0
$\begingroup$

The question is as follows:

The length of segment AB is 20 cm. Find the distance from C to AB, given that C is a point on the circle that has AB as a diameter, and that
(a) $AC$ = $CB$; (b) $AC$ = 10 cm; (c) $AC$ = 12 cm

According the Thales' Theorem, if $A$, $B$, and $C$ are distinct points on a circle where the line $AB$ is the diameter of the circle, then $\angle ABC$ is a right angle.

For (a) I was thinking of right triangle $ABC$, where $\angle ACB$ is a right angle and the diameter $AB$ is the hypotenuse of the right angle. Since $AC = BC$, it shows that it is a 45-45-90 triangle because of its side lengths $x- x- x\sqrt{2}$, and I got $x = 10\sqrt{2}$cm. However, I was thinking that $10\sqrt{2}$cm is only the measure from $C$ to the points $A$ and $B$, so to actually get from $C$ to $AB$, I thought of creating another right triangle where the right angle would be on the intersection from the line drawn from $C$ to the diameter and the hypotenuse would be $10\sqrt{2}$cm and the measure of a leg would be $10$cm (half of the diameter) and I would solve for the other leg, which I got to be $10$cm as well. Is this approach correct?

Also, for (b), I was only able to get to finding the measure of $BC$ to be $10\sqrt{3}$cm, but I don't know if I should do additional steps to find the distance from $C$ to the diameter. I thought of doing $sin(60) = \frac{d}{10}$, where $d$ is the measure of the distance from $C$ to $AB$, and I got $d \approx 8.66$cm.

$\endgroup$
  • $\begingroup$ For part (a), if AC = AB, doesn't that just mean B = C and thus the distance is 0? Because AB is a diameter. $\endgroup$ – paulinho Sep 17 '17 at 15:56
2
$\begingroup$

I believe part (a) is simpler than you think. It is given that $AB$ is a diameter. If $AC=AB$, then $AC$ must be a diameter as well. For any point on a circle, there is only one other point on the circle such that the line segment between the two points is a diameter. Thus, for part (a), the distance is 0.

For part (b), you are on the right track - it is indeed a right triangle. In fact, it is a special right triangle - a 30-60-90. So you just need to find the altitude of the triangle. There is an easy way to do this. The area is half the product of an altitude and a base. So if the altitude from the hypotenuse to the base is $\alpha$, we have $$\frac{\alpha \cdot AB}{2} = \frac{AC \cdot BC}{2}$$ and therefore $$\alpha = AC \cdot BC / AB.$$ And of course since it is a 30-60-90 triangle, $$BC = 10\sqrt{3}.$$

Part (c) is essentially the same, except with a 12-16-20 (which is a 3-4-5 triangle) right triangle. So in this case we have $AC = 12, BC = 16, AB = 20$, and the same formula from above applies for the closest distance (altitude) $\alpha$.

Hope this helps.

$\endgroup$
  • $\begingroup$ Ahh I just realized I made a typo! So sorry about that! It may have thrown off your answer. It's actually $AC = CB$. $\endgroup$ – geo_freak Sep 17 '17 at 16:10
  • $\begingroup$ Oh I see. In that case, you know that it is a 45-45-90 triangle with a hypotenuse of 20 cm. $\endgroup$ – paulinho Sep 17 '17 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.