17. Two parallel chords of length 8 cm and 6 cm
of a circle are separated by a distance of 1
cm. The radius of the circle (in cm) is
4
2. 4 √2
3. 5
4. 5√2
Answers
Step-by-step explanation:
Let O be the center and r cm. be the radius of the circle. Two parallel chords AB= 8
cm. and CD= 6 cm . Mid points of AB and CD are E and F respectively.
In right angled triangle OEB:- OE^2 + EB^2 = OB^2 or. OE^2+ (8/2)^2= r^2.
or. OE^2+16 = r^2……………………(1).
In right angled triangle OFD :- OF^2+FD^2= OD^2. or . OF^2+(6/2)^2= r^2.
or. OF^2 + 9= r^2……………………..(2).
Subtracting eqn.(1) from eqn. (2)
OF^2-OE^2 = 7. or. (OF+OE).(OF-OE) = 7……………………(3)
Condition (1). chords lie on the opposite side of the center O. which means
OF+OE = 1…………….(4) , putting OF+OE =1 .from eqn.(4) in eqn.(3).
OF-OE =7……………..(5).
by solving eqn. (4) and (5) OF= 4. and OE =-3 (not possible).
Condition (2). Chords lie on the same side of the center O. which means
OF-OE =1………………(6). , putting OF-OE =1 from eqn. (6) in (3)
OF+OE=7…………………(7)
By solving eqn. (6) and (7) OF=4 and OE = 3. , putting OE=3 cm. in eqn. (1).
(3)^2+ 16 =r^2. or. r^2 =25 =>. r =√25 = 5 cms
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