Question

in the diagram, the length of (overline{yz}) is twice the length of (overline{az}). (overline{ya}) is an altitude of (\triangle xyz). what is the length of (overline{ya})? (\bigcirc) (5sqrt{3}) units (\bigcirc) (10sqrt{3}) units (\bigcirc) 15 units (\bigcirc) 20 units

Explanation:

Step1: Analyze triangle angles

In right triangle \( \triangle XYZ \), \( \angle Y = 90^\circ \), \( YZ = 10 \), and \( YZ = 2 \cdot AZ \)? Wait, no—wait, \( YZ \) is twice \( AZ \)? Wait, the problem says "the length of \( \overline{YZ} \) is twice the length of \( \overline{AZ} \)". Wait, \( YZ = 10 \), so \( AZ=\frac{YZ}{2}=5 \). Wait, no, maybe \( \triangle YAZ \) is a 30-60-90 triangle? Wait, \( YA \) is an altitude. Wait, first, let's find \( AZ \). Given \( YZ = 10 \), and \( YZ = 2 \cdot AZ \), so \( AZ = 5 \), \( YZ = 10 \), so \( AY \) is the altitude. Wait, in right triangle \( \triangle XYZ \), \( \angle Y = 90^\circ \), \( YA \perp XZ \), so \( \triangle YAZ \sim \triangle XYZ \)? Wait, maybe \( \angle Z \) is 30 degrees? Wait, if \( YZ = 10 \), and \( YZ = 2 \cdot AZ \), then \( AZ = 5 \), so in \( \triangle YAZ \), right-angled at \( A \), \( YZ = 10 \) (hypotenuse), \( AZ = 5 \) (one leg), so by Pythagoras, \( YA = \sqrt{YZ^2 - AZ^2} = \sqrt{10^2 - 5^2} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3} \)? Wait, no, wait—wait, the problem says "the length of \( \overline{YZ} \) is twice the length of \( \overline{AZ} \)". Wait, \( YZ = 10 \), so \( AZ = 5 \). Then \( YA \) is the altitude. Wait, but maybe \( \triangle XYZ \) is a 30-60-90 triangle. Wait, \( YZ = 10 \), \( AZ = 5 \), so \( XZ = AZ + AX \)? Wait, no, maybe \( \angle Z = 30^\circ \), so in \( \triangle YAZ \), \( \cos Z = \frac{AZ}{YZ} = \frac{5}{10} = \frac{1}{2} \), so \( \angle Z = 60^\circ \)? Wait, no, \( \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \), so if \( AZ = 5 \), \( YZ = 10 \), then \( \cos Z = \frac{AZ}{YZ} = \frac{5}{10} = \frac{1}{2} \), so \( \angle Z = 60^\circ \), then \( \sin Z = \frac{YA}{YZ} \), so \( YA = YZ \cdot \sin Z = 10 \cdot \sin 60^\circ = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \). Wait, no, that's not right. Wait, \( YA \) is the altitude, so \( \triangle YAZ \) is right-angled at \( A \), so \( YA^2 + AZ^2 = YZ^2 \). \( AZ = 5 \), \( YZ = 10 \), so \( YA = \sqrt{10^2 - 5^2} = \sqrt{75} = 5\sqrt{3} \). Wait, but the options include \( 5\sqrt{3} \), \( 10\sqrt{3} \), 15, 20. So the answer should be \( 5\sqrt{3} \)? Wait, no, maybe I made a mistake. Wait, the problem says "the length of \( \overline{YZ} \) is twice the length of \( \overline{AZ} \)". Wait, \( YZ = 10 \), so \( AZ = 5 \), so \( YZ = 2 \cdot AZ \), so \( AZ = 5 \), \( YZ = 10 \). Then \( YA \) is the altitude, so in right triangle \( YAZ \), \( YA = \sqrt{YZ^2 - AZ^2} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3} \). So the answer is \( 5\sqrt{3} \) units.

Step2: Verify using 30-60-90 triangle properties

Alternatively, if \( YZ = 10 \) and \( AZ = 5 \), then \( \angle Z = 60^\circ \) (since \( \cos Z = \frac{AZ}{YZ} = \frac{1}{2} \)), so in \( \triangle YAZ \), \( YA = YZ \cdot \sin 60^\circ = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \).

Answer:

\( 5\sqrt{3} \) units