Classify the shapes based on their volumes.
a cone with a radius of 12 units
and a height of 6 units,
a cylinder with a diameter of 12 units
and a height of 27 units,
a sphere with a diameter of 12 units,
a cylinder with a diameter of 8 units
and a height of 18 units,
a sphere with a radius of 9 units,
a cone with a diameter of 18 units
and a height of 36 units

1.  volume of cone = 288π cube units

2.  volume of cylinder = 972π cube units

3.  volume of sphere = 288π cube units

4.  volume of cylinder = 288π cube units

5.  volume of sphere = 972π cube units

6.  volume of cone = 972π cube units

Step-by-step explanation:

1.  Radius of cone (r) = 12units

height of cone(h)  = 6units

volume of cone = π/3r²h  

=π/3.12.12.6

=π.12.12.2

=288π cube units

2.  diameter of cylinder(d) = 12units

radius of cylinder (r) = 6units

height of cylinder (h) = 27 units

volume of cylinder = πr²h

= π.6.6.27

= 972π cube units

3.   diameter of sphere = 12 units

radius of sphere = 6 units

volume of sphere = 4π/3.r³

= 4π.2.6.6

= 288π cube units

4.  diameter of cylinder (d) = 8 units

radius of cylinder (r) = 4 units

height of cylinder(h) = 18units

volume of cylinder = πr²h

= π.4.4.18

= 288π cube units

5.   radius of sphere(r) = 9 units

volume of sphere = 4π/3.r³

= 4π/3.9.9.9

= 4π.3.9.9

= 972π cube units

6.  diameter of cone (d) = 18 units

radius of cone (r) = 9 units

height of cone (h) = 36 units

volume of cone = π/3.r².h

= π/3.9.9.36

= π.3.9.36

=972π cube units

the figures below are made out of circles, semicircles, quarter … find the area and the perimeter of each figure and give your answers as a completely simplified exact value in terms of π (no approximations). remember that we'll also have to cut whatever areas we get in half, since we're only dealing

step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.