Guys please solve.....
Answers
Step-by-step explanation:
Given: Chord AB parallel to CD and AB = 10cm and CD = 24cm
To find: Radius of circle.
Procedure:
According to a theorem, the line joining the center to the chord perpendicularly bisect the chord.
As the distance between the parallel chords are perpendicular to the chords.
AE = EB and CF = FD
BE = AB/2 = 10cm/2 = 5cm
FD = CD/2 = 24cm/2 = 12cm
Since, FE is given 17cm
So, let OF be 'x'cm
Then, OE will be (17-x)cm
According to Pythagoras Theorem,
In ∆OFD,
FD²+OF² = OD²
(12)²+(x)² = OD²
OD² = 144+x² .........(i)
In ∆OEB,
OE²+EB² = OB²
(17-x)²+(5)² = OB²
289+x²-34x + 25 = OB²
314+x²-34x = OB² ........... (ii)
Equating equation (i) and (ii) as OB²= OD² since both are radius.
144+x² = 314+x²-34x
144-314+34x = 0
34x-170 = 0
34x = 170
x = 170/34 = 5cm.
Putting x = 5cm in equation (i)
OD² = 144+(5)²
= 144+25
= 169cm²
OD = √169cm² = 13cm
Hence, radius is 13cm
Hope it helps......please mark it as the brainliest answer..