The diagonals of trapezoid ABCD with bases AD and BC meet at point O. The perimeters of triangles BOC

| April 25, 2021 | education

The diagonals of trapezoid ABCD with bases AD and BC meet at point O. The perimeters of triangles BOC and AOD are 3: 5, BD = 24. Find the lengths of the segments BO and OD.

In triangles BOC and AOD, the angle BOC is equal to AOD as the vertical angles at the intersection of the diagonals AC and BD. The OBC angle is equal to the ODA angle as the intersecting angles at the intersection of parallel straight lines AC and DD of the secant BD.

Then the BOC triangle is similar to the AOD triangle in two angles.

The perimeters of such triangles are referred to as the coefficient of their similarity. Then K = 3/5.

Let the length of the segment BO = X cm, then OD = (24 – X).

BO / DO = X / (24 – X) = 3/5.

5 * X = 72 – 3 * X.

8 * X = 72.

X = BO = 9 cm.

OD = 24 – 9 = 15 cm.

Answer: The length of the BO is 9 cm, the length of OD is 15 cm.



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