Question

what is the length of bc, rounded to the nearest tenth? 13.0 units 28.8 units 31.2 units 33.8 units

Explanation:

Step1: Find the length of \(AB\)

Use the Pythagorean theorem \(AB=\sqrt{AD^{2}+BD^{2}}\). Given \(AD = 5\) and \(BD=12\), then \(AB=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13\).

Step2: Prove \(\triangle ABD\sim\triangle BCD\)

Since \(\angle ADB=\angle BDC = 90^{\circ}\) and \(\angle A+\angle ABD = 90^{\circ}\), \(\angle ABD+\angle CBD=90^{\circ}\), so \(\angle A=\angle CBD\). By AA (angle - angle) similarity criterion, \(\triangle ABD\sim\triangle BCD\).

Step3: Use the similarity ratio

From \(\triangle ABD\sim\triangle BCD\), we have \(\frac{AB}{BC}=\frac{BD}{CD}\). Also, from \(\triangle ABD\sim\triangle BCD\), we can use another ratio \(\frac{AB}{BC}=\frac{AD}{BD}\). Let's use the formula \(BC=\frac{AB\times BD}{AD}\) (derived from \(\frac{AB}{BC}=\frac{AD}{BD}\)). Substitute \(AB = 13\), \(BD = 12\), \(AD = 5\).
\(BC=\frac{13\times12}{5}=\frac{156}{5}=31.2\)

Answer:

31.2 units