In the diagram, $A B C D$ is a rectangle. If the area of triangle $A B P$ is 40 , then the area of the shaded region is
(A) 20
(B) 40
(C) 60
(D) 50
(E) 80

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-040.jpg?height=287&width=333&top_left_y=1637&top_left_x=1362)
Solution 1

Construct $P H$ perpendicular to $A B$, as shown.

The area of $\triangle A B P$ is 40 and so $\frac{1}{2} \times A B \times P H=40$, or

$A B \times P H=80$, and since $A B=10$, then $P H=8$.

Since $C B=P H=10$, the area of $A B C D$ is $10 \times 8=80$.

The shaded area equals the area of $\triangle A B P$ subtracted from the area of $A B C D$, or $80-40=40$.

## Solution 2

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-197.jpg?height=328&width=336&top_left_y=649&top_left_x=1469)

As in Solution 1, we construct $P H$ perpendicular to $A B$.

Since both $D A$ and $C B$ are perpendicular to $A B$, then $P H$ is parallel to $D A$ and to $C B$.

That is, $D A H P$ and $P H B C$ are rectangles.

Diagonal $P A$ divides the area of rectangle $D A H P$ into two equal areas, $\triangle P A H$ and $\triangle P A D$. Diagonal $P B$ divides the area of rectangle $P H B C$ into two equal areas, $\triangle P B H$ and $\triangle P B C$. Therefore, the area of $\triangle P A H$ added to the area of $\triangle P B H$ is equal to the area of $\triangle P A D$ added to the area of $\triangle P B C$.

However, the area of $\triangle P A H$ added to the area of $\triangle P B H$ is equal to the area of $\triangle A B P$, which is 40 .

So the area of $\triangle P A D$ added to the area of $\triangle P B C$ is also 40 .

Therefore, the area of the shaded region is 40 .

ANSWER: (B)