Question

Find the value or measure. Assume all lines that appear to be tangent are tangent. X=

Answer 1

According to the secant-tangent theorem, we have the following expression:

`(x+3)^2=10.8(19.2+10.8)`

Now, we solve for x.

`\begin{gathered} x^2+6x+9=10.8(30) \\ x^2+6x+9=324 \\ x^2+6x+9-324=0 \\ x^2+6x-315=0 \end{gathered}`

Then, we use the quadratic formula:

`x_(1,2)=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}`

Where a = 1, b = 6, and c = -315.

`\begin{gathered} x_(1,2)=\frac{-6\pm\sqrt[]{6^2-4\cdot1\cdot(-315)}}{2\cdot1} \\ x_(1,2)=\frac{-6\pm\sqrt[]{36+1260}}{2}=\frac{-6\pm\sqrt[]{1296}}{2} \\ x_(1,2)=(-6\pm36)/(2) \\ x_1=(-6+36)/(2)=(30)/(2)=15 \\ x_2=(-6-36)/(2)=(-42)/(2)=-21 \end{gathered}`

Hence, the answer is 15 because lengths can't be negative.