Question

BC is tangent to circle A at B and to circle D at C. What is AD to the nearest tenth?

Answer 1

Answer:

(C) 18.1

Step-by-step explanation:

Given: BC is tangent to circle A at B and to circle D at C.

Construction:Join DE such that DE=BC=18.

Solution: Since the radius of the circle A is BA=7 and the radius of the circle D is CD=5, then EA=7-5=2.

Now, from the ΔDEA, we have

$(AD)^{2}=(ED)^{2}+(AE)^{2}$

⇒$(AD)^{2}=(18)^{2}+(2)^{2}$

⇒$(AD)^{2}=324+4$

⇒$AD=\sqrt{328}$

⇒$AD=18.1$

Therefore, the value of AD is 18.1.

Answer 2

$Look\ at\ the\ picture.\\\\Use\ Pythagorean\ theorem:\\\\|AD|^2=2^2+18^2\\\\|AD|^2=4+324\\\\|AD|^2=328\\\\|AD|=\sqrt{328}\\\\\boxed{|AD|\approx18.1}\to\fbox{C.}$