In rhombus ABCD, the side is √10 and one of the diagonals is 2. Find the area of the rhombus.

1. Known

a) the area of the rhombus is equal to half the product of its diagonals;

b) the diagonals of the rhombus at the point of intersection are perpendicular and divided in half.

2. In the rhombus ABCD, at the intersection of the diagonals at point O, a right-angled triangle ABO was obtained, in which, according to the condition of the problem, AB = 10 ^ 1/2, BO = BD: 2 = 2: 2 = 1 cm.

By the Pythagorean theorem, in the ABO triangle, we find the AO leg.

AO ^ 2 = AB ^ 2 – BO ^ 2 = 10 – 1 = 9;

AO = 9: 1/2 = 3 cm.

3. Calculate the area S of the rhombus.

S = AO * BD = 3 cm * 2 cm = 6 cm ^ 2.

Answer: The area of a rhombus is 6 square centimeters.