Rectangle and midpoints

GRE Quant Problem Solving
1

Hi guys, so I tried solving several ways but I get different results. I’m just trying to understand where my mistake is.

So we’ll take x als the width and y as the length.
Total area rectangle= xy

Area of one shaded triangle
1/2 x . 1/2 y . 1/2 = 1/4x . 1/4y = 1/16xy
There are four triangles so 1/16xy . 4 = 4/16xy = 1/4xy

Area unshaded is xy - 1/4xy = 3/4

I know this is wrong but which mistake am I making?
Also I learned somewhere that when there are midpoints within a rectangle, that will produce a rhombus. How will that help solve this question? Thanks

Screenshot 2022-11-22 at 10.29.28
Screenshot 2022-11-22 at 10.29.28693×358 19.5 KB

2
\dfrac{1}{2} \times (b \times h) = \dfrac{1}{2}\times(\dfrac{1}{2}x\times \dfrac{1}{2}y)= \dfrac{1}{2}\times (\dfrac{1}{4}xy)
3

so the error is that the bxh should be in between brackets and thus should be performed before multiplying by 1/2 again?

4
\text{Area of one shaded Triangle: } \dfrac{1}{2} \times (b \times h) = \dfrac{1}{2}\times(\dfrac{1}{2}x\times \dfrac{1}{2}y)= \dfrac{1}{2}\times (\dfrac{1}{4}xy) = \boxed {\dfrac{1}{8}xy}

Whereas you did: