Question

two parallel chords 18 centimetre and 24 cm long are 21 CM apart and the radius of the circle is?

Answer 1

Answer:

15cm

Step-by-step explanation:

let the two parallel chords be AB(24cm) and CD(18cm) , the center of circle be O, E and F be the mid points of AB and CD respectively  ,and OE be x cm and OF be (21-x)cm and the radius be r

FC = FD = CD/2 = 18/2 = 9

AE=EB=AB/2=24/2 = 12

in triangle OFC (OC (r) is hypotenuse)

OC^2 = OF^2 + FC^2

r^2 = (21-x)^2 + 9^2

r^2 = 441 + x^2 -42x + 81 = 522 +x^2 -42x ...................................(1)

in triangle OAE (OA(r) is hypotenuse)

OA^2 = AE^2 + EO^2

r^2 = 12^2 + x^2 = 144 + x^2   ........................(2)

joining (1) and (2) we get that,

r^2 = 522 +x^2 -42 =  144 + x^2

522 +x^2 -42x =  144 + x^2

522 -42x = 144

522-144 = 42x

378 = 42x

x = 378/42 = 9  ........................(3)

form (1)

522 +x^2 -42x =  r^2

r^2 = 522 + (9)^2 - 42(9) .......................using(3)

r^2 = 522 + 81 - 378 = 225

r = root 225 = 15 cm

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