The two diagonals of rectangle ABCD intersect at point O. given that the angle AOD is equal to 120 ° and ab is equal to 2.5cm, find the length of the diagonal of rectangle?

The two diagonals of rectangle ABCD intersect at point O. given that the angle AOD is equal to 120 ° and ab is equal to 2.5cm, find the length of the diagonal of rectangle?

It is suggested that if OE ⊥ ad draws a vertical line and the foot is e, then AB = 2oe and OE = 1.25
It is easy to prove that ∠ AOE = 60 ° and double the hypotenuse with sin

As shown in the figure, the diagonals AC and BD of rectangle ABCD intersect at O, and the area of rectangle ABCD is calculated when ∠ AOD = 120 ° AB = 4 is known

The area of rectangular ABCD: 4 × 43 = 163

The circumference of rectangle ABCD is 20cm, the diagonal intersects at point O, the difference between the circumference of △ AOD and △ AOB is 1cm, and the length of AB is calculated

"Cold → wound": let AB be wide and ad be long, because the diagonals of the rectangle are equal to each other, so ob = OD; and AO = Ao △ AOD is 1cm larger than the perimeter of △ AOB, that is to say, length ad is 1cm longer than width AB, length + width = 20cm △ 2 = 10cm, length width = 1cm, length = (10 + 1) cm △ 2 = 5.5cm, width = (10-1) cm △ 2 = 4.5cm

The perimeter of rectangle ABCD is 14cm, the diagonal intersects o, the perimeter difference between △ AOD and △ AOB is 1cm, what is the area of rectangle

12 square cm