Find the area of the rhombus CDEF if the side CF = 4 dm and the diagonal DF = 4 dm, and the sides DF and CD are equal.

The area of the rhombus is equal to the product of the diagonals. The intersection point of the diagonals in the rhombus divides the diagonals in half, we denote it by the letter O. Then:

FO = OD = 4/2 = 2 dm.

The diagonals in the rhombus are perpendicular, so they form 4 right-angled triangles. Let’s find the CO leg in the right-angled triangle COF according to the Pythagorean theorem:

a² + b² = c²;

FO² + CO² = CF²;

2² + CO² = 4²;

4 + CO² = 16;

CO² = 16 – 4;

CO² = 12;

CO = √12 dm = 3√3 dm.

CE = 2 * CO = 2 * 3√3 = 6√3 dm – large diagonal of the rhombus.

Find the area of the rhombus by multiplying the diagonals:

S = CE * FD = 6√3 * 4 = 24√3 dm².

Answer: 24√3 dm²