Can someone explain why for problem two line BO is included in solving the problem while in problem 1 BO is left out? I understand that for problem 1 using the pythagorean theorem shows its not perpendicular but using that same method for problem 2 doesn't work and thus adding line BO is needed.

You can find the length of BO in either question, using just the radius. This is what you use to find out if it is a right triangle and thus, you need BO. Hope this answers your question...

why is the perpindicularity contradicting it is confusing because they look the same except for line length in both problems.

Since we know all of the lengths in this triangle, we can check if Pythagorean theorem will agree with our assumption that these are right triangles. Pythagorean theorem c² = a² + b² Problem 1: 13² = 5² + 11² 169 = 25 + 121 169 ≠ 146 (These would be equal if we had 90° angle) Problem 2: 20² = 12² + 16² 400 = 144 + 256 400 = 400 (This is a right triangle)

how is angle AOC not a right angled triangle in problem 1

The the first example is not a right triangle because it does not follow the Pythagorean Theorem of a^2 + b^2 = c^2. AC^2+OC^2 doesn't equal AO^2. (11^2 + 5^2 = 13^2, which turns out to be 146 = 169, not true). Both 45-45-90 and 30-60-90 triangles follow this rule.

The first question is vague and doesn't explain how they found the length of AO. Where did y'all even get 8?

8 was given as the length of AB. BO is a radius of the circle and therefore has length of 5. So, AO (which equals AB + BO) is 8 + 5 which is 13. They also had to test whether there was a right angle at point C, so they used the Pythagorean theorem to see whether that was true. It wasn't, so in other words, AC wasn't perpendicular to OC, so therefore line AC was NOT tangent to circle O.

how can we draw 2 common transverse tangents for 2 congruent circles if they have any distance between their centres?

you dont that is something different you are using Pythagorean theorem here.

what is the converse Pythagorean theorem?

Normally we use the Pythagorean Theorem on a Right Triangle to find the length of a missing side measurement. •The *Converse* Pythagorean Theorem is when *we already have the side lengths* of a triangle, but we _*don't know if it's a Right Triangle*_… so it's when we *use the formula _to check if a triangle has a 90° angle*_ through Comparison. We must… •Plug two of the triangle's side measurements into the corresponding variables of the Pythagorean Theorem. •Calculate to see what the third side would be _if it were a Right Triangle_. •Compare the triangle's given measurement to the calculated. •If the corresponding values are equal then it's a Right Triangle and has a 90° angle. If not, it is not. So… •To use the Pythagorean Theorem as its Converse, *is to apply the formula to _prove or disprove_ if a triangle has a 90° angle*, and therefore is, or is not a Right Triangle. (ㆁωㆁ) Hope this helps someone!

what if one has the diameter would it still work?

Yes because you would divide the diameter by 2 to get the radius

Can someone explan #2 to me? I dont understand because 2,3,4 is not a right triangle??

The sides of the triangle in problem 2 are 12, 16, and 20 (12+8), which does make it a right triangle, since 20² = 12²+16².

How do we know that the rest of the line is 12?

cause it touches the edge of the circle. line between the center of the circle that touches the edge equals the radius

Main content

Course: High school geometry > Unit 8

Lesson 10: Properties of tangents

Determining tangent lines: lengths

Google Classroom

Solve two problems that apply properties of tangents to determine if a line is tangent to a circle.

Problem 1

Segment

\overset{―}{O C}

is a radius of circle

O

Note: Figure not necessarily drawn to scale.

Is line $\overset{\leftrightarrow}{A C}$ ‍ tangent to circle $O$ ‍?

(Choice A)
Yes, because $\overset{―}{A C}$ ‍ intersects circle $O$ ‍ at point $C$ ‍
(Choice B)
Yes, because $\overset{―}{A C}$ ‍ is perpendicular to $\overset{―}{O C}$ ‍
(Choice C)
No, because $\overset{―}{A C}$ ‍ is not perpendicular to $\overset{―}{O C}$ ‍
(Choice D)
No, because $△ A O C$ ‍ intersects circle $O$ ‍ at more than one point

line

that is tangent to a circle at a particular point is perpendicular to the

radius

at that point.

Let's check to see whether

\overset{\leftrightarrow}{A C}

is tangent by using the converse pythagorean theorem to determine if

△ A O C

is a right triangle.

