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 each other. Find the length of the string of each phone.
Answers
Answer:
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
Answer:
20√3
Step-by-step explanation:
Given, radius = 20 m.
From figure:
AM and CN are median, O is the centre of the circle.
AO = 2OM = 20.
→ OM = 10 m.
→ AM = OA + OM
= 20 + 10
= 30 m.
Now,
Let BM = x.
∴ BM = MC = x.
∴ BM = (1/2) * BC
= (1/2) * BC
→ BC = AB = 2x.
In ΔAMB,
→ AB² = AM² + BM²
→ (2x)² = (30)² + (x)²
→ 4x² = 900 + x²
→ 3x² = 900
→ x² = 300
→ x = 10√3
Then, AB = 2x
= 2(10√3)
= 20√3
Therefore, Length of the string of each phone = 20√3.
Hope this helps!
