Question

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

Answer 1

A tangent line to a circle is perpendicular to the radius drawn to the tangent point   ⇒
AB ⊥ BC and CD ⊥ BC  ⇒ ABCD is a right trapezoid. 

You can find the AD using formula:

$AD= \sqrt{BC^2+(AB-CD)^2} \\ \\ AD= \sqrt{19^2+(10-7)^2} = \sqrt{361+9}= \sqrt{370} \approx 19.2$

Answer 2

Answer:

1. 19.2

Step-by-step explanation:    

Please find the attachment.

Since we know that radius is perpendicular to tangent of a circle. So AB will be perpendicular to BC at c and DC is perpendicular to CB at C.

Now we will construct a perpendicular line to radius AB at point E from the center of our small circle. Since we have two right angles at point B and C so we will also have right angles at point E and D as well.    

Length of CD is is 7 so length of BE will be 7 as well and length of EA will be 10-7=3. Length of DE will be equal to length BC that is 19.

Now we have formed a right triangle and now we will use Pythagoras theorem to find the length of AD.  

$(AD)^{2}=(DE)^{2}+(EA)^{2}$    

Upon substituting our values in above formula we will get,

$(AD)^{2}=(19)^{2}+(3)^{2}$

$(AD)^{2}=361+9$

$(AD)^{2}=370$

Upon taking square root of both sides of our equation we will get,

$AD=\sqrt{370}$  

$AD=19.2353840616713448\approx 19.2$

Therefore, the length of AD will be 19.2 and 1st option is the correct choice.