Point K is marked on the AD side of parallelogram ABCD so that AK = 4cm, KD = 5cm, BK = 12cm.

Point K is marked on the AD side of parallelogram ABCD so that AK = 4cm, KD = 5cm, BK = 12cm. The diagonal BD is 13 cm. A) Prove that triangle BKD is right-angled. b) Find the area of triangle ABK and parallelogram ABCD.

Let us check the work of the Pythagorean theorem in the triangle ВDК.

DB ^ 2 = 132 = 169.

ВK ^ 2 + DK ^ 2 = 144 + 25 = 169.

169 = 169, then the triangle ВDК is right-angled, as required.

Since ВDK is a right-angled triangle, then the ABK triangle is also right-angled, then Savk = AK * ВK / 2 = 4 * 12/2 = 24 cm2.

Since ВK is perpendicular to AD, then ВK is the height of the parallelogram, then Savsd = AD * ВK = 9 * 12 = 108 cm2.

Answer: The area of the AВK triangle is 24 cm2, the parallelogram area is 108 cm2.