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# Circle theorems — Basic example

Watch Sal work through a basic Circle theorems problem.

## Want to join the conversation?

• Couldn't you find the side of tiangle using the probeties of a 3-4-5 triangle?
• Yes, you can always do that if you encounter a right triangle with a hypotenuse of 5 and one leg measuring 3. Here we DON'T know that the small leg is 3 at first. The reason we can use the 3, 4, 5-triangle AFTER we know for sure that BC = 3 is that we know from using the Pythagorean Theorem (once or dozens of times) that our result will be 4 IF the hypotenuse is 5 and the other leg is 3. Then we learn that 3, 4, 5 is a Pythagorean triplet like 12, 13 , 5 and 24, 7, 25 and 6, 8, 10

HOWEVER, the point with this question is that we don't know without using some other geometry that the small leg is actually 3.

In geometry, we cannot use the fact that it `seems like` about half of 6. We need to use the fact that the radius is the hypotenuse of both right triangles. Then we can use the HL Congruency to show that the triangles are congruent and then using CPCTC that the AC and BC are congruent. Then we know that OC bisects AB, and BC is 3 units long. WHEW!

OR, as Sal did here, we can use the `great shortcut--thanks to one of the circle theorems--`that a radius bisects chord AB if it is perpendicular to it, which is given.

BOOM! We then can be confident that the leg BC is 3 units long and use the other shortcut of the Pythagorean Triple 3, 4, 5 to answer.
• Is the Principal Root just another name for the square root?
• No, a principal root is only the positive answer of the square root. Example: the square root of 25 is 5 and -5 but the principal root is only 5.
• Why can't we use the formula: arc length = Theta * radius
• You could, and that would be a lot faster than the process Sal uses. I think he didn't because to apply that formula your angle value has to be in radians, which is very, very easy to just forget about in the heat of the moment, and so using a simpler slightly more intuitive formula would be better for most students. But if you're nailing questions with arc length = theta * radius, go for it. It's every bit as valid, and all you have to do is convert to radians every now and then.
• As soon as you found the 3 you would notice that its a 3 4 5 right traiangle and done
• At first I thought that the arc AB equals 6 , because it said chord.
• As far as I know, "chord" refers to the line, not the arc.
• I thought the answer would be 135 because I read (in the PWN the SAT book) that the central angle of a circle is equivalent to the arc length...can someone explain in what cases this rule/theorem is valid? is it cause we don't know if O is the center or the circle or not?
• You might have misunderstood what you were reading. The central angle of a circle is equivalent to the angle measure of the arc, not its length. Think about it. Angle by itself cannot be the only factor that determines length, because bigger circles will have bigger arcs than smaller circles would, even if the central angle is the same.
It's weird that the problem doesn't explicitly say that the circle's center is O, probably because it's just for practice. On the actual SAT, you'd be right to be careful of assuming that a given point is the center of a circle, or that two lines are parallel, for example, unless the problem tells you.
• What is a sector?
• but doest 9/4 pi equal 405? why cant 405 be the right answer?