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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Find the length of the common chord of two intersecting circles of rad

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Intern  B
Joined: 30 Dec 2019
Posts: 6
Find the length of the common chord of two intersecting circles of rad  [#permalink]

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1 00:00

Difficulty:   55% (hard)

Question Stats: 52% (03:02) correct 48% (02:13) wrong based on 23 sessions

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Q. Find the length of the common chord of two intersecting circles of radii 15 cm and 20 cm, given that the distance between the centers of the circles is 25cm.

A. 9
B. 18
C. 12
D. 24
E. 15
VP  V
Joined: 19 Oct 2018
Posts: 1308
Location: India
Find the length of the common chord of two intersecting circles of rad  [#permalink]

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1
1
AC=15cm; BC=20cm; and AB=25cm

We can notice that (15,20,25) is a pythagorean triplet; hence, ACB is a right angle triangle.

Also, Line from the center perpendicularly bisects the chord.

Area of triangle ABC= $$\frac{1}{2}*AC*BC= \frac{1}{2}*CO*AB$$

CO= $$\frac{AC*BC}{AB}= \frac{15*20}{25}=12$$

CD= 2*12=24

PiyushPanda1 wrote:
Q. Find the length of the common chord of two intersecting circles of radii 15 cm and 20 cm, given that the distance between the centers of the circles is 25cm.

A. 9
B. 18
C. 12
D. 24
E. 15

Attachments Untitled.png [ 4.21 KiB | Viewed 308 times ]

Director  D
Joined: 22 Feb 2018
Posts: 530
Re: Find the length of the common chord of two intersecting circles of rad  [#permalink]

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This is bit tricky problem but fortunately got correct as I saw the Pythagoras triplet & chose the correct answer, I.e. 24.

@experts or moderators,
Still, I need to find any other alternative visual approach. Is it possible?

Posted from my mobile device
Intern  B
Joined: 30 Dec 2019
Posts: 6
Find the length of the common chord of two intersecting circles of rad  [#permalink]

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Raxit85,

This can also be solved by using variables as follows: Refer to the figure/attachment by nick1816 for below explanation.

Let AO = x. So, BO = 25 - x.
As we know, when two circles intersect, the common chord is perpendicular to the line joining the centers of the circles.
Hence, CD is perpendicular to AB.

So, $$CO^2$$ = $$AC^2$$ - $$AO^2$$ = $$BC^2$$ - $$BO^2$$.
We get, $$15^2$$ - $$x^2$$ = $$20^2$$ - $$(25-x)^2$$, which gives x = 9.

Now, $$OC^2$$ = $$15^2$$ - $$9^2$$ = 225 - 81 = 144
So, OC = 12.

As we know OC = OD, OD = 12.
CD = OC + OD = 12 + 12 = 24. Find the length of the common chord of two intersecting circles of rad   [#permalink] 02 Jan 2020, 10:38
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# Find the length of the common chord of two intersecting circles of rad  