In trapezoid ABCD, the base of AD is 5 times the base of BC. The diagonals of the trapezoid intersect at point O

| September 25, 2021 | education

In trapezoid ABCD, the base of AD is 5 times the base of BC. The diagonals of the trapezoid intersect at point O. The midline of the trapezoid intersects the diagonals at points M and N. Find the ratio of the area of the triangle MON to the area of the trapezoid.

Let the length of the base BC = X cm, then the length AD = 5 * X cm.

Let’s draw a straight line BT parallel AC. ACBT parallelogram, then AT = BC = X cm.

Then the length of the segment is DТ = AT + AD = X + 5 * X = 6 * X cm.

The length of the middle line of the trapezoid is: KP = (BC + AD) / 2 = (X + 5 * X) / 2 = 3 * X cm.

The segments KM and PN are the middle lines of the triangles ABC and DBC, then KM = PN = BC / 2 = X / 2.

Then МN = КР – КМ – РN = 3 * X – X = 2 * X cm.

The area of ​​the trapezoid is (BC + AD) * BE / 2.

Triangle area TBD = TD * BE / 2 = (AT + AD) * BE / 2 = (BC + AD) * BE / 2.

Savsd = Stvd.

Triangles TBD and MON are similar, since all their sides are parallel in pairs. Then МN / ТD = 2 * X / 6 * / = 1/3.

Then Smon / Savsd = (1/3) 2 = 1/9.

Answer: The area ratio is 1/9.



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