Question

An equilateral triangle and a regular hexagon have the same area. The ratio of the side length of the regular hexagon to the side length of the equilateral triangle can be written as m/√n. What is m+n?

Help me plz!

Answer 1

7

Explanation:

Let a and b be the sides of the equilateral triangle and the regular hexagon respectively as shown in figures (1) and (2).

The area of the equilateral having side a is

`A_T= \frac {\sqrt3}{2}a^2\cdots(i)`

Now, join all the 6 corner points of the regular hexagon to the center of the hexagon as shown in figure (3).

As there are 6 equal sides, so angle subtended by each side at the center is 360/6=60 degrees.

Considering the triangle POR:

`\angle POR= 60` degree [angle at center]

As sides PO=RO, so
`\angle OPR = \angle ORP = 60` degree.

Hence, triangle POR is an equilateral triangle.

Similarly, all the remaining 5 triangles are equilateral triangles.

So, the area of hexagon = 6 x Area of triangle POR

As the length of sides of the triangle POR is b,

so the area of the triangle POR
`= \frac {\sqrt3}{2}b^2`.

Hence, the area of the regular hexagon,

`A_H = 6 * \frac {\sqrt3}{2}b^2\cdots(ii)`

As the equilateral triangle and a regular hexagon have the same area, so from the equations (i)and (ii), we have

`A_T=A_H\\\\\frac {\sqrt3}{2}a^2=6 * \frac {\sqrt3}{2}b^2\\\\\Rightarrow a^2=6b^2\\\\\Rightarrow \frac {b^2}{a^2}=\frac 1 6\\\\\Rightarrow \frac {b}{a}=\frac {1}{\sqrt 6}\cdots(iii)`

Given that the ratio of the side length of the regular hexagon to the side length of the equilateral triangle, b/a= m/√n.

On comparing with the equation (iii), we have

m=1 and n=6

Hence, m+n=1+6=7.