Find the area of a kite with diagonals of 28ft and 13ft?

A kite is called a quadrilateral that has two adjacent sides of equal length and the other two sides of equal in length. If the kite ABCD has AB = AD and CB = CD, then diagonals AC and BD are perpendiculars and AC bisects BD.

Let AC = 28 ft, and BD = 13 ft. Let say that the two diagonals intersect each other at the point E. In the kite ABCD, we have two congruent triangle, the triangle ABC and the triangle ADC, where the diagonal AC is the common base, BE and DE are their altitudes. Since AC bisect BD, we are able to find the area of the kite, which is equal to 2 times the area of one of these congruent triangles.

Let's find it:

Area of the triangle ABC:

AC = 28 ft and BE = 6.5 ft (13/2)

A = (1/2)(AC)(BE) = (1/2)(28)(6.5) = 91 ft^2

Thus the area of the kite is 182 ft^2 (2 x 91).