Question

practice 1
1
in the right triangles shown, the measure of angle abc is the same as the measure of angle ebd.
∠abc ≅ ∠ebd
what is the length of side be?
type your answer in the box.
be =

Explanation:

Step1: Identify similar triangles

Since \(\angle ABC\cong\angle EBD\) and \(\angle BCA=\angle BDE = 90^{\circ}\), \(\triangle ABC\sim\triangle EBD\) (by AA similarity criterion).

Step2: Set up proportion

For similar triangles \(\triangle ABC\) and \(\triangle EBD\), \(\frac{BC}{BD}=\frac{AB}{BE}\). First, find \(BC\) using Pythagorean theorem in \(\triangle ABC\): \(BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{5^{2}-3^{2}}=\sqrt{25 - 9}=\sqrt{16} = 4\).

Step3: Find \(BD\)

Let \(BD=x\). In \(\triangle BDE\), using Pythagorean theorem \(BD=\sqrt{BE^{2}-DE^{2}}\). But from similarity \(\frac{BC}{BD}=\frac{AB}{BE}\). Substitute \(BC = 4\), \(AB = 5\), \(DE=2\). Let \(BE=y\), then \(BD=\sqrt{y^{2}-4}\). So \(\frac{4}{\sqrt{y^{2}-4}}=\frac{5}{y}\). Cross - multiply: \(4y = 5\sqrt{y^{2}-4}\). Square both sides: \(16y^{2}=25(y^{2}-4)\). Expand: \(16y^{2}=25y^{2}-100\). Rearrange: \(25y^{2}-16y^{2}=100\), \(9y^{2}=100\), \(y^{2}=\frac{100}{9}\), \(y=\frac{10}{3}\).

Answer:

\(\frac{10}{3}\)