Question

Ad is tangent to circle o at d. find ab. round to the nearest tenth if necessary

Answer 1

Answer:

(J) 8.4

Step-by-step explanation:

Given: From the given figure, it is given that AD=11, CB=6.

To find: The value of AB.

Solution: Using the tangent-secant segment length theorem that is If a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.

That is:

$(AD)^2=(AC)(AB)$

Now, substituting the values, we have

$(11)^2=(AB+BC)(AB)$

$(11)^2=(AB+6)(AB)$

$121=(AB)^2+6AB$

$(AB)^2+6AB-121=0$

Using the Quadratic formula, we have

$AB=\frac{-6{\pm}\sqrt{6^2-4(1)(-121)}}{2(1)}$

$AB=\frac{-6{\pm}\sqrt{520}}{2}$

$AB=\frac{-6{\pm}\sqrt{4{\cdot}130}}{2}$

$AB=\frac{-6{\pm}2\sqrt{130}}{2}$

$AB=-3{\pm}\sqrt{130}$

Since, the value of AB cannot be negative, thus

$AB=-3+\sqrt{130}$

$AB=8.4$

which is the required value of AB.

Hence, option J is correct.

Answer 2

Check the picture below.

$\bf 0=AB^2+6AB-121\implies \stackrel{\textit{using the quadratic formula}}{AB=\cfrac{-6 \pm \sqrt{6^2-4(1)(-121)}}{2(1)}} \\\\\\ AB=\cfrac{-6 \pm \sqrt{520}}{2}\implies AB=\cfrac{-6 \pm \sqrt{4\cdot 130}}{2} \\\\\\ AB=\cfrac{-6 \pm \sqrt{2^2\cdot 130}}{2}\implies AB=\cfrac{-6 \pm 2\sqrt{130}}{2} \\\\\\ AB=-3\pm\sqrt{130}\implies AB= \begin{cases} \boxed{-3+\sqrt{130}}\\\\ -3-\sqrt{130} \end{cases}$

since the distance AB cannot be a negative value, thus is not -3-√(130).