A circular park of radius m 10m is situated in a colony. Three girls Ankita, Shireen and Deepti are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone
Answers
Given : A circular park of radius 10m is situated in a colony
Three girls Ankita, Shireen and Deepti are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other.
To Find : the length of the string of each phone
Solution:
sitting at equal distance
=> Equilateral Triangle
Let say ABC is triangle
and O is center
then AO = BO = CO = Radius = 10 m
as Triangle is equilateral
Hence ∠AOB = ∠BOC = ∠COA = 360°/3 = 120°
AB² = AO² + BO² - 2AO. BO Cos∠AOB
=> AB² = 10² + 10² - 2* 10 * 10 Cos120
=> AB² = 200 - 200(-0.5)
=> AB² = 200 + 100
=> AB² = 300
=> AB = 10√3
length of the string of each phone = 10√3 m
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Class 10
>>Maths
>>Triangles
>>Pythagoras Theorem and its Converse
>>A circular park of radius 20m is situate
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A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.
Hard
Solution
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Let Ankur be represented as A, Syed as S and David as D.
The boys are sitting at an equal distance.
Hence, △ASD is an equilateral triangle.
Let the radius of the circular park be r meters.
∴OS=r=20m.
Let the length of each side of △ASD be x meters.
Draw AB⊥SD
∴SB=BD=
2
1
SD=
2
x
m
In △ABS,∠B=90
o
By Pythagoras theorem,
AS
2
=AB
2
+BS
2
∴AB
2
=AS
2
−BS
2
=x
2
−(
2
x
)
2
=
4
3x
2
∴AB=
2
3
x
m
Now, AB=AO+OB
OB=AB−AO
OB=(
2
3
x
−20) m
In △OBS,
OS
2
=OB
2
+SB
2
20
2
=(
2
3
x
−20)
2
+(
2
x
)
2
400=
4
3
x
2
+400−2(20)(
2
3
x
)+
4
x
2
0=x
2
−20
3
x
∴x=20
3
m
The length of the string of each phone is 20
3
m.
solution