Question

In a circle of radius 11cm, two parallel chord of
lenght 16cm and 30cm
respectively are drawn
on the same side of center find the distance
betwin them.​

Answer 1

Step-by-step explanation:

Let O be the centre of the circle and AB and CD be the two parallel chords of length 30 and 16 cm respectively.

Drop OE and OF perpendicular on AB and CD from the O

OE⊥AB and OF⊥CD

∴OE bisects AB and OF bisects CD

(pependicular drawn from the centre of a circle to a chord bisects it)

⇒AE=

2

30

=15 cm;CF=

2

16

=8 cm

In triangle △OAE,

OA

2

=OE

2

+AE

2

⇒OE

2

=OA

2

−AE

2

=(17)

2

−(15)

2

=64

∴OE=8 cm

In triangle △OCF,

OC

2

=OF

2

+CF

2

⇒OF

2

=OC

2

−CF

2

=(17)

2

−(8)

2

=225

∴OF=15 cm

The chords are on the same sides of the centre:

∴EF=OF−OE=(15−8)=7 cm

solution

PLEASE FOLLOW ME

Answer 2

Answer:

jajehebmrkqloeifjfif

Step-by-step explanation:

jdjndbfjekkas

dnxnhxkxk

xjdjjjd