Question

question 3 (multiple choice worth 2 points) (03.06a lc) segment ad is an altitude of triangle abc. if ad = 15, dc = 8, and ac = 17, find ba. round to the tenths place if necessary 31.9 28.1 9.1 7.1

Explanation:

Step1: Use the Pythagorean theorem in right - triangle ADC

In right - triangle ADC, by the Pythagorean theorem \(AD^{2}+DC^{2}=AC^{2}\) (which is already verified as \(15^{2}+8^{2}=225 + 64=289=17^{2}\)). In right - triangle ADB, \(\angle ADB = 90^{\circ}\).

Step2: Use the Pythagorean theorem in right - triangle ADB

Let \(BD=x\). First, note that \(\triangle ADB\sim\triangle ADC\) (by AA similarity since \(\angle ADB=\angle ADC = 90^{\circ}\) and \(\angle B+\angle BAD = 90^{\circ}\), \(\angle C+\angle CAD=90^{\circ}\) and \(\angle BAD+\angle CAD = 90^{\circ}\), so \(\angle B=\angle CAD\)). Also, we can use the geometric mean theorem or the Pythagorean theorem directly. In right - triangle ADB, \(BA^{2}=AD^{2}+BD^{2}\). Since \(\triangle ADB\sim\triangle ADC\), we know that \(\frac{BD}{AD}=\frac{AD}{DC}\), so \(BD=\frac{AD^{2}}{DC}\). Substituting \(AD = 15\) and \(DC = 8\), we get \(BD=\frac{15^{2}}{8}=\frac{225}{8}=28.125\). Then, by the Pythagorean theorem \(BA=\sqrt{AD^{2}+BD^{2}}=\sqrt{15^{2}+28.125^{2}}=\sqrt{225 + 790.90625}=\sqrt{1015.90625}\approx31.9\)

Answer:

31.9