Question

$\huge \frak{Question}$
A circular park has a radius of 40 m is situated in à colony. Three boy Aman, Amit and Shivam were talking by phone. Find the length of string of each phone.
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Answer 1

Given :

• Radius of a circular park = 40m.
• Three boys Aman, amit and shivam were talking on phone.

To find :

• The length of string of each phone = ?

Solution :

Given that, AB = BC = CA.

So,

∆ABC is an equilateral triangle.

The radius be OA that is of 40cm.

Construction :

Draw AD ⊥ BC.

Now,

AD is the median of ∆ABC and It passes throught the centre O.

Therefore, $\sf OA \ = \ \dfrac {2}{3} D$

• The centeroid of an equilateral triangle be $\sf BD \ = \ \dfrac {a}{2} m$.

In ∆ABD,

By Pythagoras theorem,

$\implies \sf AD^2 \ = \ BD^2 \ + \ AD^2$

$\implies \sf AD^2 \ = \ AB^2 \ - \ BD^2$

$\implies \sf AD^2 \ = \ a^2 \ - \ \bigg( \dfrac {a}{2} \bigg)^2$

$\implies \sf AD^2 \ - \ a^2 \ - \ \dfrac {a^2}{4}$

$\implies \sf AD^2 \ = \ \dfrac {4a^2 \ - \ a^2}{4}$

$\implies \sf AD^2 \ = \ \dfrac {3a^2}{4}$

$\implies \sf AD \ = \ \dfrac {\sqrt{3a}}{2}$

$\implies \sf OA \ = \dfrac {2}{3} AD$

$\implies \sf 40 m \ = \ \dfrac {2}{3} \times \dfrac {\sqrt {3a}}{2}$

$\implies \sf 40 \ = \ \dfrac {2 \sqrt {3a}}{6}$

$\implies \sf 40 \ = \ \dfrac {\sqrt {3a}}{3}$

$\implies \sf 40 \times 3 \ = \ \sqrt {3a}$

$\implies \sf a \ = \ \dfrac {120}{\sqrt {3}}$

$\implies \sf a \ = \ \dfrac {(120 \times \sqrt{3})}{\sqrt {3} \times \sqrt {3}}$

Here, Rationalising the denominator,

We get,

$\implies \sf a \ = \ \dfrac {120 \sqrt {3}}{3}$

$\implies \sf a \ = \ 40 \sqrt {3} m$

$\therefore$ The length of the string of each phone is $\sf 40 \sqrt {3}$ m.

Answer 2

$\; \; \; \;{\large{\sf{\bold{\underbrace{\underline{Let's \; understand \; the \; Question\: 1^{st}}}}}}}$

${\bullet}$ This question says that a circular park has radius as 40 m is situated in a colony. Three boys named Aman, Amit and Shivam were talking by phone ( string ) We have to find the length of string of each the phone.

${\large{\sf{\bold{\underline{Given \; that}}}}}$

${\bullet}$ Circular park has radius = 40 m.

${\bullet}$ 3 boys are taking by ( string ) phone.

${\large{\sf{\bold{\underline{To \; find}}}}}$

${\bullet}$ Length of string of phone.

${\large{\sf{\bold{\underline{Solution}}}}}$

${\bullet}$ Length of string of phone = 40√3 m

${\large{\sf{\bold{\underline{Full \; Solution}}}}}$

~ According to diagram,

➥ There is a ∆XYZ

~ Here, is an Equilateral triangle that's why,

➥ XY = XZ = ZX

~ And,

➥ OX = 40 cm

~ Now,

➢ As we see that the median of triangle ( equilateral ) passes through the circum centre. Circum centre = 0.

➢ As we already know that at 2:1 the medians intersect each other.

~ Now as we see that

➢ Median of given triangle XYZ is XB.

~ So, we write this as,

➢ ${\sf{\dfrac{OA}{OB}}}$ = ${\sf{\dfrac{2}{7}}}$

➢ ${\sf{\dfrac{40 \: m}{OB}}}$ = ${\sf{\dfrac{2}{7}}}$

➢ OB = 20 m

➢ Therefore, XB = OX + OB = (40+20) m

➢ Henceforth, 60 m

~ Now in ∆XBC we have to use phythagoras theorm,

➨ (AC)² = (AD)² + (DC)²

➨ (AC)² = (60)² + (AC)²/2

➨ (AC)² = 3600 + (AC)²/4

➨ 3/4AC² = 3600

➨ AC² = 4800

➨ AC = 40√3 m

➨ Therefore, 40√3 m is length of string of each phone.

${\green{\frak{Henceforth, \: 40 \sqrt{3} m \: is \: length \: of \: string}}}$

${\Large{\bold{\sf{\bf{\underbrace{Additional \; knowledge}}}}}}$

• ${\large{\pink{\bold{\sf{Area \; of \; circle = \pi r^{2}}}}}}$

• ${\large{\pink{\bold{\sf{Circumference \; of \; circle = 2 \times \pi \times r}}}}}$

• ${\large{\pink{\bold{\sf{Diameter \; of \; circle = 2 \times r}}}}}$

• ${\large{\pink{\bold{\sf{Radius \; of \; circle = \dfrac{Diameter}{2}}}}}}$