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

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Math Expert
Joined: 02 Sep 2009
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Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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01 Jun 2015, 07:44
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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 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. $$\sqrt{89}$$

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

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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 = $$2r = 2 \sqrt{16.25}$$ = $$\sqrt{4*16.25} = \sqrt{65}$$ .

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2015-08-01_10-20-37.jpg [ 17.98 KiB | Viewed 21902 times ]

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

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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

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

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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
Math Expert
Joined: 02 Aug 2009
Posts: 7764
Re: Two perpendicular chords intersect in a circle. The segments of one ch  [#permalink]

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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
Attachments

New Microsoft Office Word Document (6).docx [10.93 KiB]

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

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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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08 Apr 2019, 12:49
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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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