A circular park is situated in a colony. Four childrenAadya, Sree, Bharat are sitting at equal distance of 20√3on its boundary and Harsh sitting at the centre is passing ball of radius 7 cm to all of them one by one. Answer the following questions (any three)(3)(i)Find the central angle inscribed by Aadya and Sree (ii) Findthe distance between Harsh and Bharat. (iii) Paarth comes and sit exactly between Sree and Bharat. Find the shortest distance between Harsh and Paarth.(iv) What is the surface area of the ball.
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Answered by
0
Answer:
angle will be 20
30 n degree
Answered by
1
Step-by-step explanation:
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.
please mark my answer as a brainly
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