GMAT CLUB OLYMPICS: AB, BC and CD are tangent to the yellow semicircle

Question

https://gmatclub.com/forum/download/file.php?id=62911 AB, BC and CD are tangent to the yellow semicircle shown above. If O is the center of the circle (part of which the yellow semicircle is), the length of AB is 9 units, the length of CD is 16 units, and AO = OD, then what is the length of AD? A. 15 B. 18 C. 21 D. 24 E. 28 M37-27 Untitled2.png

hnguye10 · Accepted Answer

Let AB, BC and CD intersects the semicircle at I, H and J, respectively

Since AB, BC and CD are tangent to the semicircle, we have:
OI, OH and OJ are the three radii of the semicircle
AB and OI are perpendicular 
BC and HO are perpendicular 
CD and OJ are perpendicular 
BH = BI
CH = CJ

Since AO = OD, OI = OJ and ∠AIO = ∠DJO = 90°, we have △AIO = △DJO => ∠AOI = ∠DOJ
Since BH = BI, OH = HO and ∠BIO = ∠BHO = 90°, we have △BIO = △BHO => ∠BOI = ∠BHO
Similarly we would have ∠JOC = ∠HOC

We have ∠AOI + ∠IOH + ∠HOJ + ∠DOJ = 180°
=> 2*(∠AOI + ∠BOI + ∠JOC) = 180°
=> ∠AOI + ∠BOI + ∠JOC = 90°
=> ∠AOB = 90° - ∠JOC = ∠OCJ

We have ∠AOB = ∠OCJ and ∠IAO = ∠JDO => △AOB and △DCO are similar
Therefore AB/AO = DO/DC or AO*DO = AB*DC
=> AO^2 = 9*16 = 144 => AO =12
=> AD = 12*2 = 24

The answer is D

rvgmat12 · Answer

Triangle AOB is similar to triangle DCO 
9/AO = OD/16
AO = OD 
9*16 = AO^2 
AO = 3*4 = 12
AD = AO + OD = 2*AO = 2*12 = 24

Bunuel · Answer

Official Solution:

https://gmatclub.com/tests-beta/resources/import/img/M37-26.png 
 
AB, BC and CD are tangent to the yellow semicircle shown above. If O is the center of the circle (part of which the yellow semicircle is), the length of AB is 9 units, the length of CD is 16 units, and AO = OD, then what is the length of AD?

A.  15 
B.  18 
C.  21 
D.  24 
E.  28

The question has five exact answer choices (no "Cannot be determined" or "None of the above") so one of them must be correct. If one of the answers is correct, the answer must be correct no matter how we draw the diagram (of course we should not violate info that is given). 
 
Re-draw the diagram so that points A and D coincide with the endpoints of the diameter. In this case, AB and DC will be perpendicular to the diameter. Check the new image below:
 
  https://gmatclub.com/tests-beta/resources/import/img/M37-26-3.png 
 
Draw perpendicular BF from point B to DC. CF will be  DC - DF = 16-9=7 
 
Next, when two tangent segments are drawn to one circle from the same external point, then they are equal, so  BA = BG = 9  and  CD = CG = 16 . Therefore,  BC = 9+16=25 .
 
In triangle, BCF,  BC^2 = BF^2 + CF^2 
 
 25^2 = BF^2 + 7^2 
 
 BF=24 . 
 
 BF=AD=24 .

Answer: D

joohwangie · Answer

we cut them into them to 3 triangles.
triangle ABO
triangle OBC
triangle DOC
all are isoceles triangle
triangle OBC and DOC have two legs that are 16 each 
triangle ABO have 9 base
lineAO=lineOD=lineBO =12
therefore line AD=24

SohGMAT2020 · Answer

AB, BC and CD are tangent to the yellow semicircle shown above. If O is the center of the circle (part of which the yellow semicircle is), the length of AB is 9 units, the length of CD is 16 units, and AO = OD, then what is the length of AD?

A. 15
B. 18
C. 21
D. 24
E. 28

Using the picture, AD must be greater the 16 but less than 16+ 9 = 25.

Hence eliminate A and E.

Also 18 and 21 are highly improbable since the are very close to the upper and lower limit. Hence going with option D. ( Guesstimating this one)

chaos91 · Answer

Please see the attached picture for solution

Reviper · Answer

D) 24
 Best case scenario: AB parallel to CD (Doesn't violate the info given in the question)

BC = 9+ 16 = 25 (Lengths of 2 tangents are equal - known property)
Using Pythogoras' theorem (AD)^2 + (16-7)^2 = 25^2

=> AD = 24

rajatchopra1994 · Answer

Explanation:

Explained well in attached image. Pls check

Given: AB=9 , DC=16
OE=OG=OF=r
Let AE=x , EB=9-x
Triangle AOE and Triangle DOF are right angles Triangles, in which 2 side and one angle are equal. Therefore Congruent Triangle.
AE=DF=x , FC=16-x
Similarly Triangle GOC and Triangle COF ; Triangle BEO and Triangle BOG are Congruent Triangles
BE=BG=9-x
FC=CG=16-x
Also, by Congruent Triangles AEO and DFO, Angle EAO = Angle ODF & Angle AOD = Angle DCO

therefore, AB/AO = OD/CD
9/r = r/16
r^2 = 144
r=12

AD = 2r = 24

IMO-D

Archit3110 · Answer

for the semicircle and the lines tangents along the semicircle the length of side AD
AD^2 = 4* AB*CD
AD = 2√AB*CD
AD = 2 √9*16
AD= 2*3*4
AD= 24
option D

bobnil · Answer

Please follow the attached image file...

ashwini2k6jha · Answer

ANSWER:D
Please refer to diagram attached
Draw a line from o to line AB at point E, draw a line from Centre O to CD meeting at F, and also draw a line from center O to BC meeting at point G.
From SSA for congruent,Triangle AEO IS CONGRUENT TO TRIANGLE DFO
Hence AE=DF=x let.
From property of tangent at circle we know EB=BG=9-x
Simillarly GC=CF=16-x
IN diagram 3 pairs of triangle are congruent. Accordingly let angles are theta,alfa, and beta.
Since sum of all angles are 180 due to straight line
2θ + 2α + 2β = 180°
θ + α + β = 90°
Therefore angle OAE=α + β 
IN triangle OAE, tan ( α + β )=r/x
tan α = (9 – x)/r
tan β = (16 – x)/r
From formula
tan (α + β) = (tan α + tan β)/(1 – (tan α)(tan β))
substituting value :r/x = r(25 – 2x)/ 
r2 + x2 = 144 = 12^2
AO^2 = OD^2 = r2 + x2 = 144 = 12^2
AD=AO+OD=24

Shahalikhan · Answer

Triangles AOB and DCO are similar

thus we have

AB/AO = OD/CD
9/r = r/16
9(16) = AO(OD)
144 = AO(AO)
144 = AO2
12 = AO

and

AD=2*12=24

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GMAT CLUB OLYMPICS: AB, BC and CD are tangent to the yellow semicircle

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GMAT CLUB OLYMPICS: AB, BC and CD are tangent to the yellow semicircle

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