# The radii of two concentric circles are 13 cm and 8 cm . AB is a diameter of the bigger circle BD is tangent to the smaller circle touching it at D.Find the length of AD  We draw a line parralel to BD from O on AD at M such that OM || BD

By thales' theorem or basic proportionality theroem , or similarity

AO/BO = AM/DM
But, AO = BO
=> AM/DM = 1 or AM = DM

Also, LMOD = LODM = 900 (alternate angles)

Now, BD2 = OB2 - OD2 = 169 - 64 =105
or, BD = √105 cm
Again, since OM || BD
so, by similarity

OM/BD = AO/AB
=> OM = BD X AO / AB = √105 x 13 / 26 = √105 / 2
or, OM = √105 / 2
So in triangle ODM
DM = √OM2 + OD2 = √105/4 + 64 = √105 + 256 / 2 = √361 / 2
or, DM = √361 / 2
Hence, AD = 2 DM = √361

hope this helps you frnd........

• 69

the lenght of ad is 19 cm

in triangle BOD USE THE TANGENT THEORM AND PYTHAGORAS THEOREM TO GET BD = UNDERROOT 105 THEN BE = 2* UNDERROOT 105

THEN IN TRIANGLE ABD USE PYTHAGORAS THEOREM AND GET THE VALUE OF AE THEN TO GET THE VALUE OF AD USE PYTHAGORAS THEOREM IN TRIANGLE ADE AND THE VALUE OF DE IS  UNDERROOT 105 THEN THE LENGHT OF AD IS 19

HOPE THIS HELP IF IT HELPS A THUMBS UP !!!

• 31

@ NAREN NARENDA √361 IS 19 ....

• 16

dude,...

i 4got to rite v361 is 19....

bt my entire ans is correct , .

and i think u hav given thumbs down ,

• -2

yep crct bro!!!

these guys expect ans and then if a small mistake is found..kaldi se thumbs down de dete hain!!!

they think we are all perfect!!!

• -1

@naren ur method is absolutely rite bt u hav done it very long and 2 ur knowledge i tell u there is no need for all such constructions u did it can be done directly and for 2 marks v hav to get the ans in  few steps......... :p

• 19
thank u prachi patel ur ans is easy
• 4
Thank you soo much
• -2
1+1=2
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in triangle  ABE, angle AEB =900  since angle in the semicle.
in triangle BOD  angle BOD =900  radius divides the drawn to the point of contact of tangent
there fore ,OB //  AE  this implies  that  angle BOD=angle AEB=900   and triagles BDO and  BEA are similar
AE=2OD by midpoint theorem  but OD =8   given   there fore AE=2x8=16

in right angle triangle BOD
BD2=OB​​2 +OD2

BD = square root of 105
in triangle ade DE2 +AE2=AD2
AD2=162+105
AD =square root of 361

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thank u
• -4

Given OD=8cm
OB=13cm

Firstly find DB.
Using Pythagoras Theorem, DB^2=OB^2 - OD^2. This gives DB = sqrt(105) = DE.

Without going into the complexity of angles, as in some of the existing answers, I would suggest using the property of Similar triangles. Clearly, AEB and ODB are similar righttriangles. And as, DB~EB, therefore, AE~OD.

DB = sqrt(105), and EB = 2(sqrt(105))
and
as OD = 8cm, therefore AE = twice(8cm) = 16cm.

Finally, using Pythagoras Theorem in right triangle AED,
AD^2 = AE^2 + ED^2= 256 + 105= 361
Therefore, AD = sqrt(361) = 19cm.

• 39
Prachi , your answer is wrong.... In triangle ABD, using pythagoras theorem is not possible as angle ADB is not 90 degree, it is more than 90 degree
Guys you should use Naren's answer, it is correct and more deatiled
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yes it helped me..
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thank u all of my friends
• -4
Awesome answers meritnation
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sqr root of 361 is 19 right???
• 0
ans is 19 cm

• 6
thankk you
• -3
Thank you @akhshay nair :)
• -3
The ans is 19. And there is no wrong in what naren did. Even if it lengthy it is clear.
• 11
amazing answer naren it really helped..!!
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Thankyou for the answer :)
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19
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thnks....
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AD=19 cm.
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Parchi patel ur solution is easy......thnx
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Thank u Naren brother.. it's really useful
• 0
ans 19
• -3 :)
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Akshay nair's answer is easy to understand
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YES 19
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19 is the correct answer
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The radius of two concentric circles are 13 cm and 8 cm. AB is the diameter of the bigger circle and BD is tangent to the smaller circle touching at D and the bigger circle at E. Point A is joined to D. What is the length of AD? GIVEN: 2 Concentric circles with centre O. Radius OB of bigger circle = 13cm, Radius OD of smaller circle = 8cm. EB is tangent to the inner circle at point D. So, OD is perpendicular to EB.( as tangent is perpendicular to the radius segment through the point of contact.) Construction: Join AD & Join AE. PROOF: In right triangle ODB DB² = OB² — OD² ( by pythagoras law) DB² = 13² — 8² = 169 — 64 = 105 => DB = √105 => ED = √105 ( as perpendicular OD from the centre of the circle O, to a chord EB , bisects the chord) Now in triangle AEB, angle AEB = 90°( as angle on a semi circle is a right angle). And, OD // AE & OD = 1/2 AE ( as, segment joining rhe mid points of any 2 sides of a triangle is parallel to the third side & also half of it. So, AE = 16cm Now, in right triangle AED, AD² = AE² + ED² => AD² = 16² + (√105)² => AD² = 256 +105 => AD² = 361 => AD = √361 = 19cm
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19

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