Question

3
type the correct answer in the box. use numerals instead of words.
find the missing side length, x.
the value of the missing side length labeled as x is
reset next

Explanation:

Step1: Identify the theorem

We can use the Angle - Bisector Theorem. The Angle - Bisector Theorem states that if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the adjacent sides. In triangle \(ABC\), \(AD\) bisects \(\angle BAC\) (since \(\angle BAD=\angle CAD = 18^{\circ}\)), so by the Angle - Bisector Theorem, we have \(\frac{BD}{DC}=\frac{AB}{AC}\).

Step2: Substitute the known values

We know that \(AB = 12\), \(AC=20\), and \(DC = 12\). Let \(BD=x\). Substituting these values into the proportion from the Angle - Bisector Theorem: \(\frac{x}{12}=\frac{12}{20}\).

Step3: Solve for \(x\)

Cross - multiply to get \(20x=12\times12\). Then \(20x = 144\). Divide both sides by 20: \(x=\frac{144}{20}=\frac{36}{5} = 7.2\)? Wait, no, wait. Wait, maybe I made a mistake. Wait, let's re - check. Wait, the Angle - Bisector Theorem: In \(\triangle ABC\), if \(AD\) bisects \(\angle BAC\), then \(\frac{AB}{AC}=\frac{BD}{DC}\). Wait, \(AB = 12\), \(AC = 20\), \(DC=12\), \(BD=x\). So \(\frac{AB}{AC}=\frac{BD}{DC}\) gives \(\frac{12}{20}=\frac{x}{12}\)? No, wait, no. Wait, the sides adjacent to \(\angle BAD\) and \(\angle CAD\): \(AB\) is adjacent to \(\angle BAD\), \(AC\) is adjacent to \(\angle CAD\), and \(BD\) and \(DC\) are the segments of \(BC\). So the correct proportion is \(\frac{AB}{AC}=\frac{BD}{DC}\). So \(AB = 12\), \(AC = 20\), \(BD=x\), \(DC = 12\). So \(\frac{12}{20}=\frac{x}{12}\)? Wait, that would be \(x=\frac{12\times12}{20}=\frac{144}{20} = 7.2\)? But that seems odd. Wait, maybe the triangle is isoceles? Wait, no, wait, maybe I mixed up the sides. Wait, let's look at the triangle again. The side \(AB = 12\), \(AC = 20\), \(DC = 12\), and \(AD\) is the angle bisector. Wait, maybe the correct proportion is \(\frac{BD}{DC}=\frac{AB}{AC}\), so \(BD=\frac{AB\times DC}{AC}\). So \(AB = 12\), \(DC = 12\), \(AC = 20\). Then \(BD=\frac{12\times12}{20}=\frac{144}{20}=7.2\)? But that doesn't seem right. Wait, maybe I made a mistake in the Angle - Bisector Theorem. Wait, the Angle - Bisector Theorem: In a triangle, the angle bisector divides the opposite side into segments proportional to the adjacent sides. So in \(\triangle ABC\), with angle bisector \(AD\) (where \(D\) is on \(BC\)), \(\frac{AB}{AC}=\frac{BD}{DC}\). So \(AB = 12\), \(AC = 20\), \(BD=x\), \(DC = 12\). So \(\frac{12}{20}=\frac{x}{12}\), so \(x=\frac{12\times12}{20}=\frac{144}{20}=7.2\). But let's check again. Wait, maybe the triangle has \(AB = 12\), \(AD\) is the angle bisector, \(\angle BAD=\angle CAD = 18^{\circ}\), \(AC = 20\), \(DC = 12\). So according to the Angle - Bisector Theorem, \(\frac{AB}{AC}=\frac{BD}{DC}\), so \(BD=\frac{AB\times DC}{AC}=\frac{12\times12}{20}=\frac{144}{20}=7.2\). Wait, but maybe I misread the diagram. Wait, the side \(BC\) is composed of \(BD=x\) and \(DC = 12\). The sides \(AB = 12\), \(AC = 20\). So the calculation gives \(x = 7.2\)? But let's do the calculation again. \(12\times12 = 144\), \(144\div20=7.2\). So the value of \(x\) is \(7.2\)? Wait, no, wait, maybe the Angle - Bisector Theorem was applied incorrectly. Wait, maybe the triangle is such that \(AB = 12\), \(AD\) is the angle bisector, and \(AC = 20\), \(DC = 12\), so \(BD=x\). So \(\frac{AB}{AC}=\frac{BD}{DC}\) implies \(x=\frac{AB\times DC}{AC}=\frac{12\times12}{20}=7.2\).

Wait, but let's check with another approach. Let's use the formula for the Angle - Bisector Theorem: \(BD=\frac{AB\times BC}{AB + AC}\)? No, that's not the Angle - Bisector Theorem. The Angle - Bisector Theorem is \(\frac{BD}{DC}=\frac{AB}{AC}\). So if \(DC = 12\), \(AB = 12\), \(AC = 20\), then \(\frac{BD}{12}=\frac{12}{20}\), so \(BD=\frac{12\times12}{20}=7.2\). So the missing side length \(x = 7.2\)? Wait, but the problem says "use numerals instead of words". So the answer is \(7.2\)? Wait, no, wait, maybe I made a mistake in the diagram. Wait, maybe \(DC = 12\), \(AB = 12\), \(AC = 20\), and the angle bisector. Wait, maybe the correct answer is \(7.2\).

Wait, no, wait, let's re - derive the Angle - Bisector Theorem. Let \(\triangle ABC\) with angle bisector \(AD\), \(\angle BAD=\angle CAD\). Then by the Law of Sines in \(\triangle ABD\) and \(\triangle ACD\):

In \(\triangle ABD\): \(\frac{BD}{\sin18^{\circ}}=\frac{AB}{\sin\angle ADB}\)

In \(\triangle ACD\): \(\frac{DC}{\sin18^{\circ}}=\frac{AC}{\sin\angle ADC}\)

Since \(\angle ADB+\angle ADC = 180^{\circ}\), \(\sin\angle ADB=\sin\angle ADC\). So \(\frac{BD}{\sin18^{\circ}}\times\sin\angle ADB=\frac{AB}{\sin\angle ADB}\times\sin18^{\circ}\) (wait, no, cross - multiplying: \(\frac{BD}{\sin18^{\circ}}=\frac{AB}{\sin\angle ADB}\) and \(\frac{DC}{\sin18^{\circ}}=\frac{AC}{\sin\angle ADC}\), and since \(\sin\angle ADB=\sin\angle ADC\), we can divide the two equations: \(\frac{BD}{DC}=\frac{AB}{AC}\), which is the Angle - Bisector Theorem. So that confirms the theorem. So with \(AB = 12\), \(AC = 20\), \(DC = 12\), we have \(\frac{BD}{12}=\frac{12}{20}\), so \(BD=\frac{12\times12}{20}=7.2\).

Answer:

\(7.2\)