Question

(a)
in the given diagram 'O' is the centre of the circle. Chord AB is parallel to chord
CD. AB = 64 cm, CD = 48 cm and radius of the circle is 40 cm. Find the distance
between the two chords.

​

Answer 1

Answer:

(16)

Step-by-step explanation:

o is a centre of the circle

Answer 2

The distance   between the two chords is 56 cm.

Step-by-step explanation:

consider AB and CD are the chords parallel to each other

AB = 64 cm

CD = 48 cm

radius of the circle = 40 cm

construct OL ⊥  AB and  OM⊥ CD

join the diagonals OA and  OC

we know that  OA = OC=  40 cm

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

therfore

AL = $\frac{1}{2}AB$

AL= $\frac{1}{2}\times 64$

AL = 32 cm

similarly,

CM = 24 cm

consider $\bigtriangleup OLA$

using pythagoras theorem

OA²= AL²+ OL²

(40)²= (32)²+OL²

OL²= (40)²-(32)² = 1600 - 1024 = 576

OL²= 576

OL = 24 cm

similarly

consider OMC

using again pythagoras theorem

OC²= CM² + OM²

(40)² = (24)² + OM²

OM² = (40)²- (24)²= 1600 - 576 = 1024

OM²= 1024

OM= 32 cm

distance between the chords = OM+OL

= 24 + 32 cm

= 56 cm

hence,

The distance   between the two chords is 56 cm.

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