Balancing Problem - Part 3

Let us represent triangle as t, square as s, and hexagon as h.

Therefore, $4t=3s$

Or, $\displaystyle 2t=\frac{3s}{2}$

And, $2h=3s$

Or, $\displaystyle h=\frac{3s}{2}$

Therefore the no. of squares they will balance one hexagon and two triangles is given by,

$\frac{3s+3s}{2}$

Or, $\boxed{6}$

If T-Traingles S -circles and h-Hexagon 4T=3S=2H dividing by 2 it gives 2T=3/2 S=1H 1H+2T= 3/2S+3/2S = 3S

Triangle = T Square = S Hexagon = H

4t = 3s 2t = "1.5" 2h = 3s h = "1.5"

"1.5"+"1.5"= 3

Therefore 3 squares would balance 1 hexagon and 2 triangles