Question

1. in right triangle abc, altitude cd with length 6 is drawn to its hypotenuse. we also know ad = 12. what is the length of db?

a. \\(\frac{1}{2}\\)
b. 3
c. 4
d. 6

Explanation:

Step1: Recall the geometric mean theorem (altitude-on-hypotenuse theorem) for right triangles. The theorem states that in a right triangle, the altitude drawn to the hypotenuse is the geometric mean between the segments into which it divides the hypotenuse. Mathematically, for right triangle \(ABC\) with right angle at \(C\) and altitude \(CD\) to hypotenuse \(AB\), we have \(CD^{2}=AD\times DB\).

Given \(CD = 6\) and \(AD = 12\), we substitute these values into the formula. So, \(6^{2}=12\times DB\).

Step2: Solve the equation for \(DB\). First, calculate \(6^{2}\), which is \(36\). So the equation becomes \(36 = 12\times DB\). To find \(DB\), we divide both sides of the equation by \(12\): \(DB=\frac{36}{12}\).

Step3: Simplify the fraction \(\frac{36}{12}\). Dividing numerator and denominator by their greatest common divisor (which is 12), we get \(DB = 3\). Wait, but let's check again. Wait, maybe I misread the triangle. Wait, the triangle is labeled with \(A\), \(B\), \(C\), \(D\). Wait, maybe the right angle is at \(C\), and \(AB\) is the hypotenuse, \(CD\) is the altitude. Then the geometric mean theorem is \(CD^{2}=AD\times DB\). Wait, but if \(AD = 12\), \(CD = 6\), then \(6^{2}=12\times DB\) => \(36 = 12DB\) => \(DB = 3\). But the options are \(\frac{1}{2}\), \(3\), \(4\), \(6\). So option b is 3. Wait, but let's confirm the theorem again. In a right triangle, the altitude to the hypotenuse: \(CD\perp AB\), \(\angle ACB = 90^{\circ}\), then \(\triangle ACD\sim\triangle CBD\sim\triangle ABC\). So from similarity, \(\frac{CD}{AD}=\frac{DB}{CD}\), which is the same as \(CD^{2}=AD\times DB\). So yes, that formula is correct. So \(6^{2}=12\times DB\) => \(DB = \frac{36}{12}=3\).

Answer:

b. 3