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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)?
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 the AA (angle - angle) similarity criterion.
Step3: Find the ratio of sides
The ratio of the sides of similar triangles is given by \(\frac{AC}{AE}\). We know \(AC = 3\) and \(AE=AC + CE=3 + 21=24\). The ratio of similarity is \(\frac{AC}{AE}=\frac{3}{24}=\frac{1}{8}\). Then \(\frac{BC}{DE}=\frac{1}{8}\), and since \(BC = 5\), \(DE=40\).
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 \(A=\frac{1}{2}\times24\times40 = 480\).
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