In square $A B C D, P$ is the midpoint of $D C$ and $Q$ is the midpoint of $A D$. If the area of the quadrilateral $Q B C P$ is 15 , what is the area of square $A B C D$ ?
(A) 27.5
(B) 25
(C) 30
(D) 20
(E) 24

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-062.jpg?height=347&width=352&top_left_y=1732&top_left_x=1339)
Solution 1

Draw line segment $Q R$ parallel to $D C$, as in the following diagram. This segment divides square $A B C D$ into two halves. Since triangles $A B Q$ and $R Q B$ are congruent, each is half of rectangle $A B R Q$ and therefore one quarter of square $A B C D$. Draw line segment $P S$ parallel to $D A$, and draw line segment $P R$. Triangles $P D Q, P S Q, P S R$ and $P C R$ are congruent. Therefore each is one quarter of rectangle $D C R Q$ and therefore one eighth of square $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-277.jpg?height=452&width=466&top_left_y=1346&top_left_x=884)

Quadrilateral $Q B C P$ therefore represents $\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{5}{8}$ of square $A B C D$. Its area is therefore $\frac{5}{8}$ of the area of the square.

Therefore, $\frac{5}{8}$ of the area of the square is equal to 15 . Therefore, $\frac{1}{8}$ of the area of the square is equal to 3 . Therefore the square has an area of 24 .

## Solution 2

Draw a line segment from $Q$ to $R$, the midpoint of $B C$.

Draw a line segment from $P$ to $S$, the midpoint of $Q R$.

Let the area of $\triangle Q S P$ equal $x$. Thus, the area of $\triangle Q D P$ is also $x$ and $Q D P S$ has area $2 x$.

Square $S P C R$ is congruent to square $Q D P S$ and thus has area $2 x$.

Rectangle $Q D C R$ has area $4 x$, as does the congruent rectangle $A Q R B$.

Also, $\triangle A Q B$ and $\triangle B R Q$ have equal areas and thus, each area is $2 x$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-278.jpg?height=452&width=466&top_left_y=159&top_left_x=884)

Quadrilateral $Q B C P$ is made up of $\triangle B R Q, \triangle Q S P$ and square $S P C R$, and thus has area $2 x+x+2 x=5 x$.

Since quadrilateral $Q B C P$ has area 15 , then $5 x=15$ or $x=3$.

Therefore, the area of square $A B C D$, which is made up of quadrilateral $Q B C P, \triangle A Q B$ and $\triangle Q D P$, is $5 x+2 x+x=8 x=8(3)=24$.

ANSWER: (E)