For the first question :
the angle of A^BC is 60, so the angle of A^OC is 120 ( inscribed angles to a chord in a circle have half the value the angle from the center O to that same chord)
In triangle ACO: OA = OC = radius, then I^AO = I^CO = (180 - A^OC)/2 = 30 ( I is midpoint of AC, and OI is perpendicular to AC)

Triangle AIO is both half an equilateral triangle and a right triangle at I: then OI = 1/2 OA = 1/2*4 = 2
AI = sqrt ( OA^2 - OI^2) = sqrt (4^2 - 2^2) = sqrt(12) = 2*sqrt(3)
AC = 2 AI ( as I is the midpoint of AC) = 4*sqrt(3), the perimeter of ABC = AC*3 = 12*sqrt(3)
Correct answer is D.

For the second question:


Chord AB = radius, it means AOB is an equilateral triangle, then A^OB = 60, which suggest that A^CB = 30
and A^BC = 180 - 30 -50 = 100
so x can either be A^BC or A^CB, but from the list we can only find 30, so the answer choice must be B.

hi i dont have much background on chords or how you got the angle measures... could you please give a detailed explanation with the reasoning for why you got the angle measures?

thanks a million!

awd5045 wrote:hi i dont have much background on chords or how you got the angle measures... could you please give a detailed explanation with the reasoning for why you got the angle measures?

thanks a million!

Property of an equilateral triangle

1)Every altitude is also a median and a bisector. (in figure AD & BO)

This means BO bisects the angle ABC.
hence angle ABO =angle OBC =30.

Now look at the triangle OBD

we have angle OBD = 30 and angle BDO=90 ( AD is an altitude).
So, the third angle DOB= 60.

This gives us a 30-60-90 right angle triangle.

Ratio of sides of a 30-60-90 triangle
OD:DB:BO= x: sqrt(3)x: 2x


As OB is radius of the circle
BO =2x = r
=>x=r/2
=>DB = sqrt(3)*r/2 (as shown in figure)
DB=DC
BC= DC+DB = sqrt(3)*r
Hence the perimeter of the triangle ABC = 3*sqrt(3)*r

Since r=4

perimeter of ABC = 12*sqrt(3)
Hence option D is the correct answer.

Q2)
Property of a chord (Inscribed angle theorem)
If a central angle and an inscribed angle of a circle are subtended by the same chord and on the same side of the chord, then the central angle is twice the inscribed angle.


So according to the property angle ACB = 1/2 of angle AOB
Since triangle AOB is equilateral triangle
angle AOB = 60
and hence angle ACB = 30

Alternate method

Angle OAB = angle OAC+angle CAB
60 = OAC + 50
angle OAC = 10

Now Consider the triangle AOC
OA=OC (radius)
Hence the AOC is an isosceles triangle
So angle OAC = OCA = 10

Similarly in triangle OBC
angle OBC = angle OCB = x+10

Consider the triangle ABC
sum of interior angles of the triangle = 180
angle CAB + (angle OBA+angle OBC) + angle ACB = 180
50+60+(x+10)+x = 180
2x=60
x=30

lizonnat wrote:There is a picture of a triangle inside a circle:
If the circle above has a radius of 4, what is the perimeter of the inscribed equilateral triangle?

(A) 6 sqrt(2)
(B) 6 sqrt(3)
(C) 12 sqrt(2)
(D) 12 sqrt(3)
(E) 24

The figure above shows 3 congruent triangles, each with two sides of 4.
In each of these triangles, the third side -- in other words, the sides of the equilateral triangle -- must be less than the sum of the other two sides (4+4=8).
Thus:
Each side of the equilateral triangle ≈ 6.
Perimeter ≈ 18.

Eliminate A and B, which are way too small, and E, which is too big.

The correct answer must be C or D.
√2 implies a 45-45-90 triangle; √3 implies a 30-60-90 triangle.
Since each angle of the equilateral triangle = 60, √2 makes no sense here.
Eliminate C.

The correct answer is D.

Always look for opportunities to ballpark.
Always look at the answer choices.
The approach above requires little insight and allows us to determine the correct answer in mere seconds.