In the adjoining figure AB and CD are two parallel chords

In the adjoining figure, AB and CD are two parallel chords and O is the bet centre. If the radius of the circle is 15 cm, find the distance MN ween the two chords of length 24 cm and 18 cm respectively.

Question 19 Circles Exercise 12A

Answer:

Construct OL ⊥ AB and OM ⊥ CD

Consider △ OLB and △ OMC

We know that ∠OLB and ∠OMC are perpendicular bisector

∠OLB = ∠OMC = 90

We know that AB || CD and BC is a transversal

From the figure we know that ∠OBL and ∠OCD are alternate interior angles

∠OBL = ∠OCD

So we get OB = OC which is the radii

By AAS congruence criterion

△ OLB ≅ △ OMC

OL = CM (c. p. c. t)

We know that the chords equidistant from the centre are equal

So we get

AB = CD

Therefore, it is proved that AB = CD.

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Given – AB = 24 cm, cd = 18 cm⇒ AM = 12 cm, CN = 9 cm Also, OA = OC = 15 cmLet MO = y cm, and ON = x cm In right angled ∆AMO

`(OA)^2 = (AM)^2 + (OM)^2`

⇒ `15^2 = 12^2 + y^2`

⇒ `y^2 = 15^2 -12^2`

⇒ `y^2 = 225 -144`

⇒` y^2 = 81`⇒ y = 9 cmIn right angled ΔCON

`(OC)^2 = (ON)^2 + (CN)^2`

⇒ `15^2 = x^2 + 9^2`⇒ `x^2 =15^2 - 9^2`

⇒` x^2 = 225 - 81`

⇒ `x^2 = 144` ⇒ y = 12 cmNow, MN = MO + ON= y + x= 9 cm +12 cm

= 21 cm

In the adjoining figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.

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