A circular park of radius 20cm is situated in a colony.Three boys are sitting at equal distance on its boundary each having o toy telephone in his hands to talk to each other. Find the length of the string of each phone.
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It is given that AS = SD = DA
Therefore, ΔASD is an equilateral triangle.
OA (radius) = 20 m
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ASD. We also know that medians intersect each other in the ratio 2: 1. As AB is the median of equilateral triangle ASD, we can write :
OA / 0B =2/1
20m/OB =2/1
OB =(20/2)m = 10m ∴ AB = OA + OB = (20 + 10) m = 30 m
In ΔABD,
AD2 = AB2 + BD2
AD2 = (30)2 +( Ad /2)whole square
Ad ka square = 900+1/4 ad ka square
3/4 Ad ka square = 900
Ad ka square = 1200
Ad = 20 under the root 3
Therefore, the length of the string of each phone will be = 20 under the root 3 I hope the answer was helpful to you! If it was please mark it as brainlist ❤️
Therefore, ΔASD is an equilateral triangle.
OA (radius) = 20 m
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ASD. We also know that medians intersect each other in the ratio 2: 1. As AB is the median of equilateral triangle ASD, we can write :
OA / 0B =2/1
20m/OB =2/1
OB =(20/2)m = 10m ∴ AB = OA + OB = (20 + 10) m = 30 m
In ΔABD,
AD2 = AB2 + BD2
AD2 = (30)2 +( Ad /2)whole square
Ad ka square = 900+1/4 ad ka square
3/4 Ad ka square = 900
Ad ka square = 1200
Ad = 20 under the root 3
Therefore, the length of the string of each phone will be = 20 under the root 3 I hope the answer was helpful to you! If it was please mark it as brainlist ❤️
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