The bisectors of angles A and B of the parallelogram ABCD intersect at point E.

| August 7, 2021 | education

The bisectors of angles A and B of the parallelogram ABCD intersect at point E. Find the area of the parallelogram if BC = 12 cm, and the distance from point E to side AB is 9 cm.

Through point E we draw a straight line MK parallel to the bases of the BC and AD.

Consider triangles EBH and EBM. Both triangles are right-angled, the hypotenuse BE of the triangles is common, and the angle EBH = EBM since BE is the bisector of the angle ABC, then the triangles EBH and EBM are equal in hypotenuse and acute angle, and therefore ME = NOT = 9 cm.

Similarly, right-angled triangles AEN and AEK are equal in hypotenuse and acute angle, and therefore KE = HE = 9 cm.

Then the height of the parallelogram MK = ME + KE = 9 + 9 = 18 cm.

Determine the area of ​​the parallelogram. Savsd = BC * MK = 12 * 18 = 216 cm2.

Answer: The area of ​​the parallelogram is 216 cm2.



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