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Two perpendicular chords intersect in a circle. The segments of one ch

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Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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New post 01 Jun 2015, 07:44
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

52% (02:29) correct 48% (02:43) wrong based on 98 sessions

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Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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New post 01 Aug 2015, 07:35
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Swaroopdev wrote:
Bunuel, not able to understand above solutions, could you please explain in your words ?


Look at the attached image for a explanation of the variables and dimensions.

Given,

GF = 3, GE = 4, BG = 6 and CG = 2

O is the center of the circle. Chords BC and EF are perpendicular to each other.

Now, in triangle OBC, OB = OC = radius of the circle, thus triangle OBC is an isosceles triangle and OA \(\perp\) BC (OA is perpendicular to BC). Thus OA will bisect BC into 2 equal parts (a property of an isosceles triangle that the perpendicular drawn from one of the vertex to the base (assuming base is NOT one of the 2 equal sides), bisects the base).

---> AB = AC = (6+2)/2 = 4 ....(1)

Similarly in triangle OFE, OF = OE = radius of the circle. OD \(\perp\) EF ---> FD = DE = (3+4)/2 = 3.5 ---> FD = DG + FG ---> DG = FD - FG = 3.5 - 3 = 0.5 ...(2)

Also, OAGD form a rectangle such that GD = OA = 0.5 and AG = OD = 2.

Thus, in right triangle OAB,\(OA^2+AB^2 = OB^2\) --->\(r^2 = 0.5^2+4^2 = 16.25\) ---> \(r = \sqrt{16.25}\) ---> diameter = \(2r = 2 \sqrt{16.25}\) = \(\sqrt{4*16.25} = \sqrt{65}\) .

C is the correct answer.
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Re: Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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New post 01 Jun 2015, 08:14
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Bunuel wrote:
Two perpendicular chords intersect in a circle. The segments of one chord are 3 and 4; the segments of the other are 6 and 2. Then the diameter of the circle is:

A. \(\sqrt{56}\)
B. \(\sqrt{61}\)
C. \(\sqrt{65}\)
D. \(\sqrt{75}\)
E. \(\sqrt{89}\)


option C

Half Length of chord 1 = (3+4)/2 = 3.5
Similarly half length of chord 2 = 4
Ow we can form two right angles with radii as hypotenuse:
1) with legs 3.5 and 4-2 = 2. Radius = sqrt(3.5^2 + 2^2) = sqrt(16.25)
2) with legs 4 and 3.5-3 = .5 radius = sqrt(4^2 + .5^2) = sqrt(16.25)

In each case the diameter obtained is 2* sqrt(16.25) = sqrt(65)
Option C

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Re: Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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New post 02 Jun 2015, 02:29
Rule: Perpendicular from centre to the chord bisects the chord.

Given that, we can find out that the point of intersection of the chords is 0.5[4-(4+3)/2] units and 2[6-(6+2)/2] units away from the centre.

Once the centre distance from point of intersection is received we can use distance formula to calculate the radius:
(4*4+0.5*0.5)^2=sqrt(16.25)

Diameter = 2 * sqrt(16.25) = sqrt(65). Option C
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Re: Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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New post 06 Jun 2015, 08:28
Bunuel wrote:
Two perpendicular chords intersect in a circle. The segments of one chord are 3 and 4; the segments of the other are 6 and 2. Then the diameter of the circle is:

A. \(\sqrt{56}\)
B. \(\sqrt{61}\)
C. \(\sqrt{65}\)
D. \(\sqrt{75}\)
E. \(\sqrt{89}\)


pl refer sketch..
radius=\(\sqrt{2^2+3.5^2}\)=\(\sqrt{16.25}\)
dia=2*\(\sqrt{16.25}\)=\(\sqrt{4}\)*\(\sqrt{16.25}\)=\(\sqrt{65}\)
ans C
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Re: Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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New post 09 Jun 2015, 10:09
Hi,
VeritasPrepKarishma, can you please explain this with diagram.
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Re: Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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Re: Two perpendicular chords intersect in a circle. The segments of one ch   [#permalink] 08 Apr 2019, 12:49
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