the second to first ratio.
Let the radius of the red sphere be r . The volume of it will then be V = 4/3 x π x r^{3}.For further help have please a look at ry1.
Then, if you still have not found the result, look at the second y.
The volume of the solid part = 256/24 x π x r^{3} - 28/24 x π x r^{3} = 57/6 x π x r^{3}.
The ratio is (28/24 = 7/6) ...
Another way to look at this problem : Let the volume of the red sphere be unit volume. The volume of the whole sphere without cut is 8 unit.
The volume of the shell is 7 unit. Its cut part is 7/8 unit. Solid part is 8 - 7/8 = 57/8 unit. Since the denominators of 7/8 and 57/8 are both 8 , the ratio is ... .
The volume of the cylinder: π x r^{2} x 2 x r = 2 x π x r^{3} .
The volume of the sphere: 4/3 x π x r^{3}
This is already solved indeed.
The cylinders surface area is greater than of the sphere because of two reasons:
1. The volume of the cylinder is the greater volume.
Even if they had the same volumes the sphere is the compacter .
The ratio is supposed to be greater than 1 .
The surface area of the cylinder = 2 x π x r^{2} + 2 x π x r x 2 x r .
The surface area of the sphere = 4 x π x r^{2}
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