The perimeter of the rhombus is 40, one of its diagonals is 10. Find the corners of the rhombus.

A rhombus is a parallelogram in which all sides are equal and the angles are not right.

Since the perimeter of the rhombus is 40, and all its four sides are equal, then:

AB = BC = CD = AD = P / 4;

AB = BC = CD = AD = 40/4 = 10 cm.

Consider the triangle ΔABС.

Since the sides AB and BC are 10 cm, and the AC diagonal is also 10 cm, this triangle is equilateral.

Since in an equilateral triangle all the angles are equal, and the sum of all the angles of the triangle is 180º, then:

∠А = ∠В = ∠С = 180º / 3 = 60º.

Thus: ∠ABS = ∠ADС = 60º.

Since the sum of all the angles of the rhombus is 360º, then:

∠BAD = ∠BCD = (360º – ∠ABС – ∠ADC) / 2;

∠BAD = ∠BCD = (360º – 60º – 60º) / 2 = 240/2 = 120º.

Answer: ∠ABS = ∠ADС = 60º; ∠BAD = ∠BCD = 120º.