Question

A circular park of radius 20 m 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.​

Answer 1

$\huge{\underline{\underline{\sf{Requíred\:Solutíon:}}}}$

Let BD = x cm

Then in right triαngle ODB,

⠀⠀⠀⠀⠀⠀⠀OB² = OD² + BD²

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀| By Pythαgorαs theorem

$\implies{\sf{(20)^{2}=OD^{2} + x^{2}}}$

$\implies{\sf{(400)=OD^{2} + x^{2}}}$

$\implies{\sf{OD^{2}=400+ x^{2}}}$

$\implies{\sf{OD= \sqrt{400+x^{2}}}}$

Areα of equilαteral triαngle ABC

:$\implies{\sf{ \dfrac{ \sqrt{3}}{4}(side)^{2} = \dfrac{\sqrt{3}}{4}(BC)^{2}}}$

:$\implies{\sf{ \dfrac{ \sqrt{3}}{4}(2BD)^{2} = \sqrt{3}(BD)^{2} = \sqrt{3}x^{2}\:\:\:\:\:\:\:\:...(1)}}$

Agαin, αreα of equilaterαl triαngle ABC

= Areα of Δ OBC + Areα of Δ OCA

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ + Areα of Δ OAB

= $\displaystyle{\sf{3\:Area\:of\:Δ\:OBC\:=\:3 \dfrac{(BC)\:(OD)}{2}}}$

= $\displaystyle{\sf{ \dfrac{3(2BD)(OD)}{2} = 3(BC)(OD)}}$

= $\displaystyle{\sf{3x\: \sqrt{400-x^{2}}\:\:\:\:\:\:\:\:\:...(2)}}$

From equαtions (1) and (2),

$\displaystyle{\sf{\:\:\:\:\:\:3x\: \sqrt{400-x^{2}} = \sqrt{3}x^{2}}}$

$\implies{\sf{\:\:\:\:\:\sqrt{3}\: \sqrt{400-x^{2}} = x}}$

Squαring both sides,

$\displaystyle{\sf{\:\:\:\:\:3(400-x^{2})= x^{2}}}$

$\implies{\sf{\:\:\:\:\:1200-3x^{2}= x^{2}}}$

$\implies{\sf{\:\:\:\:\:4x^{2}= 1200}}$

$\implies{\sf{\:\:\:\:\:x^{2}= \dfrac{1200}{4}}}$

$\implies{\sf{\:\:\:\:\:x^{2}= 300}}$

$\implies{\sf{\:\:\:\:\:x= \sqrt{(300)}}}$

• Factorise 300 and simplify the equation.

$\implies{\sf{\:\:\:\:\:x= \sqrt{2\times2\times3\times5\times5}}}$

$\implies{\sf{\:\:\:\:\:x=10 \sqrt{3}}}$

$\implies{\sf{\:\:\:\:\:BD\:=\:10 \sqrt{3}}}$

$\implies{\sf{\:\:\:\:\:2BD\:=\:20 \sqrt{3}}}$

$\large\implies{\boxed{\bf{\purple{\:\:\:\:\:BC\:=\:20\sqrt{3}}}}}$

⠀⠀⠀⠀⠀⠀Hence, the length of string of eαch phone is $\displaystyle{\sf{20\sqrt{3}m.}}$

__________________________

Answer 2

Answer:

${\huge{\sf{\pink{\underline{\underline{Answer}}}}}}$

Let BD = x cm

Then in right triαngle ODB,

⠀⠀⠀⠀⠀⠀⠀OB² = OD² + BD²

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀| By Pythαgorαs theorem

\implies{\sf{(20)^{2}=OD^{2} + x^{2}}}⟹(20)

2

=OD

2

+x

2

\implies{\sf{(400)=OD^{2} + x^{2}}}⟹(400)=OD

2

+x

2

\implies{\sf{OD^{2}=400+ x^{2}}}⟹OD

2

=400+x

2

\implies{\sf{OD= \sqrt{400+x^{2}}}}⟹OD=

400+x

2

Areα of equilαteral triαngle ABC

:\implies{\sf{ \dfrac{ \sqrt{3}}{4}(side)^{2} = \dfrac{\sqrt{3}}{4}(BC)^{2}}}⟹

4

3

(side)

2

=

4

3

(BC)

2

:\implies{\sf{ \dfrac{ \sqrt{3}}{4}(2BD)^{2} = \sqrt{3}(BD)^{2} = \sqrt{3}x^{2}\:\:\:\:\:\:\:\:...(1)}}⟹

4

3

(2BD)

2

=

3

(BD)

2

=

3

x

2

...(1)

Agαin, αreα of equilaterαl triαngle ABC

= Areα of Δ OBC + Areα of Δ OCA

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ + Areα of Δ OAB

= \displaystyle{\sf{3\:Area\:of\:Δ\:OBC\:=\:3 \dfrac{(BC)\:(OD)}{2}}}3AreaofΔOBC=3

2

(BC)(OD)

= \displaystyle{\sf{ \dfrac{3(2BD)(OD)}{2} = 3(BC)(OD)}}

2

3(2BD)(OD)

=3(BC)(OD)

= \displaystyle{\sf{3x\: \sqrt{400-x^{2}}\:\:\:\:\:\:\:\:\:...(2)}}3x

400−x

2

...(2)

From equαtions (1) and (2),

\displaystyle{\sf{\:\:\:\:\:\:3x\: \sqrt{400-x^{2}} = \sqrt{3}x^{2}}}3x

400−x

2

=

3

x

2

\implies{\sf{\:\:\:\:\:\sqrt{3}\: \sqrt{400-x^{2}} = x}}⟹

3

400−x

2

=x

Squαring both sides,

\displaystyle{\sf{\:\:\:\:\:3(400-x^{2})= x^{2}}}3(400−x

2

)=x

2

\implies{\sf{\:\:\:\:\:1200-3x^{2}= x^{2}}}⟹1200−3x

2

=x

2

\implies{\sf{\:\:\:\:\:4x^{2}= 1200}}⟹4x

2

=1200

\implies{\sf{\:\:\:\:\:x^{2}= \dfrac{1200}{4}}}⟹x

2

=

4

1200

\implies{\sf{\:\:\:\:\:x^{2}= 300}}⟹x

2

=300

\implies{\sf{\:\:\:\:\:x= \sqrt{(300)}}}⟹x=

(300)

Factorise 300 and simplify the equation.

\implies{\sf{\:\:\:\:\:x= \sqrt{2\times2\times3\times5\times5}}}⟹x=

2×2×3×5×5

\implies{\sf{\:\:\:\:\:x=10 \sqrt{3}}}⟹x=10

3

\implies{\sf{\:\:\:\:\:BD\:=\:10 \sqrt{3}}}⟹BD=10

3

\implies{\sf{\:\:\:\:\:2BD\:=\:20 \sqrt{3}}}⟹2BD=20

3

\large\implies{\boxed{\bf{\purple{\:\:\:\:\:BC\:=\:20\sqrt{3}}}}}⟹

BC=20

3

⠀⠀⠀⠀⠀⠀Hence, the length of string of eαch phone is \displaystyle{\sf{20\sqrt{3}m.}}20

3

m.