Question

In trapezoid ABCD, sides AB and DC are on parallel lines. A regular hexagon PQRSTU lies inside of trapezoid ABCD so that vertices P and Q trisect the base AB (with P closer to A and Q closer to B), S and T lie on the base CD, and sides PU and QR are parallel to sides AD and BC, respectively. What fraction of the area of trapezoid ABCD lies inside of hexagon PQRSTU? Express your answer as a common fraction.

Answer 1

To solve this problem, we have to divide the trapezoid in regions where we can estimate its area, proportional to another known area.

We can draw:

We can calculate the area of the hexagon as:

`A_H=\frac{3\sqrt[]{3}}{2}s^2`

Being s the side, for example the segment PQ. This measure (s) will be our relative measure to calculate all the other areas.

Then, we can calculate the area of the parallellograms APXD and QBCY. Both have the same area.

The base is equal to s, as the segments AP and QB are equal to PQ.

The height is equal to 2 times the apothem of the hexagon, that can be calculated as:

The apothem for a regular hexagon is:

`a=\frac{\sqrt[]{3}}{2}s`

So we can calculate the area of the parallelograms as:

`A_P=b\cdot h=s\cdot(2a)=s\cdot2\frac{\sqrt[]{3}}{2}s=\sqrt[]{3}s^2`

Now, we are only left with triangles UTX and SRY.

Its side is equal to the side of the hexagon and its height is equal to the apothem, so its area is equal to 1/6 of the hexagon.

We can calculate its area as:

`A_T=(b\cdot h)/(2)=\frac{s\cdot\frac{\sqrt[]{3}}{2}s}{2}=\frac{\sqrt[]{3}}{4}s^2`

Now, we can define the area of the trapezoid as the sum of the areas of the hexagon, 2 parallellograms and 2 triangles:

`\begin{gathered} A_(Tz)=A_H+2A_P+2A_T \\ A_(Tz)=\frac{3\sqrt[]{3}}{2}s^2+2(\sqrt[]{3}s^2)+2\frac{\sqrt[]{3}}{4}s^2 \\ A_(Tz)=((3)/(2)+2+(2)/(4))\cdot\sqrt[]{3}s^2 \\ A_(Tz)=((3+2\cdot2+1)/(2))\cdot\sqrt[]{3}s^2 \\ A_(Tz)=((8)/(2))\sqrt[]{3}s^2 \\ A_(Tz)=4\sqrt[]{3}s^2 \end{gathered}`

We can now calculate the fraction of the area of trapezoid ABCD lies inside of hexagon PQRSTU dividing the area of the hexagon by the area of the trapezoid:

`(A_H)/(A_(Tz))=\frac{(3)/(2)\sqrt[]{3}s^2}{4\sqrt[]{3}s^2}=((3)/(2))/(4)=(3)/(2)\cdot(1)/(4)=(3)/(8)`

Answer: The fraction of the area of trapezoid ABCD lies inside of hexagon PQRSTU is 3/8.