In the diagram, $\triangle A B C$ is right-angled at $C$. Also, points $M, N$ and $P$ are the midpoints of sides $B C, A C$ and $A B$, respectively. If the area of $\triangle A P N$ is $2 \mathrm{~cm}^{2}$, then the area of $\triangle A B C$ is
(A) $8 \mathrm{~cm}^{2}$
(B) $16 \mathrm{~cm}^{2}$
(C) $6 \mathrm{~cm}^{2}$
(D) $4 \mathrm{~cm}^{2}$
(E) $12 \mathrm{~cm}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-099.jpg?height=271&width=458&top_left_y=1296&top_left_x=1300)
Solution 1

Since the ratio $A N: A C$ equals the ratio $A P: P B$ (each is $1: 2$ ) and $\angle A$ is common in $\triangle A P N$ and $\triangle A B C$, then $\triangle A P N$ is similar to $\triangle A B C$.

Since the ratio of side lengths between these two triangles is $1: 2$, then the ratio of areas is $1: 2^{2}=1: 4$.

Thus, the area of $\triangle A B C$ is $4 \times 2=8 \mathrm{~cm}^{2}$.

## Solution 2

Since the ratio $A N: A C$ equals the ratio $A P: P B$ (each is $1: 2$ ) and $\angle A$ is common in $\triangle A P N$ and $\triangle A B C$, then $\triangle A P N$ is similar to $\triangle A B C$.

Therefore, $\angle A N P=\angle A C B=90^{\circ}$.

Similarly, $\triangle P M B$ is similar to $\triangle A C B$, and $\angle P M B=\angle A C B=90^{\circ}$.

Therefore, $A N=N C=P M=\frac{1}{2} A C$ and $N P=C M=M B=\frac{1}{2} C B$.

Thus, $\triangle P M B$ is congruent to $\triangle A N P$, so has area $2 \mathrm{~cm}^{2}$.

Rectangle $N P M C$ has the same width and height as $\triangle A N P$, so has double the area of $\triangle A P N$ or $4 \mathrm{~cm}^{2}$.

Thus, the area of $\triangle A C B$ is $2+4+2=8 \mathrm{~cm}^{2}$.

## Solution 3

Join $C$ to $P$.

Since $C P$ is the diagonal of the rectangle, then $\triangle C P N$ and $\triangle P C M$ are congruent, so have equal areas.

Since $A N=N C$ and $\triangle P N A$ and $\triangle P N C$ have equal height $(P N)$, then the areas of $\triangle P N A$ and $\triangle P N C$ are equal.

Similarly, the areas of $\triangle P M C$ and $\triangle P M B$ are equal.

In summary, the areas of the four small triangles are equal.

Since the area of $\triangle A P N$ is $2 \mathrm{~cm}^{2}$, then the total area of $\triangle A B C$ is $8 \mathrm{~cm}^{2}$.

ANSwER: (A)