i did not understand the way to find perimiter of quadrilaterals and triangles

The perimeter of a shape is just the sum of the lengths of all the sides. So for a triangle you add up all 3 sides and for a quadrilateral you add up all 4 sides.

I am confused. How did you solve when you still haven't found x?

In both cases, the x cancels itself out. Take question 2: the perimeter is a whole bunch of numbers which I am too lazy to type out, and then there is an unknown length which I will call x. To find the other unknown length, you can just take 12-x. The perimeter will be all those numbers + x + 12 - x. The solution to x+12-x is 12, as you can rearrange it as x-x+12.

what are the answer to number 1 and/or 2. Im very confused on this subject matter.

Hi! I think I might be able to help :) So basically for #1, the main theorem you want to apply to the problem is the theorem that "2 tangent lines drawn to a circle from the same point are congruent." Let's call the point between AC "x", the point between BC "y" and the point between AB "z". Because of the theorem, you can say that YB is congruent to ZB (YB = ZB = 16). You have one piece of the puzzle right there! Now you can move on to the other pairs of tangent lines. Based on the same theorem we can say that XC is congruent to YC, and that XA is congruent to ZA. Let's set XC equal to "x" (XC = YC = "x") Because AC is equal to 14, we can call XA "14-x." (XA = ZA = "14-x") Now, let's add it all up! 2x + 2(14-x) + 2(16) = 28 + 32 = 60. #2 is extremely similar. If you want more help, don't hesitate to ask me or anyone else on here! :)

Have you watched this video? - https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-tangents/a/challenge-problems-perimeter-of-circumscribing-shapes It proves a theorem which is used in these problems. The theorems states that if two tangent segments share one point outside of the circle and end at two different points on the circle they are congruent. Using that along with the answer they gave to the problem should hopefully help. I can explain more in depth if you are still confused! :)

yes i agree that is a fantastic point

Challenge problems: circumscribing shapes Problem 1 What logic tells us that we can use the same variable x on line AB versus line AC? They visually look to be very different lengths.

we do not have to consider AB=AC=x, just considering each segment as an individual pair of segments, for example, DC= x+y and BC=y+z and AB=m+z (by the way z=9.6 and m=3.7)...AD=m+x.......... add them all and then you have the answer...

I noticed a pattern where the solution was 2*each side. Is that applicable to all problems like this or just some?

Except that you may be given a set of numbers that does not match this pattern. The reason that it doubles is the fact that the two lines formed from the outside point to the two tangents have to be equal. So if you label the point between AB as X, between BC as Y, and AC as Z, this theorem says that AX = AZ, XB=BY, and YC = ZC. Thus, two segments are 16 (XB and BY), and we also know that AZ + CZ =14 which substituting means that AX + CY =14 (even though we do not know the measures of any of these 4). Thus, adding what is known and doubling it gives the answer.

Main content

Challenge problems: circumscribing shapes

Google Classroom

Solve two challenging problems that apply properties of tangents to find the perimeter of a circumscribing shape.

Problem 1

All sides of

△ A B C

are tangent to circle

P

What is the perimeter of triangle $A B C$ ‍?

units

Any two segments tangent to a circle from a common endpoint are congruent.

With this information, we can identify another segment which is also

16

units long.

Let's use the variable

x

to represent the distance from point

C

to the circle. By the same logic as above, we can label the rest of the lengths in the perimeter.

Now we can find the perimeter by adding up the lengths of the sides:

14 - x + x + x + 16 + 16 + 14 - x

= 60

The perimeter of the triangle is $60$ ‍ units long.

Problem 2

All sides of quadrilateral

A B C D

are tangent to circle

P

What is the perimeter of quadrilateral $A B C D$ ‍?

units

Any two segments tangent to a circle from a common endpoint are congruent.

With this information, we can identify two other segments which are

3.7

and

9.6

units long, respectively.

Let's use the variable

x

to represent the distance from point

C

to the circle. By the same logic as above, we can label the rest of the lengths in the perimeter.

Now we can find the perimeter by adding up the lengths of the sides:

12 - x + 12 - x + x + x + 9.6 + 9.6 + 3.7 + 3.7

= 50.6

The perimeter of the quadrilateral is $50.6$ ‍ units long.

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rvillard
Posted 8 years ago. Direct link to rvillard's post “I am confused. How did yo...”
I am confused. How did you solve when you still haven't found x?
Button navigates to signup pageComment on rvillard's post “I am confused. How did yo...”
(5 votes)
Answer
- Felicia L.
  Posted 8 years ago. Direct link to Felicia L.'s post “In both cases, the x canc...”
  In both cases, the x cancels itself out. Take question 2: the perimeter is a whole bunch of numbers which I am too lazy to type out, and then there is an unknown length which I will call x. To find the other unknown length, you can just take 12-x. The perimeter will be all those numbers + x + 12 - x. The solution to x+12-x is 12, as you can rearrange it as x-x+12.
  Button navigates to signup page
  (36 votes)
Harmon Byerly
Posted 7 years ago. Direct link to Harmon Byerly's post “what are the answer to nu...”
what are the answer to number 1 and/or 2. Im very confused on this subject matter.
Button navigates to signup pageComment on Harmon Byerly's post “what are the answer to nu...”
(4 votes)
Answer
- Arohee Bhoja
  Posted 7 years ago. Direct link to Arohee Bhoja's post “Hi! I think I might be ab...”
  Hi! I think I might be able to help :)
  
  So basically for #1, the main theorem you want to apply to the problem is the theorem that "2 tangent lines drawn to a circle from the same point are congruent." Let's call the point between AC "x", the point between BC "y" and the point between AB "z".
  Because of the theorem, you can say that YB is congruent to ZB (YB = ZB = 16). You have one piece of the puzzle right there!
  
