The diagonals of a rectangle are 12 inches and intersect at an angle of 60 degrees. Find the perimeter of the rectangle.
Answer choices:
A-48
B-12√3
C-24√3
D-12+12√3
If the rectangle has AB=width=w and BC=height=h, and the diagonals intersect at O, then angle BOC is 60° and is isosceles, meaning angles OBC and OCB are also 60°. That makes it an equilateral triangle, with side OB=OC = 12/2 = 6
Thus BC = 6 and AB = 6√3
P = 2(AB+BC) = 12(1+√3)
To find the perimeter of the rectangle, we need to have the lengths of its sides.
Let's denote the length of the rectangle as "L" and the width of the rectangle as "W."
Since the diagonals of the rectangle intersect at a 60-degree angle and have a length of 12 inches, we can use trigonometry to find the lengths of the sides.
In a right triangle formed by one of the diagonals, the hypotenuse is 12 inches, and the angle opposite the side "L" is 60 degrees.
Using the definition of sine, we have: sin(60°) = L / 12.
Simplifying, we find: L = 12 × sin(60°) = 12 × (√3 / 2) = 6√3.
Therefore, the length of the rectangle is 6√3 inches.
Since the opposite sides of a rectangle are equal in length, the width of the rectangle is also 6√3 inches.
Now, we can calculate the perimeter by adding up the lengths of all four sides:
Perimeter = 2 × (L + W) = 2 × (6√3 + 6√3) = 2 × 12√3 = 24√3.
Therefore, the correct answer choice is C-24√3.