Answer:
7
Step-by-step 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
[tex]A_T= \frac {\sqrt3}{2}a^2\cdots(i)[/tex]
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:
[tex]\angle POR= 60[/tex] degree [angle at center]
As sides PO=RO, so [tex]\angle OPR = \angle ORP = 60[/tex] 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 [tex]= \frac {\sqrt3}{2}b^2[/tex].
Hence, the area of the regular hexagon,
[tex]A_H = 6 \times \frac {\sqrt3}{2}b^2\cdots(ii)[/tex]
As the equilateral triangle and a regular hexagon have the same area, so from the equations (i)and (ii), we have
[tex]A_T=A_H\\\\\frac {\sqrt3}{2}a^2=6 \times \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)[/tex]
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.
