solve it iq of 115

Let $O$ be the center of circle of radius 4. We note that

$\begin{aligned} AB & = A'O + B'O & \small \blue{\text{By Pythagorean theorem}} \\ & = \sqrt{\left(\frac 72\right)^2-\left(\frac 52\right)^2} + \sqrt{5^2-1^2} \\ & = \sqrt 6 + 2\sqrt 6 = \boxed{3\sqrt 6} \end{aligned}$

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Reference: Pythagorean theorem

The length of $\overline {AB}$ is the sum of two distances, namely, the horizontal distance between the centres of the left and the middle circle, which is equal to $\sqrt {(3.5)^2-(2.5)^2}=\sqrt {6}$ and the horizontal distance between the middle and the right circles, which is equal to $\sqrt {(2+3)^2-1^2}=2\sqrt {6}$ . Thus the required distance is $\boxed {3\sqrt {6}}$

The process of elimination is the most useful in this question.

We know that AB must be around 7 units long (4+3), therefore the first answer, $\sqrt 6$ , is not right.

Likewise, the second option, 3.5 is not right either, because it is much less than AB.

The third option, $\Large \frac{5 \sqrt[3]6}{3}$ , is not right, because it is almost the same length as the radius of the 6 diameter circle.

Therefore, the last option, $3 \sqrt 6$ must be right, as it is just greater than 7 units, being $\large \boxed {7.34}$ , as expected.

No Pythagoras or any fancy stuff required!