In the trapezium ABCD base BC = 24 and BP = 30, lateral side AB = 3 and angle BAD = 30 degrees. Find the area of the triangle

| January 4, 2021 | education

In the trapezium ABCD base BC = 24 and BP = 30, lateral side AB = 3 and angle BAD = 30 degrees. Find the area of the triangle COD, where O is the point of intersection of the diagonals.

Let’s build the height of the HВ trapezoid of AВСD.
In a right-angled triangle ABН, the BН leg is located opposite the angle of 300, then its length is equal to half the length AB.
ВН = AB / 2 = 3/2 = 1.5 cm.
Let’s draw the height of the MK through the point of intersection of the diagonals. The ВOС and AOD triangles are similar on two sides, with a similarity coefficient equal to: K = 24/30 = 4/5.
Then MO / KO = 4/5.
MO = 4 * KO / 5.
MK = MO + KO = 4 * KO / 5 + KO = 9 * KO / 5 = 3/2.
KO = 5 * 3/2 * 9 = 5/6 cm.
Determine the area of ​​the triangle AСD. Sasd = AD * ВН / 2 = 30 * 1.5 / 2 = 22.5 cm2.
Determine the area of ​​the triangle AOD. Saod = AD * KO / 2 = 30 * (5/6) / 2 = 12.5 cm2.
Then Ssod = Sasd – Saod = 22.5 – 12.5 = 10.5 cm2.
Answer: The area of ​​the СOD triangle is 10 cm2.



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