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

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 (altitude-on-hypotenuse theorem)

For right triangle \(ABC\) with altitude \(BD\) to hypotenuse \(AC\), \(BD^{2}=AD\times DC\) and \(AB^{2}=AD\times AC\), \(BC^{2}=DC\times AC\). First, find \(AC\) using \(AB^{2}=AD\times AC\):
\(13^{2}=5\times AC\Rightarrow AC=\frac{169}{5}=33.8\). Wait, no—wait, actually, the theorem is \(AB^{2}=AD\times AC\), so \(AC=\frac{AB^{2}}{AD}=\frac{169}{5}=33.8\)? No, that’s not right. Wait, correct approach: Let \(DC = x\). Then \(BD^{2}=AD\times DC\Rightarrow12^{2}=5x\Rightarrow x=\frac{144}{5}=28.8\). Then \(AC = AD + DC=5 + 28.8 = 33.8\). Now, in right triangle \(BDC\), \(BC=\sqrt{BD^{2}+DC^{2}}=\sqrt{12^{2}+28.8^{2}}=\sqrt{144 + 829.44}=\sqrt{973.44}\approx31.2\). Wait, no—wait, let's recalculate \(DC\): From \(BD^{2}=AD\times DC\), \(DC=\frac{BD^{2}}{AD}=\frac{144}{5}=28.8\). Then \(BC=\sqrt{BD^{2}+DC^{2}}=\sqrt{12^{2}+28.8^{2}}=\sqrt{144 + 829.44}=\sqrt{973.44}\approx31.2\). Wait, but also, using \(AB^{2}+BC^{2}=AC^{2}\) (since \(ABC\) is right? Wait, the diagram shows \(BD\perp AC\), and \(\angle ABD\) is part of it. Wait, maybe \(ABC\) is right-angled at \(B\)? Wait, the diagram has a right angle at \(D\) (BD perpendicular to AC) and a right angle at \(B\)? Wait, no, the diagram: \(BD\) is altitude to \(AC\) in right triangle \(ABC\) (right-angled at \(B\)). So by geometric mean, \(AB^{2}=AD\times AC\), \(BC^{2}=DC\times AC\), and \(BD^{2}=AD\times DC\). We found \(AB = 13\), \(AD = 5\), so \(AC=\frac{AB^{2}}{AD}=\frac{169}{5}=33.8\). Then \(DC = AC - AD=33.8 - 5 = 28.8\). Then \(BC^{2}=DC\times AC=28.8\times33.8\). Wait, no, that's the other geometric mean. Wait, \(BC^{2}=DC\times AC\)? Wait, no, in right triangle with altitude to hypotenuse, the leg squared is equal to the product of the adjacent segment and the hypotenuse. So \(BC^{2}=DC\times AC\), \(AB^{2}=AD\times AC\). So \(BC^{2}=DC\times AC=(AC - AD)\times AC=(33.8 - 5)\times33.8=28.8\times33.8\). Wait, but also, \(BC=\sqrt{BD^{2}+DC^{2}}=\sqrt{12^{2}+28.8^{2}}=\sqrt{144 + 829.44}=\sqrt{973.44}\approx31.2\). Let's compute \(28.8\times33.8\): \(28.8\times33.8 = (30 - 1.2)(30 + 3.8)=900 + 114 - 36 - 4.56 = 900 + 73.44 = 973.44\), so \(\sqrt{973.44}=31.2\). So \(BC\approx31.2\) units.

Answer:

31.2 units (corresponding to the option "31.2 units")