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 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). The ratio of the sides of similar triangles is given by \(\frac{AC}{AE}=\frac{BC}{DE}\). We know that \(AE=AC + CE=3 + 21=24\) units.

Step3: Find DE

From \(\frac{AC}{AE}=\frac{BC}{DE}\), substituting \(AC = 3\), \(AE = 24\), and \(BC = 5\), we get \(\frac{3}{24}=\frac{5}{DE}\), cross - multiplying gives \(3DE=24\times5\), so \(DE = 40\).

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

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

Answer:

480