Question

in the diagram below, $overrightarrow{qr}$ is perpendicular to $overline{bd}$.
find the length of $overline{bd}$.
if entering your answer as a decimal, round your final answer to the nearest hundredth.
$bd = \square$ units

Explanation:

Step1: Identify Similar Triangles

Triangles \(ABQ\) and \(CDQ\) are similar because \(\angle ABQ = \angle CDQ = 90^\circ\) and \(\angle AQB=\angle CQD\) (since \(QR\perp BD\) and vertical angles or complementary angles). So, the ratios of corresponding sides are equal. Let \(QD = x\). Then \(\frac{AB}{CD}=\frac{BQ}{QD}\), so \(\frac{32.0}{19.2}=\frac{15.0}{x}\).

Step2: Solve for \(QD\)

Cross - multiply: \(32.0x=19.2\times15.0\). Calculate \(19.2\times15.0 = 288\). Then \(x=\frac{288}{32.0}=9.0\).

Step3: Calculate \(BD\)

\(BD = BQ + QD\). We know \(BQ = 15.0\) and \(QD = 9.0\), so \(BD=15.0 + 9.0=24.0\)? Wait, no, wait. Wait, actually, the ratio: Wait, maybe I mixed up. Wait, \(\triangle ABQ\sim\triangle QDC\) (right triangles, and \(\angle AQB+\angle CQD = 90^\circ\), \(\angle AQB+\angle BAQ=90^\circ\), so \(\angle BAQ=\angle CQD\)). So \(\frac{AB}{QD}=\frac{BQ}{CD}\). Let's correct that. So \(AB = 32.0\), \(BQ = 15.0\), \(CD = 19.2\). So \(\frac{32.0}{QD}=\frac{15.0}{19.2}\). Then cross - multiply: \(15.0\times QD=32.0\times19.2\). \(32.0\times19.2 = 614.4\). Then \(QD=\frac{614.4}{15.0}=40.96\)? Wait, no, that can't be. Wait, the diagram: \(B\), \(Q\), \(D\) are on a straight line. \(AB\perp BD\), \(CD\perp BD\), \(QR\perp BD\). So \(\angle ABQ=\angle QDC = 90^\circ\), and \(\angle AQB+\angle CQD = 90^\circ\), \(\angle AQB+\angle BAQ = 90^\circ\), so \(\angle BAQ=\angle CQD\). So \(\triangle ABQ\sim\triangle QDC\) by AA similarity. So \(\frac{AB}{QD}=\frac{BQ}{CD}\). So \(AB = 32\), \(BQ = 15\), \(CD = 19.2\). So \(\frac{32}{QD}=\frac{15}{19.2}\). Then \(QD=\frac{32\times19.2}{15}\). Calculate \(32\times19.2 = 614.4\), then \(614.4\div15 = 40.96\)? Wait, but then \(BD=BQ + QD=15 + 40.96 = 55.96\)? Wait, no, maybe I had the ratio reversed. Let's check again. If \(\triangle ABQ\sim\triangle QDC\), then corresponding sides: \(AB\) corresponds to \(QD\), \(BQ\) corresponds to \(DC\). So \(\frac{AB}{QD}=\frac{BQ}{DC}\). So \(AB = 32\), \(BQ = 15\), \(DC = 19.2\). So \(QD=\frac{AB\times DC}{BQ}=\frac{32\times19.2}{15}=\frac{614.4}{15}=40.96\). Then \(BD=BQ + QD=15 + 40.96 = 55.96\)? Wait, that seems large. Wait, maybe the first ratio was wrong. Let's do it again. Let's denote \(BQ = x\), \(QD = y\), \(BD=x + y\). \(\triangle ABQ\) and \(\triangle QDC\) are similar. So \(\frac{AB}{CD}=\frac{BQ}{QD}\). \(AB = 32\), \(CD = 19.2\), \(BQ = 15\). So \(\frac{32}{19.2}=\frac{15}{y}\). Cross - multiply: \(32y=19.2\times15\). \(19.2\times15 = 288\). \(y=\frac{288}{32}=9\). Then \(BD=15 + 9 = 24\). Ah, here we go. I had the correspondence wrong. \(\triangle ABQ\sim\triangle CDQ\) (wait, no, \(AB\) and \(CD\) are the vertical sides, \(BQ\) and \(QD\) are the horizontal sides). So \(\frac{AB}{CD}=\frac{BQ}{QD}\). So \(32/19.2 = 15/QD\). Then \(QD=(19.2\times15)/32\). \(19.2\times15 = 288\), \(288/32 = 9\). Then \(BD=BQ + QD=15 + 9 = 24\). Wait, that makes sense. So the correct ratio is \(\frac{AB}{CD}=\frac{BQ}{QD}\) because \(\triangle ABQ\) and \(\triangle QDC\) are similar (right angles, and \(\angle AQB=\angle QCD\) or something? Wait, no, let's use the right angles. \(\angle ABQ=\angle QDC = 90^\circ\), and \(\angle AQB+\angle CQD = 90^\circ\), \(\angle CQD+\angle QCD = 90^\circ\), so \(\angle AQB=\angle QCD\). So \(\triangle ABQ\sim\triangle QDC\) (AA), so \(\frac{AB}{QD}=\frac{BQ}{CD}\) is wrong, \(\frac{AB}{CD}=\frac{BQ}{QD}\) is correct. Because \(AB\) corresponds to \(CD\) (both vertical), \(BQ\) corresponds to \(QD\) (both horizontal). So \(\frac{32}{19.2}=\frac{15}{QD}\), so \(QD=\frac{19.2\times15}{32}=9\). Then \(BD = 15 + 9=24\). Wait, but that seems too simple. Wait, maybe the diagram has \(BQ = 15\), \(AB = 32\), \(CD = 19.2\). So using similar triangles, the ratio of the heights (AB and CD) is equal to the ratio of the bases (BQ and QD). So \(\frac{AB}{CD}=\frac{BQ}{QD}\), so \(QD=\frac{CD\times BQ}{AB}=\frac{19.2\times15}{32}=9\). Then \(BD=BQ + QD=15 + 9 = 24\).

Answer:

\(24.00\)