ABCD is a rectangle. BD and AC are the diagonals that meet at point O. AB = 6 cm, BC = 8 cm.

ABCD is a rectangle. BD and AC are the diagonals that meet at point O. AB = 6 cm, BC = 8 cm. Find P of triangle ABO, and S of triangle ABO.

1.Start with the fact that the diagonals of the rectangle divide it into 4 equal triangles, then S Δ AOB we will find as 1/4 of the area of the rectangle ABCD
Therefore S Δ AOB = 1/4 * (6 * 8) = 12
2. Perimeter is the sum of the lengths of all sides.
Consider Δ AOB, let the base be AB, and the sides of the BO and AO will be equal, since the diagonals of the rectangle are equal and the intersection point is halved.
Therefore, we can find one side – subtract the base from the area and divide by two
(12-6) / 2 = 3, and now P = 3 + 3 + 6 12
Answer S = 12 P = 12