Question

A rectangle is drawn around a regular hexagon so that it forms four congruent right triangles in the corners of the rectangle.How much larger is the area of the rectangle than the area of the hexagon?

Answer 1

The difference between the area of the rectangle and the area of the hexagon can be found with the area of the 4 triangles as it must be the same value.

To find the area of the triangles we have to find the height of them. We can use the pythagorean theorem to find it, doing so, we have:

`\begin{gathered} c^2=a^2+b^2 \\ (10)^2=(5)^2+b^2\text{ (Replacing the values. Taken b as the height of the triangle)} \\ 100=25+b^2\text{ (Raising both numbers to the power of 2)} \\ 75=b^2\text{ (Subtracting 25 from both sides of the equation)} \\ \sqrt[]{75}=b\text{ (Taking the square root of both sides)} \\ 5\sqrt[]{3}=b\text{ (Simplifying)} \\ \text{The height of the triangles must be 5}\sqrt[]{3} \end{gathered}`

Then, we use the formula for the area of the triangle:

`\begin{gathered} A=(b\cdot h)/(2)\text{ (b:base, h=height)} \\ A=\frac{5\cdot5\sqrt[]{3}}{2}cm^2\text{ (Replacing)} \\ A=(43.3)/(2)\text{ cm}^2\text{ (Multiplying)} \\ A=21.65\text{ }cm^2\text{ (Dividing)} \end{gathered}`

Given that we have 4 triangles, we must multiply the last result by 4. Doing so, we have:

At= 4*21.65 cm^2

At= 86.6 cm^2

The answer would be 86.6 cm^2 or 86.6 square cm. (First option)