Given a parallelogram ABCD, AB = 12 cm, AC = 16 cm. The vertex D is 4 cm away from the diagonal AC

Given a parallelogram ABCD, AB = 12 cm, AC = 16 cm. The vertex D is 4 cm away from the diagonal AC. Calculate the distance from point D to line AB.

1) The AC diagonal divides ABCD into two equal triangles ABC and ACD. Triangle area S (ACD) = AC * DH, where AC = 16 cm; DH = 4cm – distance from point D to AC. Then S (ADC) = 16 * 4/2 = 32cm ^ 2.

2) Area S (ABCD) = 2 * S (ACD) = 32 cm ^ 2 * 2 = 64 cm ^ 2.

3) Also, the area can be defined as follows: S (ABCD) = AB * DH1, where DH1 is the height from point D to the side AB, or the distance from point D to AB. Let’s find the distance DH1:

DH1 = S (ABCD) / AB = 64 cm ^ 2/12 cm = 16/3 cm = 5 1/3 cm.

Answer: the distance from point D to AB is 5 1/3 cm.