The rhombus (AFCE) is inscribed in a rectangle (ABCD)....

The rhombus (AFCE) is inscribed in a rectangle (ABCD). The length has a width of 20 yards and a length of 25 yards, what would be the total length (the perimeter) of a fence along the sides defined by AFCE?

A. 80 yards.
B. 82 yards.
C. 84 yards.
D. 85 yards.
E. 90 yards.

The OA is B.

Please, can any expert assist me with this PS question? I don't have it clear and I appreciate if any explain it for me. Thanks.

Hi AAPL,
Let's take a look at your question.

Since AFCE is a rhombus, there it has all sides of equal length let's say 'x'.
$$CD=25$$
$$AD=20$$
Let's suppose,
$$DE\ =\ y$$
Therefore, we can write:
$$DE\ +EC=\ 25$$
$$y+x=\ 25$$
$$x=\ 25-y...\left(i\right)$$

In the right triangle ADE,
$$\left(AE\right)^2=\left(AD\right)^2+\left(DE\right)^2$$
Since AE is the side of the rhombus so it is x as we supposed in the beginning and DE is y, therefore,
$$\left(x\right)^2=\left(20\right)^2+\left(y\right)^2$$
$$\left(x\right)^2=400+\left(y\right)^2$$

Pluggingin the value of x from eq(i), we get:
$$\left(25-y\right)^2=400+\left(y\right)^2$$
$$625-50y+y^2=400+y^2$$
$$625-50y=400$$
$$50y=625-400$$
$$50y=225$$
$$y=\frac{225}{50}$$
$$y=4.5$$

We need to find the perimeter of the rhombus, for that purpose we will calculate the side length of the rhombus i.e. x.
We know from equation (i), that:
$$x=25-y$$
$$x=25-4.5$$
$$x=20.5$$

Perimeter of Rhombus = 4 * (Length of side)
$$Perimeter\ =\ 4x$$
$$Perimeter\ =\ 82$$

Therefore, Option B is correct.

Hope it helps.
I am available if you'd like any follow up.