# Find the radius of a circle inscribed in a rhombus whose side is 25 and one of the diagonals is 14.

July 22, 2021 | education

| The diagonals of the edge at the point of their intersection are halved and intersect at right angles. Then ОВ = ВД / 2 = 14/2 = 7 cm, and the triangle AOB is rectangular.

Let point H be the point of tangency between the circle and the side AB of the rhombus.

The segment OK is the radius drawn to the point of tangency, then OH is the height of the AOB triangle and the radius of the inscribed circle.

In the right-angled triangle AOB, we determine the length of the leg AO.

AO ^ 2 = AB ^ 2 – OB ^ 2 = 625 – 49 = 576.

AO = 24 cm.

Determine the area of the triangle AOB.

Saov = ОВ * AO / 2 = 7 * 24/2 = 84 cm2.

Also Saov = AB * OH / 2.

OH = 2 * Saov / AB = 2 * 84/25 = 6.72 cm.

Answer: The radius of the inscribed circle is 6.72 cm.

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