A rectangle $A B C D$ is divided into four smaller rectangles. The areas of three of them are indicated in the given figure. What is the area of rectangle $A B C D$?
(a) 80
(b) 84
(c) 86
(d) 88
(e) 91

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-011.jpg?height=312&width=396&top_left_y=735&top_left_x=1481)
The correct option is (e).

Solution 1: First, observe that the ratio between the areas of two rectangles that have the same base is equal to the ratio between their heights. Indeed, in the figure on the left, two rectangles are represented that have the same base $b$ and heights $h_{1}$ and $h_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-049.jpg?height=257&width=585&top_left_y=2190&top_left_x=290)

| 16 | $S$ |
| :---: | :---: |
| 12 | 27 |

Their areas $S_{1}$ and $S_{2}$ are given by $S_{1}=b h_{1}$ and $S_{2}=b h_{2}$, therefore,

$$
\frac{S_{1}}{S_{2}}=\frac{b h_{1}}{b h_{2}}=\frac{h_{1}}{h_{2}}
$$

Applying this observation to the two pairs of rectangles given (see the figure above, on the right) and denoting by $S$ the area of the fourth rectangle, we have

$$
\frac{S}{27}=\frac{h_{1}}{h_{2}}=\frac{16}{12}=\frac{4}{3}
$$

so that $S=\frac{1}{3}(27 \times 4)=36$. Thus, the area of rectangle $A B C D$ is

$$
12+16+27+36=91
$$

Solution 2: Let $x$ and $y$ be the sides of the rectangles with areas 12 and 27, respectively, as indicated in the figure. Then the other sides of these rectangles are $12 / x$ (rectangle with area 12), 16/x (rectangle with area 16), and 27/y (rectangle with area 27), as indicated in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-050.jpg?height=271&width=577&top_left_y=704&top_left_x=1302)

Thus, the length of rectangle $A B C D$ is $x+y$ and its width is $\frac{16}{x}+\frac{12}{x}=\frac{28}{x}$. Clearly, $\frac{12}{x}=\frac{27}{y}$, so $\frac{y}{x}=\frac{27}{12}=\frac{9}{4}$. The area of a rectangle is the product of its length and width. Therefore, the area of $A B C D$ is $A=(x+y) \times \frac{28}{x}=28+\frac{28 y}{x}=$ $28+28 \frac{y}{x}$. Thus, $A=28+28 \times \frac{9}{4}=28+7 \times 9=91$.