If a diagnol for a rectangle ABCD was measured to be 20 and a square was put in it slanted(looking like a rhombus). This rhombus is called EFGH. Prove find the perimeter of EFGH. (IT IS EASIER TO DRAW IT THEN SOLVE IT =))

I can only speak for myself of course, but I can't figure out what your diagram would look like.

What do you mean by "a square was put in it"?
Where does the square touch ?

on the sides of the square touching and bisecting each side

So the original "rectangle" was also a square ?

No the square inside touches the rectangle on the outside.

I concluded that your problem deals with the midpoint-parallel theorem

that is, "a line joining the midpoints of two sides of a triangle is parallel to the third side and one half its length
So each side of the inside rhombus is 1/2 the diagonal.
There are 4 sides, so their total length is twice the diagonal of the rectangle or
2(20) = 40

BTW, if the original is not a square, then the inside figure also cannot be a square. That is where the confusion arises in your problem, and that is whay nobody answered it. Often questions that are worded in a confusing or contradictory way are left unanswered.

k thanks Reiny

To find the perimeter of rhombus EFGH, we first need to determine the length of its sides. Since EFGH is a rhombus, all four sides have the same length.

In rectangle ABCD, the diagonal is given as 20. Let's use this information to find the length of the side of the rectangle.

We can utilize the Pythagorean theorem to solve for the sides of the rectangle. The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (in this case, the diagonal of the rectangle) is equal to the sum of the squares of the other two sides.

Let's call the length of one side of the rectangle as "a" and the width as "b".

Using the Pythagorean theorem, we have:

a^2 + b^2 = diagonal^2

Since the rectangle is a right-angled triangle, one of the sides is the height (a) and the other side is the base (b). The diagonal acts as the hypotenuse of the right triangle.

The diagonal is given as 20, so the equation becomes:

a^2 + b^2 = 20^2
a^2 + b^2 = 400

Since it is a rectangle, we know that opposite sides are equal, so a = b.

Substituting b with a in the equation above, we get:

a^2 + a^2 = 400
2a^2 = 400
a^2 = 200
a ≈ √200
a ≈ 14.14

Therefore, the length of one side of the rectangle is approximately 14.14.

In the rhombus EFGH, the diagonals bisect each other at right angles. So, EF (and GH) splits AB (and CD) into two equal parts.

As a result, the length of EF is half the length of AB/CD, which is half of the length of the side of the rectangle:

EF = AB/2 = CD/2 ≈ 14.14/2
EF ≈ 7.07

Since all sides of the rhombus are equal, the perimeter of rhombus EFGH is found by multiplying the length of one side (EF) by 4:

Perimeter of rhombus EFGH = 4 * EF = 4 * 7.07 ≈ 28.28

Therefore, the perimeter of rhombus EFGH is approximately 28.28 units.