two chords of length 40 centimetre and 48cm are parallel to each other and are on either side of a circle Find the distance between the two chords of radius is 25 cm
Answers
Step-by-step explanation:
D1 = root 25sq - 20sq
= root 625-400
= 15cm
D2 = root 25sq - 24sq
= root 625-576
= 7cm
Distance between chords = 7+15 = 22cm
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Step-by-step explanation:
follow On the basis of question we draw a figure of a circle with centre O in two different cases ,
Case I,
When chords lie on both sides of centre.
AB = 40 cm.
CD = 48 cm.
CE = DE = 24 cm.
AF = BF = 20 cm.
OA = OC = 25 cm.
In ∆ AOF,
OF = √OA² - AF²
OF = √25² - 20²
OF = √(25 + 20)(25 - 20)
OF = √45 × 5 = √5 × 3 × 3 × 5 = 15 cm.
In ∆ COE,
OE = √OC² - CE²
OE = √25² - 24²
OE = √(25 + 24)(25 - 24)
OE = √49 = 7 cm.
∴ Required distance = EF = OE + OF = (7 + 15) cm = 22 cm.
Case II
When the chords lie on the same side of centre
AF = 20 cm.
CE = 24 cm.
OC = OA = 25 cm.
In ∆ OAF
OF = √OA² - AF²
OF = √25² - 20²
OF = √625 - 400
OF = √225 = 15 cm.
In ∆ OCE,
OE = √OC² - CE² = √25² - 24²
OE = √(25 + 24)(25 - 24)
OE = √49 = 7 cm.
∴ Required distance = EF = OF – OE = 15 – 7 = 8 cm.