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# Challenge problems: circumscribing shapes

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

## Problem 1

All sides of $\mathrm{△}ABC$ are tangent to circle $P$.
What is the perimeter of triangle $ABC$?
units

## Problem 2

All sides of quadrilateral $ABCD$ are tangent to circle $P$.
What is the perimeter of quadrilateral $ABCD$?
units

## Want to join the conversation?

• I am confused. How did you solve when you still haven't found x?
(5 votes)
• 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.
(34 votes)
• what are the answer to number 1 and/or 2. Im very confused on this subject matter.
(4 votes)
• 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! :)
(16 votes)
• i did not understand the way to find perimiter of quadrilaterals and triangles
(6 votes)
• 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.
(4 votes)
• still don't get it
(5 votes)
• 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.
(3 votes)
• 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...
(4 votes)
• makes no since
(3 votes)
• 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
(4 votes)
• wut da flip
(2 votes)
• yes i agree that is a fantastic point
(3 votes)
• 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?
(3 votes)
• I noticed a pattern where the solution was 2*each side. Is that applicable to all problems like this or just some?
(2 votes)
• 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.
(2 votes)