SOLUTION: In the diagram below, EFGH is a rhombus formed by connecting the mid-points of the sides of the rectangle ABCD. The rectangle is circumscribed by the circle with center O. Find the

Algebra ->  Polygons -> SOLUTION: In the diagram below, EFGH is a rhombus formed by connecting the mid-points of the sides of the rectangle ABCD. The rectangle is circumscribed by the circle with center O. Find the      Log On


   

Question 1148559: In the diagram below, EFGH is a rhombus formed by connecting the mid-points of the sides of the rectangle ABCD. The rectangle is circumscribed by the circle with center O. Find the area, in cm2 of the rhombus if OF=15cm, and JH=6cm.
Diagram: https://ibb.co/rwnM6f9
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


FK = JH = 6, so the radius of the circle is OF+FK = OF+JH = 21.

OC is also a radius of the circle, so OC = 21.

Then triangle OFC is a right triangle and we know OF=15 and OC=21; the Pythagorean Theorem then tells us

FC+=+sqrt%2821%5E2-15%5E2%29+=+sqrt%28441-225%29+=+sqrt%28216%29+=+6%2Asqrt%286%29

Then the diagonals of the rhombus are 30 and 12*sqrt(6).

The area of a rhombus is half the product of the diagonals:

A+=+%281%2F2%29%2830%29%2812%2Asqrt%286%29%29+=+180%2Asqrt%286%29