Exercise 23

In the square \(\square ABCD \), \(E\) and \(F\) are midpoints. The ratio between the area of the trapezoid \(AFCD\) and the area of the triangle \( \triangle FEC \) is:

  1. \( 0.75 \)

  2. \( 1.33 \)

  3. \( 0.125 \)

  4. \( 6 \)

  5. \( 4 \)

Try to solve it before checking the answer.

  1. \( 6 \)

Let \(L\) be the side of the square. The bases \(B\) and \(b\), and height \(h\) of the trapezoid \(AFCD\) are the following:

\[ B = L, \quad b = \frac{L}{2} \quad \text{ and } \quad h = L \]

According Triangle properties, the area is:

\[ \begin{aligned} A_1 &= \frac{ (B + b) }{2} \\[.5em] &= \frac{ \left( \frac{L}{2} + L\right) L }{ 2 } \\[.5em] &= \frac{ (L + 2L) L }{4} \\[.5em] &= \frac{3L^2}{4} \end{aligned} \]

The base \(b\) and height \(h\) of the triangle are, respectively:

\[ b = \frac{L}{2} \quad \text{ and } \quad h = \frac{L}{2} \]

Therefore, its area is:

\[ \begin{aligned} A_2 &= \frac{b \times h}{2} \\[.5em] &= \frac{ \frac{L}{2} \times \frac{L}{2} }{2} \\[.5em] &= \frac{ \frac{L^2}{4} }{2} \\[.5em] &= \frac{L^2}{8} \end{aligned} \]

The ratio between both areas is:

\[ \begin{aligned} \frac{A_1}{A_2} &= \frac{ \frac{3 L^2}{4} }{ \frac{L^2}{8} } \\[1em] &= \frac{8 \left( 3L^2 \right)}{4 L^2} \\[1em] &= \frac{24}{4} \\[1em] &= \boldsymbol{6} \end{aligned} \]

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