In the figure below, triangle $ABC$ and rectangle $PQRS$ have the same area and the same height 1. For each value of $x$ between 0 and 1, the trapezoid $ABED$ of height $x$ is drawn, and then the rectangle $PQNM$ of area equal to that of the trapezoid, as shown in the figure. Let $f$ be the function that associates each $x$ with the height of rectangle $PQNM$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-13.jpg?height=303&width=1005&top_left_y=1182&top_left_x=431)

a) What is the ratio between $AB$ and $PQ$?

b) What is the value of $f\left(\frac{1}{2}\right)$?

c) Find the expression for $f(x)$ and draw the graph of $f$.
Solution

a) Let $m$ and $n$ be the measures of the bases of triangle $ABC$ and rectangle $PQRS$, respectively, as shown in the figure below. Since the height of these figures is 1, it follows that area $(ABC) = \frac{m}{2}$ and area $(PQRS) = n$. From the equality of these areas, we have $\frac{m}{2} = n$, hence $\frac{m}{n} = 2$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-41.jpg?height=240&width=657&top_left_y=2259&top_left_x=605)

b) When $x = \frac{1}{2}$, points $D$ and $E$ coincide with the midpoints $T$ and $U$ of sides $AC$ and $BC$, respectively. If $V$ is the midpoint of side $AB$, we can decompose triangle $ABC$ into four congruent triangles, as shown in the following figure.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-42.jpg?height=280&width=582&top_left_y=228&top_left_x=840)

Thus,

$$
\text{area}(ABUT) = \frac{3}{4} \text{ area}(ABC) = \frac{3}{4} \frac{m}{2} = \frac{3m}{8}
$$

On the other hand, we have

$$
\text{area}(PQNM) = f\left(\frac{1}{2}\right) n
$$

Thus, for the areas to be equal, we must have:

$$
f\left(\frac{1}{2}\right) n = \frac{3m}{8} = \frac{3(2n)}{8} = \frac{3n}{4}
$$

Hence, $f\left(\frac{1}{2}\right) = \frac{3}{4}$.

c) First, we will calculate the area of trapezoid $ABED$ in terms of $x$. Since $DE$ is parallel to $AB$, triangles $DEC$ and $ABC$ are similar; the ratio of similarity is the ratio of their heights, which is $\frac{1-x}{1} = 1-x$. Since the areas of similar figures are in the ratio of the square of the similarity ratio, it follows that

$$
\text{area}(DEC) = (1-x)^2 \text{ area}(ABC) = \frac{(1-x)^2 m}{2}
$$

Thus,

$$
\text{area}(ABED) = \text{area}(ABC) - \text{area}(DEC) = \frac{m}{2} - \frac{(1-x)^2 m}{2} = \left(2x - x^2\right) n
$$

From the equality of the areas of $ABED$ and $PQMN$, it follows that

$$
\left(2x - x^2\right) n = f(x) n
$$

We conclude that $f(x) = 2x - x^2$. The figure below shows the graph of $f(x)$ for $0 \leq x \leq 1$.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-42.jpg?height=288&width=297&top_left_y=1895&top_left_x=979)