ex 10.5 and 10.4 handwritten
Answers
Step-by-step explanation:
Given:-
Radius of circular park = 20m
Let 3 boys be denoted by points A, S and D
AS = SD = AD
ToFind:-
Length of string of each phone
Solution:-
Let AS = SD = AD = 2x
In ASD all sides are equal
So, ASD is equilateral triangle
We draw OP ⊥ SD
\rm\implies \: SP = DP = \frac{1}{2}SD⟹SP=DP=
2
1
SD
\rm\implies \:SP = DP = \frac{2x}{2} = x⟹SP=DP=
2
2x
=x
Join OS and AO
★ In ∆OPS
(By phythagoras theorem)
\rm\implies \:OS^2 = OP^2 + PS^2⟹OS
2
=OP
2
+PS
2
\rm\implies \:20^2 = OP^2 + x^2⟹20
2
=OP
2
+x
2
\rm\implies \:400 - x^2 = OP^2⟹400−x
2
=OP
2
\rm\implies \:OP^2 = 400 - x^2⟹OP
2
=400−x
2
\rm\implies \:OP = \sqrt{400 - x^2}⟹OP=
400−x
2
★ In ∆APS
(By phythagoras theorem)
\rm\implies \:AS^2 = AP^2 + PS^2⟹AS
2
=AP
2
+PS
2
\rm\implies \:2x^2 = AP^2 + x^2⟹2x
2
=AP
2
+x
2
\rm\implies \:4x^2 - x^2 = AP^2⟹4x
2
−x
2
=AP
2
\rm\implies \:AP = \sqrt{3}x⟹AP=
3
x
★ Now,
\rm\implies \:AP = AO + OP⟹AP=AO+OP
\rm\implies \:\sqrt{3}x=20-\sqrt{400-x^2}⟹
3
x=20−
400−x
2
\rm\implies \:\sqrt{400-x^2} = \sqrt{3}x-20⟹
400−x
2
=
3
x−20
★ Squaring on both sides
\rm\implies \:(\sqrt{400-x^2} )^2 = (\sqrt{3}x-20)^2⟹(
400−x
2
)
2
=(
3
x−20)
2
\rm\implies \:400-x^2 = (\sqrt{3}x)^2+(20)^2 - 2 \times (\sqrt{3}x)\times 20⟹400−x
2
=(
3
x)
2
+(20)
2
−2×(
3
x)×20
\rm\implies \:4x^2 = 40\sqrt{3}x⟹4x
2
=40
3
x
\rm\implies \:x = 10\sqrt{3}m⟹x=10
3
m
\rm\therefore\:AS = SD = AD = 2x=2\times 10\sqrt{3}= {\rm{\pink{20\sqrt{3}m}}}∴AS=SD=AD=2x=2×10
3
=20
3
m
★ Hence,
\rm{Length \:of \:string \:of \:each \:phone\: is \:20\sqrt{3}m}Lengthofstringofeachphoneis20
3
m