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High school geometry
Course: High school geometry > Unit 8
Lesson 10: Properties of tangents- Proof: Radius is perpendicular to tangent line
- Determining tangent lines: angles
- Determining tangent lines: lengths
- Proof: Segments tangent to circle from outside point are congruent
- Tangents of circles problem (example 1)
- Tangents of circles problem (example 2)
- Tangents of circles problem (example 3)
- Tangents of circles problems
- Challenge problems: radius & tangent
- Challenge problems: circumscribing shapes
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Challenge problems: radius & tangent
CCSS.Math:
Solve two challenging problems that apply properties of tangents to find the radius of a circle with a tangent.
Problem 1
Problem 2
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Hey guys, do you guys have any tricks that I could use on a test when doing these types of problems? Any way to do it quicker without all the calculations?(9 votes)
Depending on the test (ACT for example) you will be able to use a calculator. On problem 2, I skipped a few steps by leaving the square integers as they were; as in I wrote 30^2 instead of bothering to calculate that it equals 900. Proceeded with my algebra, then came up with (OC)= ((30^2)-(18^2))/36, which I could just type into my calculator to get 16. As you practice more and more you'll get little nuances like that which work for you to speed up your game, and it all comes from having fun with the numbers and the math and practice.(11 votes)
why is there a 36 i didn't get a 36 when i did my answer(1 vote)
We're just squaring binomials (x+a)^2 = x^2+2ax+a^2(2 votes)
how do you get 36OC from the second problem. wouldn't you get 324 again because you are putting 18 to the power of 2(3 votes)
Lets say OC = a
(a + 18)^2
(a + 18) * (a + 18)
a^2 + 36a +324
http://www.khanacademy.org/math/algebra/introduction-to-polynomial-expressions/multiplying-binomials-2/v/multiplying-binomials(2 votes)
Are they sure it's actually solvable?
Because to me it looks like (talking about the first problem) they're pulling that 4OC out of nowhere, along with adding together OC and 2 in their squared forms, which would not yield the same result as adding them together and THEN squaring them.
Okay, was a bit frustrated writing that, but all that said, could anyone please better explain where some of those extra numbers come from, and how everything ends up working like a regular Pythagorean problem?(1 vote)
We're finding the square of two things added together, so we're using the fact that (a+b)^ 2 = a^2 + b^2 + 2ab
so (OC + 2)^2 = (OC)^2 + (2)^2 + 2(OC)(2)
= OC^2 + 4OC + 4
yay!(4 votes)
- You can use the Pythagorean theorem:
Set OC and OB as x, they are equal because all radii are congruent.
OC is perpendicular to AC (line tangent to a circle is perpendicular to the radius drawn to the point of tangency), making OAC a right triangle.
OA is the hypotenuse, OC and AC are the legs. So you can set the equations for the Pythagorean Theorem:
1) 4^2+x^2=(2+x)^2
16+x^2=4+4x+x^2
16=4+4x
12=4x
x=3
2) 30^2+x^2=(18+x)^2
900+x^2=324+36x+x^2
900=324+36x
576=36x
x=16(1 vote)
Does anyone have a simple explanation for this PLEASE(3 votes)
which problem do you need help on. I would advise rewatching the videos(0 votes)
- Does anyone know a simpler way to explain this? It's a tad bit confusing(2 votes)
- why there is (OC+18) cant be (18+OC)(0 votes)
if the internal secant isnt a radius but a chord, can i still use the pythagorean theorem to find a missing triangle side?(0 votes)- I dont understand this(0 votes)
If you have a radius and a tangent, these will always form a right angle, and if you have a right angle, you can then form a right triangle. You can use the Pythagorean Theorem on any right triangle which is what is being done. If you want to generalize, you have one leg that is the radius (r) and another that we can call b. The hypotenuse is formed by adding r+c. Thus, you get b^2+r^2=(r+c)^2. Solving for r, we square to get b^2 + r^2 = r^2 +2rc + c^2. If we subtract r^2, they cancel on both sides. So we subtract c^2 and divide by 2c to get r=(b^2-c^2)/2c.(0 votes)
