Question

what is the length of \\(\overline{bc}\\), rounded to the nearest tenth?\
\\(\bigcirc\\) 13.0 units\
\\(\bigcirc\\) 28.8 units\
\\(\bigcirc\\) 31.2 units\
\\(\bigcirc\\) 33.8 units\
(image of triangle with points a, b, c, d; ad = 5, bd = 12, right angles at d and at b between ab and bc)

Explanation:

Step1: Find length of AB

In right triangle \(ABD\), \(AD = 5\), \(BD = 12\). By Pythagorean theorem, \(AB=\sqrt{AD^{2}+BD^{2}}=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13\).

Step2: Use geometric mean theorem

In right triangle \(ABC\) with altitude \(BD\), we have \(AB^{2}=AD\times AC\) and \(BC^{2}=CD\times AC\), also \(BD^{2}=AD\times CD\). First, find \(CD\) from \(BD^{2}=AD\times CD\), so \(12^{2}=5\times CD\), \(CD=\frac{144}{5}=28.8\). Then \(AC = AD + CD=5 + 28.8 = 33.8\). Now, use \(AB^{2}=AD\times AC\) (we can also use similar triangles or Pythagorean theorem for \(BC\)). Alternatively, in right triangle \(BDC\), \(BD = 12\), \(CD = 28.8\), so \(BC=\sqrt{BD^{2}+CD^{2}}=\sqrt{12^{2}+28.8^{2}}=\sqrt{144 + 829.44}=\sqrt{973.44}\approx31.2\). Wait, correction: Wait, first, from \(BD^{2}=AD\times CD\), \(CD=\frac{BD^{2}}{AD}=\frac{144}{5}=28.8\). Then \(AC=AD + CD = 5+28.8 = 33.8\). Then, since \(\triangle ABD\sim\triangle ACB\) (right triangles with common angle at \(A\)), so \(\frac{AB}{AC}=\frac{AD}{AB}\), so \(AB^{2}=AD\times AC\) (which we saw \(13^{2}=5\times33.8\), \(169 = 169\), correct). Then, for \(BC\), since \(\triangle BDC\sim\triangle ABC\), or use Pythagorean in \(\triangle BDC\): \(BC=\sqrt{BD^{2}+CD^{2}}=\sqrt{12^{2}+28.8^{2}}=\sqrt{144 + 829.44}=\sqrt{973.44}\approx31.2\).

Answer:

31.2 units (the option C: 31.2 units)