  Now you can move on to the other pairs of tangent lines. Based on the same theorem we can say that XC is congruent to YC, and that XA is congruent to ZA. Let's set XC equal to "x" (XC = YC = "x") Because AC is equal to 14, we can call XA "14-x." (XA = ZA = "14-x")
  
  Now, let's add it all up! 2x + 2(14-x) + 2(16) = 28 + 32 = 60.
  
  #2 is extremely similar. If you want more help, don't hesitate to ask me or anyone else on here! :)
  Comment on Arohee Bhoja's post “Hi! I think I might be ab...”
  (17 votes)
Abhi Reddy Mulukuri
Posted 7 years ago. Direct link to Abhi Reddy Mulukuri's post “i did not understand the ...”
i did not understand the way to find perimiter of quadrilaterals and triangles
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(6 votes)
Answer
- Amelia Brose
  Posted 7 years ago. Direct link to Amelia Brose's post “The perimeter of a shape ...”
  The perimeter of a shape is just the sum of the lengths of all the sides. So for a triangle you add up all 3 sides and for a quadrilateral you add up all 4 sides.
  Button navigates to signup page
  (4 votes)
eperez3463
Posted a year ago. Direct link to eperez3463's post “still don't get it”
still don't get it
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JasmineC
Posted a year ago. Direct link to JasmineC's post “makes no since”
makes no since
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- strawberry killer
  Posted a year ago. Direct link to strawberry killer's post “Have you watched this vid...”
  Have you watched this video? - https://www.khanacademy.org/math/geometry/hs-geo-circles/hs-geo-tangents/a/challenge-problems-perimeter-of-circumscribing-shapes
  It proves a theorem which is used in these problems.
  The theorems states that if two tangent segments share one point outside of the circle and end at two different points on the circle they are congruent.
  Using that along with the answer they gave to the problem should hopefully help.
  
  I can explain more in depth if you are still confused! :)
  Comment on strawberry killer's post “Have you watched this vid...”
  (6 votes)
Zander Loren
Posted a year ago. Direct link to Zander Loren's post “wut da flip”
wut da flip
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(4 votes)
Answer
- #1
  Posted a year ago. Direct link to #1's post “yes i agree that is a fan...”
  yes i agree that is a fantastic point
  Button navigates to signup page
  (5 votes)
b fish
Posted 7 years ago. Direct link to b fish's post “Challenge problems: circu...”
Challenge problems: circumscribing shapes Problem 1
What logic tells us that we can use the same variable x on line AB versus line AC? They visually look to be very different lengths.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- atharva2001b
  Posted 7 years ago. Direct link to atharva2001b's post “we do not have to conside...”
  we do not have to consider AB=AC=x, just considering each segment as an individual pair of segments, for example, DC= x+y and BC=y+z and AB=m+z (by the way z=9.6 and m=3.7)...AD=m+x..........
  add them all and then you have the answer...
  Button navigates to signup page
  (4 votes)
shaikh sharfuddin
Posted 7 years ago. Direct link to shaikh sharfuddin's post “Interior and exterior bis...”
Interior and exterior bisector of angle A of triangle ABC met BC and BC produce point P&Q result.if point O is mid point of PQ .show that OA is tangent to the circle circumscribed about triangle ABC
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(4 votes)
Answer
💎Chυcκ Lørrε💎
Posted 6 years ago. Direct link to 💎Chυcκ Lørrε💎's post “Hey is this sentence true...”
Hey is this sentence true:"Any two segments tangent to a circle from a common endpoint are congruent." Is that because of they both have a 90° angle?
And do you know why I can't sending messages nor asking questions but I can only edit something?
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(3 votes)
Answer
Will Crosswhite
Posted 4 years ago. Direct link to Will Crosswhite's post “I noticed a pattern where...”
I noticed a pattern where the solution was 2*each side. Is that applicable to all problems like this or just some?
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(2 votes)
Answer
- David Severin
  Posted 4 years ago. Direct link to David Severin's post “Except that you may be gi...”
  Except that you may be given a set of numbers that does not match this pattern. The reason that it doubles is the fact that the two lines formed from the outside point to the two tangents have to be equal. So if you label the point between AB as X, between BC as Y, and AC as Z, this theorem says that AX = AZ, XB=BY, and YC = ZC. Thus, two segments are 16 (XB and BY), and we also know that AZ + CZ =14 which substituting means that AX + CY =14 (even though we do not know the measures of any of these 4). Thus, adding what is known and doubling it gives the answer.
  Button navigates to signup page
  (3 votes)

Geometry (Eureka Math/EngageNY)

Course: Geometry (Eureka Math/EngageNY) > Unit 5

Problem 1

Problem 2

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