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2 years ago

Folding a rectangle along one diagonal results in a shape whose area is 2/3 that of the original

Mathematics

1 answer:

Charra [1.4K]2 years ago

3 0

Answer:

60°

Step-by-step explanation:

Refer to attached picture

Given rectangle ABCD. Its folded shape is given as well.

It has sides a and b and diagonal d, with the angle BCD marked as α.

We need to find the value of angle AOC or BOD.

This is going to be 2α, which is easy to find out. BOD is cut by the angle bisector, each of the angles is α.

Let the area of the rectangle is A, then the folded shape has area 2/3A.

The area of rectangle ABCD:

A = ab

From the triangle CBD we get:

a/d = sin α ⇒ a = d sin α
b/d = cos α ⇒ b = d cos α

Then

A = d sin α * d cos α = d²sin α cos α

Now, lets find the area of the folded shape.

It is the area of the rectangle minus the area of the triangle CMB as this part is lost as part of overlap.

Area of the triangle:

A = 1/2dh

Since h/(d/2) = tan α ⇒ h = 1/2d tan α, the area is:

A = 1/2d*1/2d tan α = 1/4d² tan α

We have the difference of the areas which is 1/3 of the area of the rectangle:

1/3(d²sin α cos α) = 1/4d² tan α
sin α cos α = 3/4 tan α
sin α cos α = 3/4 sin α / cos α
cos² α = 3/4
cos α = √(3/4)
cos α = √3/2
α = arccos (√3/2)
α = 30°

The acute angle formed by the two diagonals of the rectangle is:

2α = 2*30° = 60°

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