Points A and B are on a circle of radius 5 and AB = 6. Point C is the

Question

Points A and B are on a circle of radius 5 and AB = 6. Point C is the midpoint of the minor arc AB. What is the length of the line segment AC?

(A)  \sqrt{10}

(B) 7/2

(C)  \sqrt{14}

(D)  \sqrt{15}

(E) 4

itsanujsingh · Answer

Let O be centre of the circle

As AB is the arc and it's midpoint is given as C. If we draw a perpendicular from to AB it will bisect the chord AB at D(assumed point) and meet C such that OC=5=OD+DC
NOW, TRIANGLE OAD is rt angled triangle 
And we can find OD using Pythagoras OD=rt(5^2-3^2)=4

Similarly triangle ADC is also a rt angled triangle and we can find AC using Pythagoras, AC=rt(3^2+1^2)=rt10

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Shreshtha55 · Answer

chetan2u  can you please explain this?

chetan2u · Answer

Let join AB, and CO. Now, CO will bisect AB in two halves of 6/2 or 3..
Now since C is in the middle, the other thing will be that the chord AB and radius OC will be perpendicular to each other.

So, take triangle ADO with hypotenuse as 5 and one side as 3, so sides are 3:4:5, so OD = 4.

Now let us take triangle ADC, again a right angled triangle. CD=5-4=1, and AD = 3, so AC =  \sqrt{1^2+3^2}=\sqrt{10} .

stne · Answer

Hi  chetan2u ,
Thank you for your solution,but some portions are not clear, can you check , in the last line did you mean triangle ADC rather than ADB also did you mean CD rather than BD,

Thanks.

v12345 · Answer

Let AC = x
OA = OB = OC = 5 (Radius)
AD = DB = 3
OD  = \sqrt{5^2 - 3^2} = 4 
DC = 5 - OD = 1

In triangle ADC, 
AC  =\sqrt{DC^2 + AD^2} = \sqrt{10} 
Hence, OA is (A).

stne

Points A and B are on a circle of radius 5 and AB = 6. Point C is the

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Points A and B are on a circle of radius 5 and AB = 6. Point C is the

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