Question

use the geometric mean (leg) theorem. what is the value of ( a )?
options: ( 7sqrt{2} ), ( 2sqrt{70} ), ( 20sqrt{5} ), ( 70sqrt{5} ).
diagram: triangle with points ( a ), ( c ), ( b ), ( d ); ( ad = 6 ), ( db = 14 ), right angles at ( c ) (between ( a ) and ( b )) and ( d ) (between ( c ) and ( b )); ( ac = b ), ( bc = a ), ( cd = h ).

Explanation:

Step1: Recall Geometric Mean (Leg) Theorem

The geometric mean (leg) theorem states that in a right triangle, the length of a leg is the geometric mean of the length of the hypotenuse segment adjacent to that leg and the length of the entire hypotenuse. For right triangle \( \triangle ABC \) with right angle at \( C \) and altitude \( CD \) to hypotenuse \( AB \), we have \( a^2 = AD \times AB \). Here, \( AD = 6 \), \( DB = 14 \), so \( AB=AD + DB=6 + 14 = 20 \).

Step2: Apply the Theorem to Find \( a \)

Using the formula \( a^2=AD\times AB \), substitute \( AD = 6 \) and \( AB = 20 \):
\( a^2=6\times20=120 \)? Wait, no, wait. Wait, actually, the geometric mean (leg) theorem: the leg \( a \) (which is \( BC \)) is the geometric mean of the adjacent segment \( DB \) and the hypotenuse \( AB \)? Wait, no, let's correct. The correct formula is: In right triangle \( ABC \), right-angled at \( C \), and \( CD \perp AB \), then \( BC^2=BD\times BA \), and \( AC^2 = AD\times AB \). So here, \( a = BC \), \( BD = 14 \), \( BA=6 + 14=20 \). So \( a^2=14\times20 = 280 \)? Wait, no, that's not matching. Wait, maybe I mixed up. Wait, \( AD = 6 \), \( DB = 14 \), so \( AB=20 \). Then for leg \( a \) (BC), the geometric mean theorem says \( a^2=DB\times AB \)? Wait, no, let's rederive.

In \( \triangle ABC \sim \triangle CBD \) (by AA similarity, since \( \angle B \) is common and \( \angle ACB=\angle CDB = 90^\circ \)). So the ratio of corresponding sides: \( \frac{BC}{BD}=\frac{AB}{BC} \), so \( BC^2=BD\times AB \). So \( a^2=14\times(6 + 14)=14\times20 = 280 \)? Wait, but 280 is not a perfect square. Wait, maybe I made a mistake. Wait, the other leg: \( AC^2=AD\times AB=6\times20 = 120 \), but we need \( a \), which is \( BC \). Wait, but the options are \( 7\sqrt{2} \), \( 2\sqrt{70} \), \( 20\sqrt{5} \), \( 70\sqrt{5} \). Let's compute \( \sqrt{280}=\sqrt{4\times70}=2\sqrt{70} \). Ah, there we go. So \( a^2 = 14\times20 = 280 \), so \( a=\sqrt{280}=\sqrt{4\times70}=2\sqrt{70} \).

Answer:

\( 2\sqrt{70} \) (corresponding to the option "2√70")