Question

In a circle of radius 17cm, two parallel chords are drawn on opposite sides of a diameter. the distance between the chords is 23cm. if the length of one chord is 16 cm, then the length of the other is

Answer 1

Answer:

30 cm

Step-by-step explanation:

Given: Radius of the circle = 17 cm

Distance between the two parallel chords = 23 cm

AB||CD and LM = 23 cm

Join OA and OC.

∴OA=OC=17cm

Let OL = x cm, then OM = (23 - x) cm

AB = 16 cm

Now in right

ΔOAL,$OA^{2}$=$OL^{2}$+$AL^{2}$

⇒$17^{2}$=$x^{2}$+$AL^{2}$

⇒289=$x^{2}$+$AL^{2}$

⇒$x^{2}$=289-$AL^{2}$=289-$8^{2}$

=289-64

=225

=$15^{2}$

∴x=15 cm

and OM = 23 - x = 23 - 15 = 8 cm

Now in right ΔOCM, $OC^{2}$=$OM^{2}$+$CM^{2}$

⇒$17^{2}$=$8^{2}$+$CM^{2}$

⇒289=64+$CM^{2}$

⇒$CM^{2}$

=289−64

=225

=$15^{2}$

∴ CM=15cm

CD=2×CM

=2×15

=30cm

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Answer 2

According the question,

In a circle of radius 17cm, two parallel chords are drawn on opposite sides of a diameter.

Let,s consider two chords are,

Chord MP = 16cm,

Chord NR = Xcm

The distance between two chords = 23cm= SQ

In Δ POQ

PO²= OQ² + PQ²           ( Using Pythagoras law)

( PO = 17cm, PQ= 8cm)

OQ² =PO² - PQ²

        = ( 17)² - ( 8)²

       =( 289 -64)

       =225

OQ = √225= 15 cm

Hence OS = ( 23-15)cm = 8cm

Now in ΔROS

OR² = OS² + SR²                  ( Using Pythagoras law)

SR² = OR² - OS²

      = (17)²- (8)²

      = (289-64)

     = 225

SR =√225= 15cm

RN = 2RS=( 2×15)cm

Hence  the length of the other chord ( RN) will be 30cm.

Ans :-  The length of the other chord ( RN) will be 30cm.

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