### Video Transcript

A conical vessel with base radius
five centimeters and height 24 centimeters is full of water. The water is emptied into a
cylindrical vessel of base radius 10 centimeters. Find the height of the water in the
cylindrical vessel. Use 𝜋 is equal to 22 over
seven.

Let’s start by drawing the two
vessels. Here is the conical vessel which
has a base radius of five centimeters and a height of 24 centimeters. This is the cylindrical vessel
which has a base radius of 10 centimeters and a height, which we do not know.

Now, we are told the water is
emptied into the cylindrical vessel. So what this means is that all of
the water that was in the conical vessel has been poured into the cylindrical
vessel. So let’s let this line represent
the water level in our cylinder. And we can call the length from the
base of the cylinder to the top of the water level ℎ.

Now, we can see that the water in
the cylindrical vessel actually creates a cylinder of its own. It has a base radius which is the
same as the cylinder of 10 centimeters and a height of ℎ. And now, since all of the water was
emptied into the cylindrical vessel from the conical vessel, we know that the volume
of water in the conical vessel is equal to the volume of water in the cylindrical
vessel.

Now, let’s recall the equations for
the volume of a cone and the volume of a cylinder. We have that the volume of a cone
is one-third 𝜋𝑟 squared ℎ, where 𝑟 is the base radius of the cone and ℎ is the
height of the cone. And the volume of a cylinder is
equal to 𝜋𝑟 squared ℎ, where 𝑟 is the base radius of the cylinder and ℎ is the
height.

Now, to help us distinguish between
the base radius and height of the cone and the base radius and height of the
cylinder, let’s call the base radius and height of the cone 𝑟 one and ℎ one and the
base radius and height of the cylinder 𝑟 two and ℎ two. Now that we have these two
equations, we can use them in our equation for the volume of water.

For the volume of water in the
conical vessel, we have one-third 𝜋 𝑟 one squared ℎ one. And we have that this is equal to
the volume of water in the cylindrical vessel which is 𝜋 𝑟 two squared ℎ two. Now, we’re ready to substitute in
𝑟 one ℎ one, 𝑟 two and ℎ two.

Our value of 𝑟 one is the base
radius of the cone, which is five centimeters. Our value of ℎ one is the height of
the cone, which is 24 centimeters. Our 𝑟 two is the base radius of
the water in the cylinder, which is 10 centimeters. And our ℎ two is the height of the
water in the cylindrical vessel or ℎ.

And so we can simply substitute
these values into our equation, which gives us one-third 𝜋 times five squared times
24 is equal to 𝜋 times 10 squared times ℎ. Now, we notice that we have a 𝜋 on
both sides of the equation. And so we can cancel them out. And now, in order to get our
equation in terms of ℎ, we can divide both sides by 10 squared. And this gives us that five squared
times 24 over three times 10 squared is equal to ℎ.

This can also be written as ℎ is
equal to five times five times 24 over 10 times 10 times three. And since 10 is equal to five times
two, we can cancel the two fives from the 10s with the two fives in the
numerator. This leaves us with 24 over two
times two times three.

Now, 24 is also equal to eight
times three. And so the three from the 24 can
cancel with a three in the denominator. And this leaves us with eight over
two times two. Now since eight is two cubed or two
times two times two, we can cancel two of the twos from the eight with the two twos
in the denominator. And so this leaves us with just
two.

And since ℎ represents the height
of water in the cylindrical vessel, this height of ℎ is equal to two centimeters is
our final answer.