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hey chum here is your answer check it screenshot
12 is answer because chord given and asked for distance from centre so I do half hope it help you mark as brainlist
12 is answer because chord given and asked for distance from centre so I do half hope it help you mark as brainlist
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hey mate_______________
☆Welcome☆
Let AB be a chord of the given circle with Centre O and radius 13 cm.
Then, OA =13 cm and AB= 10 cm.
from O, draw OL perpendicular AB.
we know that the perpendicular from the centre of a circle to be a chord bisects the chord.
therefore, AL =1/2AB = (1/2×10) =5 cm.
FROM, the right triangle OLA, we have
OA^2 = OL^2 + AL^2
=__OL^2 =( 13^2 - 5^2) = 144 cm.
OL = 12cm
THEREFORE,,
the distance of the code from the centre is 12 cm.
☆Welcome☆
Let AB be a chord of the given circle with Centre O and radius 13 cm.
Then, OA =13 cm and AB= 10 cm.
from O, draw OL perpendicular AB.
we know that the perpendicular from the centre of a circle to be a chord bisects the chord.
therefore, AL =1/2AB = (1/2×10) =5 cm.
FROM, the right triangle OLA, we have
OA^2 = OL^2 + AL^2
=__OL^2 =( 13^2 - 5^2) = 144 cm.
OL = 12cm
THEREFORE,,
the distance of the code from the centre is 12 cm.
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