Since every figure repr. an equil. triangle, each of the angles shown is 60 deg.
Thus, the height of one of the shaded triangles is s sin 60 deg, or s/sqrt(3).
The base of one such triangle is s/3.
Thus, the area of ONE shaded triangle is A = bh/2, which here is
(s/3)(s/sqrt(3) ) (s/3)(s)
A = ----------------------- = --------------------
2 2sqrt(3)
Rationalize the denom.: Mult numerator and denominator both by sqrt(3), and then mult the resulting numerator and den by 3:
sqrt(3) s^2/3 s^2*sqrt(3) s^2*sqrt(3)
----------- * ----------- = ----------------- = ----------------
sqrt(3) 2sqrt(3) 3*2*3 18
Unfortunately, this does not agree with any of the four answer choices.
Let's take a slightly different approach here:
Each (horizontal) base of each shaded triangle has length s/3. Each has height (s/3)sin(60 deg) = s*sqrt(3)/(3*2) = s*sqrt(3)/6.
[s/3]*[s*sqrt(3)]/6
Then each shaded triangle has area A = bh/2 = ----------------------------
2
and the total shaded area is three times that:
s^2 * sqrt(3) s^2*sqrt(3)
Total area = -------------------- = ------------------ (answer)
3*2 * 6 36
