In trapezoid ABCD, the diagonals intersect at point O, and their midpoints form a segment MN = 6 cm

In trapezoid ABCD, the diagonals intersect at point O, and their midpoints form a segment MN = 6 cm. The height of the trapezoid is 10 cm. The lower base is 36 cm. Find the area of the triangle MON.

Since, by condition, the segment MN connects the midpoints of the diagonals, it lies on the midline of the trapezoid. Let us extend the segment MN to the intersection with the lateral sides of the trapezoid, then КР is the middle line of the trapezoid.

In triangle ВСD the segment NL is its middle line and is equal to half of AD. NL = АD / 2 = 36/2 = 18 cm.Then NL = 18 – 6 = 12 cm.

In the ABC triangle, the KM segment is its middle line and is equal to half BC. KM = BC / 2.

KM = NL since they are the midlines of triangles that have one base BC. Then the middle line of the trapezoid is: KP = KM + MN + NL = 12 + 6 + 12 = 30 cm.

BC = 2 * KR – AD = 60 – 36 = 24 cm.

Then Strapezium = KР *ВН = 30 * 10 = 300 cm2.

From point C of the trapezoid, draw a line parallel to the lateral side of CD until it intersects AD at point E.

In the formed triangle ACE, the base length is equal to AD + DE, and since ВСЕ is a parallelogram, then DE = BC = 24 cm.

AE = AD + DE = 36 + 24 = 60 cm.

Then the area of ​​the triangle ADE is equal to the area of ​​the trapezoid 300 cm2.

In triangles ADE and MOН, all angles are equal, as are the corresponding angles at the intersection of parallel straight lines, then these triangles are similar.

K = AE / MН = 60/6 = 10.

The ratio of the areas of such triangles is equal to the square of the similarity coefficient.

Sade / Smon = 100.

S mon = 300/100 = 3 cm2.

Answer: S mon = 3 cm2.