In the figure below, $ABCD$ is a rectangle, $M$ and $N$ are points on the sides $BC$ and $AD$, respectively, and the numbers represent the areas of triangles $ABQ$, $BQM$, $MPC$, and $CPD$ in square centimeters.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-12.jpg?height=228&width=591&top_left_y=2256&top_left_x=641)

a) What is the area of triangle $AMD$? Why?

b) Calculate the sum of the areas of triangles $AQN$ and $NPD$.

c) Calculate the area of quadrilateral $MPNQ$.
Solution

We recall that the area of a triangle is given by the formula $\frac{1}{2}$ base $\times$ height and the area of a rectangle by base $\times$ height. In the figure below, we conclude that the area of the shaded triangle is half the area of the rectangle, as both have the same base and the same

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-36.jpg?height=132&width=215&top_left_y=702&top_left_x=1503)
height. Therefore, the sum of the areas of the two white triangles is also half the area of the rectangle, that is, equal to the area of the shaded triangle.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-36.jpg?height=194&width=508&top_left_y=974&top_left_x=680)

a) As seen above, we have area $(A M D)=$ area $(A B M)+$ area $(M D C)=16 \mathrm{~cm}^{2}+8 \mathrm{~cm}^{2}+27 \mathrm{~cm}^{2}+9 \mathrm{~cm}^{2}=60 \mathrm{~cm}^{2}$. b) Since area $(A M D)=$ area $(B N C)$, we have area $(A Q N)+$ area $(N D P)=$ area $(A M D)-$ area $(M Q N P)=$ area $(B N C)-$ area $(M Q N P)=$ area $(B Q M)+$ area $(M P C)=8 \mathrm{~cm}^{2}+27 \mathrm{~cm}^{2}=35 \mathrm{~cm}^{2}$

c) We have

$$
\text { area }(M Q N P)=\text { area }(B N C)-\text { area }(B Q M)-\text { area }(M P C)=60 \mathrm{~cm}^{2}-8 \mathrm{~cm}^{2}-27 \mathrm{~cm}^{2}=25 \mathrm{~cm}^{2}
$$

Note: The areas of the triangles in this problem were not chosen at random. We leave it as an exercise for the reader to show that it is possible to construct the figure below, where $a, b, c$ and $d$ represent the areas of the corresponding triangles, if and only if $\frac{a^{2}}{b}+\frac{d^{2}}{c}=b+c$.

![](https://cdn.mathpix.com/cropped/2024_05_01_984603b876c0bc28095bg-36.jpg?height=208&width=363&top_left_y=1678&top_left_x=755)

Alternative solution for items b) and c): We can solve item $c$) first, as follows. Since area $(A M D)=$ area $(B N C)=60 \mathrm{~cm}^{2}$, we have

$$
\text { area }(M N P Q)=\text { area }(B N C)-27 \mathrm{~cm}^{2}-8 \mathrm{~cm}^{2}=25 \mathrm{~cm}^{2} \text {, }
$$

then for item $b$)

$$
\text { area }(A Q N)+\text { area }(N D P)=\text { area }(A M D)-\text { area }(M N P Q)=60 \mathrm{~cm}^{2}-25 \mathrm{~cm}^{2}=35 \mathrm{~cm}^{2}
$$