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

September 25, 2021 | education

| **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.

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