In the rectangle ABCD, the diagonals intersect at point O, find the perimeter of the triangle AOB

In the rectangle ABCD, the diagonals intersect at point O, find the perimeter of the triangle AOB, if the angle CAD = 30 °, AC = 14cm.

A rectangle is a quadrangle in which all angles are right and equal to each other.

The diagonals of the rectangle are equal and the intersection point is halved.

Based on this:

AO = BO = OC = OD = AC / 2;

AO = BО = OC = OD = 14/2 = 7 cm.

In order to calculate the perimeter of the triangle ΔAOB, we find the length of the side AB.

To do this, consider the triangle ΔCAD. This triangle is right-angled, since the angle ∠СDА = 90 °. Using the sine theorem, we calculate the length of the side CD.

The sine of an acute angle of a right triangle is the ratio of the opposite leg to the hypotenuse:

sin A = CD / AC;

CD = AC ∙ sin A;

sin 30 ° = 1/2;

CD = 14 1/2 = 7 cm.

Since in a rectangle the opposite sides are equal, then AB = CD = 7 cm.

The perimeter of a triangle is the sum of all its sides:

P = AB + BO + OA = 7 + 7 + 7 = 21 cm.

Answer: the perimeter of the triangle ΔAOB is 21 cm.