Two sheets of $11 \mathrm{~cm} \times 8 \mathrm{~cm}$ paper are placed on top of each other, forming an overlapping $8 \mathrm{~cm} \times 8 \mathrm{~cm}$ square in the centre, as shown. The area of rectangle $W X Y Z$ is
(A) $88 \mathrm{~cm}^{2}$
(B) $112 \mathrm{~cm}^{2}$
(C) $136 \mathrm{~cm}^{2}$
(D) $121 \mathrm{~cm}^{2}$
(E) $176 \mathrm{~cm}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-030.jpg?height=263&width=418&top_left_y=167&top_left_x=1317)
In the diagram, rectangles $W Q R Z$ and $P X Y S$ are the two sheets of $11 \mathrm{~cm} \times 8 \mathrm{~cm}$ paper.

The overlapping square $P Q R S$ has sides of length $8 \mathrm{~cm}$.

That is, $W Q=Z R=P X=S Y=11 \mathrm{~cm}$ and

$W Z=Q R=P S=X Y=P Q=S R=8 \mathrm{~cm}$.

In rectangle $W Q R Z, W Q=W P+P Q=11 \mathrm{~cm}$ and so

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-132.jpg?height=216&width=331&top_left_y=239&top_left_x=1512)

$W P=11-P Q=11-8=3 \mathrm{~cm}$.

In rectangle $W X Y Z, W X=W P+P X=3+11=14 \mathrm{~cm}$.

Since $W X=14 \mathrm{~cm}$ and $X Y=8 \mathrm{~cm}$, the area of $W X Y Z$ is

$14 \times 8=112 \mathrm{~cm}^{2}$.

ANswer: (B)