Question

$\huge\rm\underline{\bold \red{ Q}{ü} \red{ê}{s} \red{t}{ï} \red{ø}{ñ}\purple{\huge{\checkmark}}}$

Two parallel chords of length 30cm and 16cm are drawn on the opposite side of the center of a circle of radius 17cm. find the distance between the chords​​

Answer 1

Step-by-step explanation:

I think it will help you

pls make me as Brainlist

Answer 2

Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the center.

AB = 30cm and CD = 16cm [ Given ]

Draw OL ⊥ AB and OM ⊥ CD.

Join OA and OC.

OA = OC = 17cm [ Radius of a circle ]

The perpendicular from the center of a circle to a chord bisects the chord.

p∴ AL = AB/2 = 30/2 = 15 cm

Now, in right angled ∆ OLA,

∴(OA)² = (AL)² + (LO)²

[ By Pythagoras theorem ]

⇒(LO)² = (OA)² – (AL)²

⇒ (LO)² = (17)² – (15)²

⇒ (LO)² = 289 - 225

⇒ (LO)² = 64

⇒ LO = 8

Similarly,

In right angled ACMO,

= (OC) = (CM)+(MO)

= (MO)² = (OC) – (CM)²

⇒ (MO)² = (17)² – (8)²

⇒ (MO)² = 289 - 64

⇒ (MO)² = 225

∴ MO 15 cm

Hence, distance between the chords = (LO + MO) = (8 + 15) cm = 23 cm

Hope it helps!

Brainliest please! :)