Exercise 45

In the triangle \( \triangle ABC \) we have that \( \overline{AB} = 6 \, cm \), \( \overline{DC} = 2 \sqrt{5} \, cm \), and \( \overline{DB} = 2 \, cm \). Therefore, the area of the triangle \( \triangle ABC \) is:

  1. \( 2 \; cm^2 \)

  2. \( 4 \sqrt{5 - 1} \; cm^2 \)

  3. \( 6 \; cm^2 \)

  4. \( 8 \; cm^2 \)

  5. \( 4 \left( 2 \sqrt{5 - 1} \right) \; cm^2 \)

Try to solve it before checking the answer.

  1. \(8 \; cm^2 \)

The triangle \( \triangle ADC \) has base \( b = \overline{AD} \) and height \( h = \overline{CB} \). But:

\[ b = \overline{AD} = \overline{AB} – \overline{DB} = 6 – 2 = 4 \]

Furthermore, \( h = \overline{CB} \) is a leg of the right triangle \( \triangle{DBC} \), whose hypotenuse and other leg are, respectively:

\[ \overline{DC} = 2 \sqrt{5} \quad \text{ and } \quad \overline{DB} = 2 \]

Then, according to Pythagoras’ theorem:

\[ \begin{aligned} h^2 + 2^2 = \left( 2 \sqrt{5} \right)^2 &\Rightarrow h^2 + 4 = 20 \\[.5em] &\Rightarrow h^2 = 16 \\[.5em] &\Rightarrow h = 4 \end{aligned} \]

Hence, the area of the triangle \( \triangle ADC \) is:

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

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