A plane is drawn through one side of the rhombus at a distance of 4m from the opposite side. The projection of the diagonals on this side of the plane is 8m and 2m. Find side projections?
From the right-angled triangle ACC1, we determine the length of the hypotenuse AC, which is the diagonal of the rhombus ABCD.
AC ^ 2 = AC1 ^ 2 + CC1 ^ 2 = 64 + 16 = 80.
AC = √80 cm.
From the right-angled triangle DBB1 we determine the length of the hypotenuse BD, which is the diagonal of the rhombus ABCD.
BD ^ 2 = DB1 ^ 2 + BB1 ^ 2 = 4 + 16 = 20.
ВD = √20 cm.
Since the diagonals of the rhombus ABCD are divided in half at point O, OS = √80 / 2, OD = √20 / 2.
Then in a right-angled triangle AOD, according to the Pythagorean theorem, DC ^ 2 = OC ^ 2 + OD ^ 2 = (√80 / 2) ^ 2 + (√20 / 2) ^ 2 = 20 + 5 = 25.
DC = 5 cm.
In a right-angled triangle DCC1, we determine the length of the leg DС1.
DC1 ^ 2 = DC ^ 2 – CC1 ^ 2 = 25 – 16 = 9.
DC1 = 3 cm.Since the triangles ABB1 and DCC1 are equal, the projection of the AB side is also 3 cm, and the projection of the BC side is 5 cm, since BB1C1C is a rectangle and BC = B1C1 = 5 cm.
Answer: The projections of the sides of the rhombus are 3 cm and 5 cm.