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fredd [130]
3 years ago
15

Recall what you know about similarity. If circle B is similar to circle A, what must exist?

Mathematics
2 answers:
Dima020 [189]3 years ago
6 0

Answer:

Circle A and circle B must exist.

That is it—no other requirement.

That is because all circles are similar; therefore, pick any two circles, and they are similar. Somewhat loosely speaking, two geometric entities are similar if and only if they are the same shape. All circles are the same shape—perfectly round.

There is not any requirement on size. In general, two arbitrarily picked circles will be of different sizes—there is nothing requiring them to be different sizes, but in general they will be. If you really want the two circles the same size, you really should specify congruent circles.

There is not any requirement on location. Location has nothing to do with shape. If you really want the two circles to have the same center, you really should specify concentric circles. If you really want the two circles to just barely touch each other and some third geometric entity at the same point, you really should specify tangent circles—internally tangent if you want the two circles on the same side of the line, plane, …, and externally tangent if you want the two circles on opposite sides. If you are dealing with 3-dimensional space and really want the circles to be the the same plane, you really should specify coplanar circles.

Without any additional specification, all circles are similar. The same is true for equilateral triangles, for squares, for rectangles with a 2:1 aspect ratio, all regular heptagons, …, as well as 3-dimensional entities such as for spheres, for cubes, for regular icosahedrons, ….

Step-by-step explanation:

Hope it helps you

kenny6666 [7]3 years ago
4 0
It's not exactly clear what your teacher or book is asking for. By the way, all circles are similar to one another. There's no need to make the statement "If circle B is similar to circle A" because it's something that can be proven for any pair of circles. There are no cases where we can find two circles that are not similar. 
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Answer:

a) 8π

b) 8/3 π

c) 32/5 π

d) 176/15 π

Step-by-step explanation:

Given lines :  y = √x, y = 2, x = 0.

<u>a) The x-axis </u>

using the shell method

y = √x = , x = y^2

h = y^2 , p = y

vol = ( 2π ) \int\limits^2_0 {ph} \, dy

     = ( 2\pi ) \int\limits^2_0 {y.y^2} \, dy  

∴ Vol = 8π

<u>b) The line y = 2  ( using the shell method )</u>

p = 2 - y

h = y^2

vol = ( 2π ) \int\limits^2_0 {ph} \, dy

     = ( 2\pi ) \int\limits^2_0 {(2-y).y^2} \, dy

     = ( 2π ) * [ 2/3 * y^3  - y^4 / 4 ] ²₀

∴ Vol  = 8/3 π

<u>c) The y-axis  ( using shell method )</u>

h = 2-y  = h = 2 - √x

p = x

vol = (2\pi ) \int\limits^4_0 {ph} \, dx

     = (2\pi ) \int\limits^4_0 {x(2-\sqrt{x}  ) } \, dx

     = ( 2π ) [x^2 - 2/5*x^5/2 ]⁴₀

vol = ( 2π ) ( 16/5 ) = 32/5 π

<u>d) The line x = -1    (using shell method )</u>

p = 1 + x

h = 2√x

vol = (2\pi ) \int\limits^4_0 {ph} \, dx

Hence   vol = 176/15 π

attached below is the graphical representation of P and h

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