Question

A regular hexagon of maximum possible area is cut off from an equilateral triangle. The ratio of area of triangle to the area of hexagon will be

• A. $\sqrt{\frac{3}{2}}$
• B. $\frac{\sqrt{6}}{2}$
• C. $\frac{3}{\sqrt{2}}$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

Each side is 4a. We can see that the area of the equilateral triangle is
twice the area of a smaller right triangles. The right triangle's area
is 1/2 base x height, and the height (b) is obtained through Pythagorus'
Theorem from the other two sides. If we put that all together, we get:
$A_T=2\times (\frac{1}{2}\times 2a\times b)=2a\sqrt{{\left(4a\right)}^2-{\left(2a\right)}^2}=2a\sqrt{12a^2}=4a^2\sqrt{3}$
Now let's look at the hexagon. You are right that it is made up of six equilateral triangles, so each side is 2a this time:
Using the same logic as before, let's calculate the area of each small equilateral
$A_t=2\times (\frac{1}{2}\times a\times c)=a\sqrt{{\left(2a\right)}^2-{\left(a\right)}^2}=a\sqrt{3a^2}=a^2\sqrt{3}$
We can see by this that the ratio of $A_t$ is not half of $A_T$ it is $\frac{1}{4}$ instead.Now when we look at the ratio of the area of the hexagon to the area of the large triangle, we get:
$\frac{6\times A_t}{A_T}=\frac{6a^2\sqrt{3}}{4a^2\sqrt{3}}{A}_t=\frac{3}{2}$