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A Circle has two parallel chords AB and CD on opposite sides of centre ,AB=10 CM CD=24 CM and the distance between them is 17 cm . Find the RADIUS of the circle?????

Answer :

Avishek
Const :- OE perpendicular AB and OF perpendicular to CD.
According to the question,
Let OE = x . Therefore, OF = 17 - x
Let OB = r = OD

Now in triangle OEB. EB = 1/2AB = 5cm
Thus, OB² = OE² + EB²
= x² + 25 ....................................... (i)
Similarly , OD² = OF² + FD²
= (17-x)² + 144 .............................. (ii)

As OB = OD or OB² = OD²
We have (i) = (ii)
= x² + 25 = (17-x)² + 144
= x² + 25 = 289 - 34x + x² + 144
= 34x = 144 + 289 - 25
= 34x = 408
= x = 12 cm

In triangle OEB,
r² = OE² + EB²
= r² = 12² + 5²
= r² = 169
= r = 13 cm

Hope that helps !!
View image Avishek
Join OD and OB
draw a line passing through O that is perpendicular to both the chords
let it meet CD at F and AB  at E
now,consider the tri.FOD
OD^2=OF^2+EB^2
        =12^2+(17-X)^2
       =144+289-34X+X^2
now,consider tri.OEB
OB^2=OE^2+EB^2
       =X^2+25
we know that OB=OD(radii)
hence,OD^2=OB^2
X^2+25=(17-X)^2+144
X=12
So,by substituting the value of X in one of the equation,
X^2+25=OB^2
OB=13
i.e radius =13

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