The area of the rhombus is 48 cm2. Find the area of a quadrilateral whose vertices are the midpoints of the sides of this rhombus.

Let ABCD be the given rhombus, points E, P, K and H – the midpoints of the sides AB, BC, CD and AD, respectively.

Consider triangle ABC: point E is the middle of AB, point P is the middle of BC, which means that EP is the middle line of triangle ABC, EP is parallel to AC and equal to half of it: EP = 1/2 AC, hence AC = 2 * EP.

Similarly, we prove that NC is the midline of the ACD triangle, EH is the midline of the ABD triangle (EH = 1/2 BD, BD = 2 * EH), RK is the midline of the BCD triangle.

Since EP and NC are parallel to AC, and EH and RK are parallel to BD, and AC is perpendicular to BD, it means that the ERNK quadrangle is a rectangle, its area is equal to Spr = EP * EH.

Since the area of ​​the rhombus is half the product of the diagonals, Sp = 1/2 * AC * BD (Sp = 48 cm², AC = 2 * EP, BD = 2 * EH), we get the equation:

48 = 1/2 * 2 * EP * 2 * EH;

48 = 2 * EP * EH.

EP * EN = 48: 2 = 24.

Answer: the area of ​​the quadrangle is 24 cm².