BH and BK are the heights of the parallelogram ABCD. The angle between the heights is 45 degrees

| February 6, 2021 | education

BH and BK are the heights of the parallelogram ABCD. The angle between the heights is 45 degrees. Height BH divides side AD into AH = 8 cm and HD = 6 cm. Find the area of the parallelogram.

Let us determine the angles of the ВKDН quadrangle.

Angle ВНD = ВKD = 900, since ВН and ВK are the heights of the parallelogram, then the angle ADС = НDK = 360 – 90 – 90 – 45 = 135.

In a parallelogram, the sum of adjacent angles is 180, the angle ВAD + ADС = 180, then the angle ВAD = 180 – 135 = 45.

In a right-angled triangle ABН, one of the acute angles is 45, then triangle ABН is rectangular and isosceles. BH = AH = 8 cm.

Side of AD = AН + DН = 8 + 6 = 14 cm.

Determine the area of the parallelogram. Savsd = AD * ВН = 14 * 8 = 112 cm2.

Answer: The area of the parallelogram is 112 cm2.



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