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two equal circles of radius 5 cm have two common tangent AB and CD which touch the circle on A, C and B, D respectively and if CD = 24 cm, find the length of AB.

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interesting :)
can i know what did you attempt here?

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Other answers:

can i know what was your attempt here?
in a traingle bcd, i apply pythagoras theorem
but how you got CD?
it is given
ok you see for getting AB we need to take one more triangle.can you tell which one that will be?
i don't know
1 Attachment
well try to think of a triangle containing AB
a right angled one
plse see this diagram..
yep i saw
do u think taking a triangle BCD is right
okay...wait ..the other triangle be ABC.
wait a sec...
to make things clear join OB
where O is centre
in btw hw did u get bcd as rt angled ? :P
that's why i m saying i m wrong , we can't take tr.BCD
ob n od shld b joined
yeh :)
join OA also
i will try my best to draw that
yea thn tk d triangle AOD
& thn apply d theorem
ad^2= oa^2+od^2
@msingh i hand over this one to her :) seeya
vat is value of AD and OA
oa is 5
OA is not a radius
well msing we need to use property of tangents from same point i guess
lemme try this on properly ...
*on paper
draw a segment joining the centers.
nice one @sirm3d
isnt OO'=AB @sirm3d ?
that's right. O'O = AB
ABOO' is a rectangle it seems
my first attempt on this question wasnt as succesful as i thought .well this one was nice :)
@msingh any doubts here?
when the triangle are congruent, vat after that
AO'=BO =5cm CD=24cm
find OO' using that !!
CP= CD/2=24/2=12
OP=12 because the triangles O'CP and ODP are congruent
oops. i meant CP=DP =12
Find OP 2OP =OO'
use pythagorean theorem to find the length of hypotenuse OP of triangle ODP
O'P =13cm
OO' = 26cm=AB
right you are!
thank u @AravindG and @sirm3d
yw :)
i liked @sirm3d novelty :P

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