Math, asked by YoshitaVishwakarma, 2 months ago

Two parallel chords are drawn on the opposite side of the centre of a

circle of radius 25 cm. If the lengths of the chords are 48 cm and 40

cm reapectively, then find the distance between them.​

Answers

Answered by kimtaehyung1730
1

Answer:

30 cm

Step-by-step explanation:

Answer:

30 cm

Step-by-step explanation:

Consider the attached figure in which Two chords AB and CD are given and O is centre of the circle. QP is the distance between two chords

Now, Given that

CD = 48 cm

∴ CQ = 1/2CD              [ Perpendicular from centre bisects a chord]

CQ = 24 cm

OC = 25 cm ( radius)

In ΔOQC

OC² = OQ² + QC²                       [By Pythagoras Theorem]

25² = OQ² + 24²

OQ² = 25² - 24²

OQ² = (25 + 24) ( 25 - 24)          [ ∵ a² - b² = ( a + b) (a - b) ]

OQ² = 49

OQ = 7 cm

Now Distance between chords (QP) = 27 cm

QP = OQ + OP

27 = 7 + QP                                 [∵ OQ = 7 cm ]

QP = 20 cm

In Δ APO

AO² = OP² + AP²                      [By Pythagoras Theorem]

25² = 20² + AP²

AP² = 25² - 20²

AP² = 625 - 400

AP² = 225

AP² = 15²                                  [ ∵ 225 = 15² ]

AP = 15 cm

∵ Perpendicular from centre bisects a chord

∴ AP = PB = 15 cm

AB = AP + PB = 30 cm

∴ Length of other chord is 30 cm

Attachments:
Answered by Ranveerx107
1

30 cm

Step-by-step explanation:

Consider the attached figure in which Two chords AB and CD are given and O is centre of the circle. QP is the distance between two chords

Now, Given that

CD = 48 cm

∴ CQ = 1/2CD          

   [ Perpendicular from centre bisects a chord]

CQ = 24 cm

OC = 25 cm ( radius)

In ΔOQC

OC² = OQ² + QC²   [By Pythagoras Theorem]

25² = OQ² + 24²

OQ² = 25² - 24²

OQ² = (25 + 24) ( 25 - 24)          [ ∵ a² - b² = ( a + b) (a - b) ]

OQ² = 49

OQ = 7 cm

Now Distance between chords (QP) = 27 cm

QP = OQ + OP

27 = 7 + QP                                 [∵ OQ = 7 cm ]

QP = 20 cm

In Δ APO

AO² = OP² + AP²    [By Pythagoras Theorem]

25² = 20² + AP²

AP² = 25² - 20²

AP² = 625 - 400

AP² = 225

AP² = 15²             [ ∵ 225 = 15² ]

AP = 15 cm

∵ Perpendicular from centre bisects a chord

∴ AP = PB = 15 cm

AB = AP + PB = 30 cm

∴ Length of other chord is 30 cm

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