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In the figure given below, O is the center and AB is the diameter of

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Bunuel
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In the figure given below, O is the center and AB is the diameter of [#permalink] New post   11 Feb 2020, 01:17
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In the figure given below, O is the center and AB is the diameter of the circle of radius 15. From point D, two tangents DC and DB are drawn to the circle. If AC is parallel to OD and AC+OD=43, find the length of CD given CD is positive integer

A. 16
B. 18
C. 20
D. 22
E. 24

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KarishmaB
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In the figure given below, O is the center and AB is the diameter of [#permalink] New post   11 Feb 2020, 04:42
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When I read this question, the first thing that struck me as odd was that they mentioned "CD is a positive integer" (which presumably means that the length of CD is a positive integer)
Length would obviously be positive but needn't be an integer. If it is specified and seeing that OCD is a right triangle (tangent makes 90 degree angle with radius), this might have something to do with pythagorean triplets.
The ones with 15 or factors of 15 (since OC = 15) are
3, 4, 5 (the sides can be 15, 20, 25)
5, 12, 13 (the sides can be 15, 36, 39)
8, 15, 17 (the sides can be 8, 15, 17)
(OC is one leg and CD is the other)

The only length given in the options is 20 (we have neither 36 nor 8)
Answer (C)
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Re: In the figure given below, O is the center and AB is the diameter of [#permalink] New post   11 Feb 2020, 15:11
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Since AC || OD, < CAO = < COD.
—> if you draw a segment CB, then triangle ACB will be right-angled triangle.
—> Also, triangle ACB and triangle OCD are similar triangles.
—> \(\frac{AB}{AC} = \frac{OD}{OC}\)

\(\frac{30}{AC} = \frac{OD}{15}\)

AC*OD = 450
AC+ OD = 43
—> AC= 18, OD = 25
(Using Pythagorean theorem)
\(CD^{2} + OC^{2} = OD^{2}\)

—> \(CD^{2} = 625 —225 = 400\)
CD = 20

The answer is C

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In the figure given below, O is the center and AB is the diameter of [#permalink] New post   Updated on: 12 Feb 2020, 08:37
we can find DE ; AC+OD=43
∆ACB ~ ∆ OCB
AB/AC= OD/OC
30/AC= OD/15
AC+OD=43
AC= 13, OD= 25
CD^2+OC^2=OD^2
CD^2 = 625-225 ; 400
CD= 20
IMO C


In the figure given below, O is the center and AB is the diameter of the circle of radius 15. From point D, two tangents DC and DB are drawn to the circle. If AC is parallel to OD and AC+OD=43, find the length of CD given CD is positive integer

A. 16
B. 18
C. 20
D. 22
E. 24

Originally posted by Archit3110 on 11 Feb 2020, 02:36.
Last edited by Archit3110 on 12 Feb 2020, 08:37, edited 1 time in total.
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Re: In the figure given below, O is the center and AB is the diameter of [#permalink] New post   11 Feb 2020, 02:54
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- Radius OA=OC=15=3*5
- CD is a positive integer and OCD is a right triangle at C, so:
OC^2+CD^2=OD^2
(3.5)^2+CD^2=OD^2

A quick and sensible value of positive integer CD to satisfy the above equation is CD=20=4*5, and thus, we easily satisfy
OC^2+CD^2=OD^2
(3*5)^2+(4*5)^2=(5*5)^2

Check AC=43-OD then AC=43-25=18.
AC^2+CB^2=AB^2
(3*6)^2+(4*6)^2=(5*6)^2 (OK!!)
We can roughly infer that angle B = angle D.

FINAL ANSWER IS (C)

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Re: In the figure given below, O is the center and AB is the diameter of [#permalink] New post   11 Feb 2020, 22:55
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From the given figure, radius, AO = OC = AC = OB = 15
AC+OD=43
OD = 43 – 15 = 28

So AOC is a equilateral triangle. ∠ACO = ∠AOC = ∠OAC = 60
As AC is parallel to OD , ∠CAO = ∠DOB = 60.
Therefore, ∠COD = 60.

As DC is tangent, ∠DOC = 90

Now, 90:60:30 = 2x : √2x :1x
2x = 28
or, x = 14
Opposite side of 60 = √2x = 1.4 *14 = 20(Appoximate)

Answer: 20(C)
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Re: In the figure given below, O is the center and AB is the diameter of [#permalink] New post   12 Feb 2020, 04:13
Archit3110 wrote:
and ∆ CDO similar ∆ DOB
so side CD= DB
find DB ; 28^2-15^2 = ~ 24
IMO E: 24 ; side CD




whats wrong in this, can someone explain this
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In the figure given below, O is the center and AB is the diameter of [#permalink] New post   Updated on: 23 Feb 2020, 17:17
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Originally posted by freedom128 on 12 Feb 2020, 05:53.
Last edited by freedom128 on 23 Feb 2020, 17:17, edited 1 time in total.
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Re: In the figure given below, O is the center and AB is the diameter of [#permalink] New post   13 Feb 2020, 02:57
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madgmat2019 wrote:
Archit3110 wrote:
and ∆ CDO similar ∆ DOB
so side CD= DB
find DB ; 28^2-15^2 = ~ 24
IMO E: 24 ; side CD




whats wrong in this, can someone explain this


It is assumed here that AC is a radius too and hence of length 15. That is not true.
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Re: In the figure given below, O is the center and AB is the diameter of [#permalink] New post   12 Jul 2023, 04:12
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