In the rectangle ABCD, the diagonals intersect at point O. Find the perimeter of the triangle AOB, if the angle CAD = 30º, AC = 12cm
Since the diagonals of the rectangle are equal and are divided by the point of intersection in half, then in the triangle AOB the lengths of the sides AO and OB are the same and equal to half of the diagonal AC:
| AС | = | ОВ | = | AC | / 2 = 12/2 = 6 cm.
Let’s find the length of the side AB.
According to the condition of the problem, the CAD angle is 30º.
Since the sides AB and CD of this rectangle are parallel, the angle ACB is equal to the angle CAD and is also 30º.
Then the ACB angle is 90 – 30 = 60º.
Consider a triangle ABC.
In this triangle, the angle ABC is right, and the diagonal AC is the hypotenuse.
Using the formulas of a right-angled triangle, we find leg AB:
| AB | = | AC | * sin (60º) = 12 * √3 / 2 = 6√3 cm.
Knowing all sides of the triangle AOB, we find its perimeter:
| AB | + | AO | + | ОВ | = 6√3 + 6 + 6 = 6√3 + 12 cm.
Answer: the perimeter of the AOB triangle is 6√3 + 12 cm.