Grades 11-12 Video Solutions 2022-2022_11-12

Video Transcription

A regular hexagonal prism has its top corner shaved off as shown. The top face becomes a smaller regular hexagon, and the six rectangular faces around the middle become 12 isosceles  triangles of two different sizes. What fraction of the volume of the original prism has been lost?  So the fraction that's lost is the sum of the volumes of these corner pieces divided by the  volume of the entire thing. So let's first find the volume of the entire thing. So here's the  hexagonal base of the original prism. Since later on each side of the base gets cut in half  by this when we shave off the corners, I'm just gonna do that already and call half of this side  x. So the area of this equilateral triangle is x times x root 3, so x squared root 3, and six of  them in total make up the base. So the area of the base is 6x square root 3, and the volume is  that multiplied by the height, which we're going to call h. So here's the volume right here.  So now let's figure out the volume of one of these triangular prisms.  So the base of the triangular prism is this triangle right here with the same size x and x,  and the area of this is one-half this side times this side times the sine of the angle between them.  This angle is 120 degrees, so the area of this base is one-half x squared sine 120.  The volume of the prism that we get, the triangular prism, is one-third times the area of the base  times the height, so it's right this right here. Now we can multiply the one-third and the one-half  and replace the sine of 120 degrees with a root 3 over 2, and this simplifies to one-twelfth x  squared root 3h. So that's the volume of one of the prisms. Now we have six of them, and so we get six  times this quantity as the volume that we lost. Now the fraction of the volume is this is that  volume divided by this guy right here. The h cancels, the root 3 cancels, the x squared cancels,  and we had six of these, so the 6 cancels as well, and we're just left with the 112 at the end of the day.

Video Summary

The problem involves determining the fraction of volume lost when the top corner of a regular hexagonal prism is shaved off. The original prism's volume is calculated from its hexagonal base, comprising six equilateral triangles, and its height \(h\). The altered shape creates six isosceles triangular prisms, each with a specific volume expressed in terms of \(x\) and \(h\). The calculation reveals each triangular prism's volume as \(\frac{1}{12} x^2 \sqrt{3} h\). Multiplying by six (for all triangular prisms), the fraction of the original volume lost is \(\frac{1}{12}\), simplifying the ratios, leading to this result.

Keywords

hexagonal prism

volume loss

isosceles triangular prism

fraction of volume

geometry problem