The perimeters of a square and an equilateral triangle are equal. If the diagonal of the square is 9 cm, what is the area of the equilateral triangle?

Let the side of square = $$s$$ cm and diagonal, $$d=9$$ cm

=> $$(s)^2+(s)^2=(d)^2$$

=> $$2s^2=9^2=81$$

=> $$s^2=\frac{81}{2}$$

=> $$s=\sqrt{\frac{81}{2}} = \frac{9}{\sqrt{2}}$$

Thus, perimeter of square = $$4s$$

= $$4 \times \frac{9}{\sqrt{2}} = 18\sqrt{2}$$ cm

Also, perimeter of square = Perimeter of equilateral triangle = $$18\sqrt{2}$$ cm

Let side of equilateral triangle = $$a$$ cm

=> $$3a=18\sqrt{2}$$

=> $$a=\frac{18\sqrt{2}}{3} = 6\sqrt{2}$$ cm

$$\therefore$$ Area of equilateral triangle = $$\frac{\sqrt{3}}{4} a^2$$

= $$\frac{\sqrt{3}}{4} \times (6\sqrt{2})^2$$

= $$18\sqrt{3}$$ $$cm^2$$

=> Ans - (A)