We will need to know the length of

\overset{―}{A O}

B O = 5

because it is a radius just like

C O

, so:

$A O = 8 + 5 = 13$ ‍

Now we can use the Pythagorean theorem to determine whether

△ A O C

forms a right triangle.

\begin{aligned} {A C}^{2} + {O C}^{2} & \overset{?}{=} A O^{2} \\ 11^{2} + 5^{2} & \overset{?}{=} 13^{2} \\ 146 & \neq 169 \end{aligned}

So,

△ A O C

is not a right triangle.

No, $\overset{\leftrightarrow}{A C}$ ‍ is not tangent to circle $O$ ‍, because $\overset{―}{A C}$ ‍ is not perpendicular to $\overset{―}{O C}$ ‍.

Problem 2

Segment

\overset{―}{O C}

is a radius of circle

O

Note: Figure not necessarily drawn to scale.

Is line $\overset{\leftrightarrow}{A C}$ ‍ tangent to circle $O$ ‍?

(Choice A)
Yes, because $\overset{―}{A C}$ ‍ intersects circle $O$ ‍ at point $C$ ‍
(Choice B)
Yes, because $\overset{―}{A C}$ ‍ is perpendicular to $\overset{―}{O C}$ ‍
(Choice C)
No, because $\overset{―}{A C}$ ‍ is not perpendicular to $\overset{―}{O C}$ ‍
(Choice D)
No, because $△ A B C$ ‍ intersects circle $O$ ‍ at more than one point

line

that is tangent to a circle at a particular point is perpendicular to the

radius

at that point.

Let's check to see whether

\overset{\leftrightarrow}{A C}

is tangent by using the converse pythagorean theorem to determine if

△ A O C

is a right triangle.

We will need to know the length of

\overset{―}{A O}

B O = 12

because it is a radius just like

C O

, so:

$A O = 8 + 12 = 20$ ‍

Now we can use the Pythagorean theorem to determine whether

△ A O C

forms a right triangle.

\begin{aligned} {A C}^{2} + {O C}^{2} & \overset{?}{=} A O^{2} \\ 16^{2} + 12^{2} & \overset{?}{=} 20^{2} \\ 400 & = 400 \end{aligned}

So,

△ A O C

is a right triangle.

Yes, $\overset{\leftrightarrow}{A C}$ ‍ is tangent to circle $O$ ‍, because $\overset{―}{A C}$ ‍ is perpendicular to $\overset{―}{O C}$ ‍.

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josha westy
Posted 8 years ago. Direct link to josha westy's post “how is angle AOC not a ri...”
how is angle AOC not a right angled triangle in problem 1
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(3 votes)
Answer
- Kali Bach
  Posted 8 years ago. Direct link to Kali Bach's post “The the first example is ...”
  The the first example is not a right triangle because it does not follow the Pythagorean Theorem of a^2 + b^2 = c^2. AC^2+OC^2 doesn't equal AO^2. (11^2 + 5^2 = 13^2, which turns out to be 146 = 169, not true). Both 45-45-90 and 30-60-90 triangles follow this rule.
  Comment on Kali Bach's post “The the first example is ...”
  (16 votes)
faithevanson09
Posted 7 years ago. Direct link to faithevanson09's post “The first question is vag...”
The first question is vague and doesn't explain how they found the length of AO. Where did y'all even get 8?
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(3 votes)
Answer
- Judith Gibson
  Posted 7 years ago. Direct link to Judith Gibson's post “8 was given as the length...”
  8 was given as the length of AB. BO is a radius of the circle and therefore has length of 5.
  So, AO (which equals AB + BO) is 8 + 5 which is 13.
  They also had to test whether there was a right angle at point C, so they used the Pythagorean theorem to see whether that was true. It wasn't, so in other words, AC wasn't perpendicular to OC, so therefore line AC was NOT tangent to circle O.
  Button navigates to signup page
  (11 votes)
Kagaku Doragon
Posted 4 years ago. Direct link to Kagaku Doragon's post “Can someone explain why f...”
Can someone explain why for problem two line BO is included in solving the problem while in problem 1 BO is left out? I understand that for problem 1 using the pythagorean theorem shows its not perpendicular but using that same method for problem 2 doesn't work and thus adding line BO is needed.
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Kevin K.
  Posted 3 years ago. Direct link to Kevin K.'s post “You can find the length o...”
  You can find the length of BO in either question, using just the radius. This is what you use to find out if it is a right triangle and thus, you need BO. Hope this answers your question...
  Button navigates to signup page
  (6 votes)
Mary
Posted 3 years ago. Direct link to Mary's post “what is the converse Pyth...”
what is the converse Pythagorean theorem?
Button navigates to signup pageComment on Mary's post “what is the converse Pyth...”
(2 votes)
Answer
- Seed Something
  Posted 2 years ago. Direct link to Seed Something's post “Normally we use the Pytha...”
  Normally we use the Pythagorean Theorem on a Right Triangle to find the length of a missing side measurement.
  
