Rhombus ABCD has perimeter 148, and one of its diagonals has length \(28\). what is the area of ABCD?

Guest Jan 20, 2022

#1**+3 **

Because ABCD is a rhombus, then AB = BC = CD = AD.

AB = BC = CD = AD = 37 units.

One of rhombus ABCD's diagonals has length 28. We can divide 28 by 2 to get 14, which is the leg of a right triangle when rhombus ABCD is split into 4 congruent triangles.

The hypotenuse is 37, and the leg is 14, so that means the other leg is:\(\sqrt{37^2 - 14^2}\) = \(\sqrt{1173}\). Ew that looks ugly, but it is correct. That means the other diagonal has length \(2\sqrt{1173}\).

Since the area of a rhombus is one diagonal length times the other diagonal length over 2, then the area of this rhombus would be \(28\sqrt{1173}\) units squared.

proyaop Jan 23, 2022