The three medians AX, BY and CZ of $$\triangle$$ABC intersect at point L. If the area of $$\triangle$$ABC is 30 $$cm^2$$, then the area of the quadrilateral BXLZ is:

Given, Area of $$\triangle$$ABC = 30 $$cm^2$$

The three medians AX, BY and CZ of $$\triangle$$ABC intersect at point L as shown in the above figure

The area of six triangles in the above figure are equal and area is equal to one-sixth of the area of triangle ABC.

Area of $$\triangle$$BLX = $$\frac{1}{6}\times$$Area of $$\triangle$$ABC

= $$\frac{1}{6}\times$$30

= 5 $$cm^2$$

Similarly, Area of $$\triangle$$BLZ = 5 $$cm^2$$

$$\therefore\ $$Area of quadrilateral = Area of $$\triangle$$BLX + Area of $$\triangle$$BLZ = 5 + 5 = 10 $$cm^2$$

Hence, the correct answer is Option A