A bucket is 20cm in diameter at the open end, 12cm in diameter at the bottom, and 16cm deep. to what depth would the bucket fill a cylindrical tin 28cm in diameter?

1 Answer
May 10, 2018

#5.33cm#

Explanation:

.

The bucket has a base diameter of #12# and top opening diameter of #20#. This makes the base radius #6# and top radius #10#.

Let's look at the figure below:

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The bucket as shown has the bottom base of radius:

#BC=6#

and the top opening of radius:

#DE=10#

The depth is:

#DB=16#

If we extend the lateral surface of the bucket down it will generate a cone that contains the bucket in it.

To calculate the volume of the bucket, we need to calculate the volume of the large cone with the depth of #DA#, then calculate the volume of the small cone with the depth of #BA#, and subtract the volume of the small cone from the volume of the large cone.

Let's let #BA=x#. The two triangles #DeltaABC# and #DeltaADE# are similar due to the Angle Angle theorem because they both share angle #/_ADE=90^@# and angle #/_DAE#.

Therefore, the ratio of their corresponding sides are equal.

#(BA)/(DA)=(BC)/(DE)#

#x/(16+x)=6/10#

#x/(16+x)=3/5#

#48+3x=5x#

#2x=48#

#x=24# cm

The volume of a cone is:

#V=1/3hpir^2#

The height #(h)#, which we are calling the depth, of the small cone is #x=24 cm# and the depth of the large cone is #x+16=24+16=40 cm#

#V_("Large Cone")=1/3(40)(pi)(10)^2=(4000pi)/3cm^3#

#V_("Small Cone")=1/3(24)(pi)(6)^2=288picm^3#

#V_("Bucket")=(4000pi)/3-288pi=(3136pi)/3cm^3#

Volume of a cylinder is:

#V_("Cylinder")=pir^2h# where #h# is the height(depth) and #r# is the radius of the base of the cylinder.

If the base of the cylinder has a diameter of #28cm# then the radius of the base is #14cm#

#(pi)(14)^2(h)=(3136pi)/3#

#196pih=(3136pi)/3#

#196h=3136/3#

#h=(3136/3)/196=3136/588=5.33cm#