Two circles with radii 30 cm and 40 cm . their centers are 50 cm apart. find the length of common chord
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sum of radii of circles ( 30cm + 40cm = 70cm) > distance between their centres (50 cm)
It means , both the given circles are intersecting . Meet the intersecting points , a line is formed . This is what ? Of course common chord of circles .
Let Length of common chord = 2P
we know, a line passing through centre , bisects the chord perpendicularly.
so, AO = OB = P [see attachment]
From ∆AOC ,
30² = x² + P² -----(1)
From ∆AOD
40² = (50 - x)² + P² ---(2)
From equation (1) and (2),
30² - x² = 40² - (50 - x)²
⇒-700 = x² - (50 - x)²
⇒ -700 = x² - 2500 - x² + 100x
⇒1800 = 100x
⇒x = 18 , put it in equation (1)
30² = 18² + P²
⇒900 - 324 = P²
P = 24 cm
Hence, common chord = 2P = 48cm
sum of radii of circles ( 30cm + 40cm = 70cm) > distance between their centres (50 cm)
It means , both the given circles are intersecting . Meet the intersecting points , a line is formed . This is what ? Of course common chord of circles .
Let Length of common chord = 2P
we know, a line passing through centre , bisects the chord perpendicularly.
so, AO = OB = P [see attachment]
From ∆AOC ,
30² = x² + P² -----(1)
From ∆AOD
40² = (50 - x)² + P² ---(2)
From equation (1) and (2),
30² - x² = 40² - (50 - x)²
⇒-700 = x² - (50 - x)²
⇒ -700 = x² - 2500 - x² + 100x
⇒1800 = 100x
⇒x = 18 , put it in equation (1)
30² = 18² + P²
⇒900 - 324 = P²
P = 24 cm
Hence, common chord = 2P = 48cm
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