Question

note: figure not drawn to scale. in the figure shown, $ab = \sqrt{34}$ units, $ac = 3$ units, and $ce = 21$ units. what is the area, in square units, of triangle $ade$?

Explanation:

Step1: Find length of BC using Pythagorean theorem

In right - triangle ABC, by the Pythagorean theorem \(AB^{2}=AC^{2}+BC^{2}\). Given \(AB = \sqrt{34}\) and \(AC = 3\), we have \((\sqrt{34})^{2}=3^{2}+BC^{2}\), so \(34 = 9+BC^{2}\), then \(BC^{2}=34 - 9=25\), and \(BC = 5\).

Step2: Prove similarity of triangles

Since \(\angle ACB=\angle AED = 90^{\circ}\) and \(\angle A\) is common to both \(\triangle ABC\) and \(\triangle ADE\), \(\triangle ABC\sim\triangle ADE\) (by AA similarity criterion).

Step3: Find the ratio of sides

The ratio of the sides of similar triangles is determined by the ratio of corresponding sides. \(AE=AC + CE=3 + 21=24\). The ratio of the sides of \(\triangle ABC\) to \(\triangle ADE\) is \(\frac{AC}{AE}=\frac{3}{24}=\frac{1}{8}\). Since the ratio of the sides of similar triangles is \(k=\frac{1}{8}\), and the ratio of the heights (BC and DE) of similar triangles is the same as the ratio of the bases. So if \(BC = 5\), then \(\frac{BC}{DE}=\frac{1}{8}\), and \(DE=40\).

Step4: Calculate the area of \(\triangle ADE\)

The area of a triangle is \(A=\frac{1}{2}\times base\times height\). For \(\triangle ADE\), the base \(AE = 24\) and the height \(DE = 40\). So \(A=\frac{1}{2}\times24\times40 = 480\).

Answer:

480