The side of the rhombus is 10, and the distance from the point of intersection of the diagonals

The side of the rhombus is 10, and the distance from the point of intersection of the diagonals to the side = 3 find the area of the rhombus

Since ABCD is a rhombus, its diagonals AC and BD intersect at right angles and are divided at the point of their intersection in half. AO = OC, BO = DO.

Then the triangles AOB, BOС, COD, AOD are rectangular and equal in two legs.

Then the area of the rhombus is:

Savsd = 4 * Sаod.

The segment OH is the distance from the point O to the side of AD, therefore, OH is the height of the right-angled triangle AOD.

Saod = AD * OH / 2 = 10 * 3/2 = 15 cm ^ 2.

Savsd = 4 * 15 = 60 cm ^ 2.

Answer: The area of the rhombus is 60 cm ^ 2.