  •The Converse Pythagorean Theorem is when we already have the side lengths of a triangle, but we don't know if it's a Right Triangle…
  
  so it's when we use the formula to check if a triangle has a 90° angle through Comparison.
  
  We must…
  
  •Plug two of the triangle's side measurements into the corresponding variables of the Pythagorean Theorem.
  
  •Calculate to see what the third side would be if it were a Right Triangle.
  
  •Compare the triangle's given measurement to the calculated.
  
  •If the corresponding values are equal then it's a Right Triangle and has a 90° angle.
  If not, it is not.
  
  So…
  •To use the Pythagorean Theorem as its Converse, is to apply the formula to prove or disprove if a triangle has a 90° angle, and therefore is, or is not a Right Triangle.
  
  (ㆁωㆁ) Hope this helps someone!
  Button navigates to signup page
  (8 votes)
joannazhu123
Posted 7 years ago. Direct link to joannazhu123's post “Can someone explan #2 to ...”
Can someone explan #2 to me?
I dont understand because 2,3,4 is not a right triangle??
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Avia
  Posted 7 years ago. Direct link to Avia's post “The sides of the triangle...”
  The sides of the triangle in problem 2 are 12, 16, and 20 (12+8), which does make it a right triangle, since 20² = 12²+16².
  Button navigates to signup page
  (10 votes)
zoya zeeshan
Posted 8 years ago. Direct link to zoya zeeshan's post “how can we draw 2 common ...”
how can we draw 2 common transverse tangents for 2 congruent circles if they have any distance between their centres?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Colin Satchie
  Posted 8 years ago. Direct link to Colin Satchie's post “you dont that is somethin...”
  you dont that is something different you are using Pythagorean theorem here.
  Button navigates to signup page
  (6 votes)
Marvyn Greco
Posted 8 months ago. Direct link to Marvyn Greco's post “why is the perpindiculari...”
why is the perpindicularity contradicting it is confusing because they look the same except for line length in both problems.
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Mark
  Posted 3 months ago. Direct link to Mark's post “Since we know all of the ...”
  Since we know all of the lengths in this triangle, we can check if Pythagorean theorem will agree with our assumption that these are right triangles.
  
  Pythagorean theorem c² = a² + b²
  
  Problem 1:
  13² = 5² + 11²
  169 = 25 + 121
  169 ≠ 146 (These would be equal if we had 90° angle)
  
  Problem 2:
  20² = 12² + 16²
  400 = 144 + 256
  400 = 400 (This is a right triangle)
  Button navigates to signup page
  (3 votes)
Omar Sidani
Posted 7 years ago. Direct link to Omar Sidani's post “how many types of tangent”
how many types of tangent
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(5 votes)
Answer
4070235
Posted a year ago. Direct link to 4070235's post “How do we know that the r...”
How do we know that the rest of the line is 12?
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(1 vote)
Answer
- Nord
  Posted a year ago. Direct link to Nord's post “cause it touches the edge...”
  cause it touches the edge of the circle. line between the center of the circle that touches the edge equals the radius
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  (5 votes)
EMILIAR
Posted 2 years ago. Direct link to EMILIAR's post “what if one has the diame...”
what if one has the diameter would it still work?
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(2 votes)
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- AgentX
  Posted 2 years ago. Direct link to AgentX's post “Yes because you would div...”
  Yes because you would divide the diameter by 2 to get the radius
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  (2 votes)