Question

in a circle of diameter 40cm the length of a chord is 20 cm find the length of minor arc of the chord​

Answer 1

$\huge\sf{Given:-}$

• $\textsf{Diameter = 40cm}$
• $\textsf{Length of a chord = 20cm}$

$\huge\sf{To\:Find:-}$

• $\textsf{Length of minor arc of the chord}$

$\huge\sf{Solution:-}$

• $\textsf{Diameter of the circle = 40cm (given)}$
• $\sf{Radius \ of \ the \ circle \ = \ \frac{40}{2} \ cm \ = \ 20cm}$
• $\textsf{Let AB be a chord of the circle}$

$\star\:\sf{In \ \triangle OAB,}$

$\to\:\sf{OA \ = \ OB \ = \ radius \ of \ circle \ (20cm)}$

$\to\:\textsf{AB = 20cm}$

$\sf{\triangle \ OAB \ is \ a \ equilateral \ triangle}$

$\to\:\sf{\theta \ = \ 60^{\circ} \ = \ \dfrac{\pi}{3} \ radian}$

$\star\:\sf{We \ know \ that \ \theta \ = \ \dfrac{arc}{radius}}$

$\mapsto\:\sf{Arc \ = \ \theta \ \times \ Radius}$

$\mapsto\:\sf{\dfrac{\pi}{3} \ \times \ 20cm}$

$\mapsto\:\sf{\dfrac{20\pi}{3} \ cm}$

$\therefore\:\sf{The \ length \ of \ minor \ arc \ is \ \dfrac{20\pi}{3} \ cm}$

Answer 2

$\red{\large{\underline{\underline{ \rm{Given: }}}}}$

$\tt{ Diameter \: of \: a\: circle = 40 \: cm}$

$\tt{ Length \: of \: a \: chord = 20 \: cm}$

$\red{\large{\underline{\underline{ \rm{To \: Find: }}}}}$

$\tt{The \: length \: of \: minor \: arc \: of \: the \: chord.}$

$\red{\large{\underline{\underline{ \rm{Solution: }}}}}$

$\tt{ Diameter \: of \: a \: circle = 40 \: cm}$

$\tt{\therefore{Radius \: of \: a \: circle \: = \dfrac{Diameter}{2} }}$

$\tt{{ \therefore CA = CB = \dfrac{40}{2} = 20 \: cm}}$

$\tt{Also, \: cord \: AB = 20 \: cm}$

Now we know that:

$\tt{CA = CB = AB = 20 \: cm}$

Now, All the three sides of ∆ ABC are equal,

So ∆ ABC is an equilateral triangle.

We know :-

$\tt{\theta = \dfrac{arc}{radius}}$

$\tt{ \theta = 60 \degree }$

Convert the degree in radian, we have :

$\tt{60 \times \dfrac{ \pi}{180}rad}$

On cutting,

$\tt{ \dfrac{ \pi}{3} rad}$

$\tt{ As, \: \theta = \dfrac{arc}{radius}}$

$\tt{ \therefore arc = \theta \times radius}$

$\tt{Arc = \dfrac{ \pi}{3} \times 20 \: cm}$

$\tt{Arc \: i.e., \: AB = \dfrac{20 \pi}{3} \: cm}$

$\tt{\small{ \therefore The \: length \: of \: minor \: arc \: of \: the \: cord }}$ $\tt = { \green { \underline{ \boxed{ \gray { \tt{ \frac{ 20 \pi}{3} \: cm}}}}}}$

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