Question

question 5 of 28 what is the value of x in the diagram below? if necessary, round your answer to the nearest tenth of a unit. a 14 b x d 30 c a. 14 b. 16 c. 30.1 d. 6.5

Explanation:

Step1: Use geometric - mean theorem

In a right - triangle \(ABC\) with right - angle at \(A\) and altitude \(AD\), we have \(AD^{2}=BD\times DC\). First, we can use the Pythagorean theorem in \(\triangle ABD\) and \(\triangle ABC\). Let \(BD = x\) and \(DC = 30\), and \(AB = 14\). In right - triangle \(ABC\), by the Pythagorean theorem \(AC^{2}=BC^{2}-AB^{2}=(x + 30)^{2}-14^{2}\). In right - triangle \(ADC\), \(AC^{2}=AD^{2}+30^{2}\). Also, from the geometric - mean theorem in right - triangle \(ABC\) with altitude \(AD\), we know that \(AB^{2}=BD\times BC\), that is \(14^{2}=x(x + 30)\).

Step2: Expand the equation

Expand \(14^{2}=x(x + 30)\) to get \(196=x^{2}+30x\). Rearrange it to the quadratic form \(x^{2}+30x - 196 = 0\).

Step3: Use the quadratic formula

The quadratic formula for \(ax^{2}+bx + c = 0\) is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 1\), \(b = 30\), and \(c=-196\). First, calculate the discriminant \(\Delta=b^{2}-4ac=(30)^{2}-4\times1\times(-196)=900 + 784 = 1684\). Then \(x=\frac{-30\pm\sqrt{1684}}{2}=\frac{-30\pm41.0366}{2}\).
We have two solutions for \(x\): \(x_1=\frac{-30 + 41.0366}{2}\) and \(x_2=\frac{-30 - 41.0366}{2}\). Since \(x\) represents a length, we discard the negative solution.
\(x=\frac{-30 + 41.0366}{2}=\frac{11.0366}{2}=5.5183\approx6.5\) (rounded to the nearest tenth).

Answer:

D. 6.5