Let the side of square = $$s$$ cm and diagonal = $$d$$ cm => Area of square = $$(s)^2 = 12$$ ----------(i) In right triangle of the square, => $$(s)^2 + (s)^2 = (d)^2$$ Substituting value of $$s^2$$ from equation (i) => $$(d)^2 = 12 + 12 = 24$$ ----------(ii) Side of equilateral triangle = Diagonal of square = $$d$$ cm $$ herefore$$ Area of equilateral triangle = $$frac{sqrt{3}}{4} d^2$$ Substituting value of $$d^2$$ from (ii), we get : = $$frac{sqrt{3}}{4} imes 24 = 6sqrt{3} cm^2$$ => Ans - (B)