Question

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) length of the arc.
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord.


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Answer 1

Step-by-step explanation:

Radius of the circle, r = 21 cm.

Angle at the centre of the circle, θ = 60° (

1) the length of the arc ?

(2) Area of the sector formed by the arc

(3) Area of the segment.

Answer 2

⠀⠀⠀⠀⠀⠀⠀⠀⠀Diagram is in the attachment

$\huge\bf\underline\mathfrak{Answer :}$

• $\text{Length of the arc}$ = $\sf\red{22 \: cm}$.
• $\text{Area of the sector}$ = $\sf\red{231 \: cm²}$.
• $\text{Area of the segment}$ = $\sf\red{231 - \frac{441}{4} \sqrt{3}\: cm²}$

$\huge\bf\underline\mathfrak{Step \: by \: step \: explanation :}$

$\huge\bf\underline\mathfrak{Given :}$

• $\text{A circle with centre 'o'}$
• $\text{radius = OP = OQ}$ = $\sf\purple{21 \: cm}$
• $\text{Angle subtended by the arc, ∠POQ}$ = $\sf\purple{60°}$

$\huge\bf\underline\mathfrak{To \: find :}$

1. $\text{Length of the arc.}$
2. $\text{Area of the sector formed by the arc.}$
3. $\text{Area of segment formed by the corresponding chord.}$

$\huge\bf\underline\mathfrak{Solution :}$

$\sf\underbrace{Finding \: length \: of \: the \: arc, \: PRQ \: :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\text{Length of PRQ}$ = $\sf \dfrac{θ}{360°} \times (2πr)$

Where,

• $\text{r = radius = OP = 21 cm}$
• $\text{π = 22/7}$
• $\text{θ = angle subtended by the arc = 60°}$

$\underline\text{Putting the above values in the formula :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf \dfrac{60°}{360°} \times 2 \: \times \frac{22}{7} \times \: 21$

⠀⠀⠀⠀⠀⠀$\implies$ $\sf \dfrac{1}{6} \times 2 \: \times \frac{22}{7} \times 21$

⠀⠀⠀⠀⠀⠀$\implies$ $\sf\red{22 \: cm}$

$\sf\underbrace{Finding \: area \: of \: sector \: formed \: by \: PRQ \: :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\text{Area of sector OPRQ}$ = $\sf \dfrac{θ}{360°} \times \pi r^{2}$

$\underline\text{Putting the suitable values :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf \dfrac{60}{360} \times \frac{22}{7} \times 21 \times 21$

⠀⠀⠀⠀⠀⠀$\implies$ $\sf \dfrac{1}{6} \times \frac{22}{7} \times 21 \times 21$

⠀⠀⠀⠀⠀⠀$\implies$ $\sf \dfrac{1}{6} \times 22 \times 3 \times 21$

⠀⠀⠀⠀⠀⠀$\implies$ $\sf\purple{231 \: cm²}$

$\sf\underbrace{Finding \: area \: of \: the \: segment \: PRQ \: :-}$

$\underline\text{In ∆OPQ :}$

$\text{OP = OQ ( Radii of the same circle )}$

$⇒$ $\text{∠OPQ = ∠OQP ( Angles opposite to equal sides)}$

$\sf\underline{Hence, \: ∆OPQ \: is \: an \: equilateral \: ∆}$ $\text{(2 of it's angles are 60°)}$

$\underline\text{As we know that,}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\text{Area of equilateral ∆OPQ}$ = $\sf \dfrac{ \sqrt{3} }{4} (s)^{2}$

Where, s = side of the ∆OPQ = 21 cm

$\underline\text{Putting the suitable value of s :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\sf \dfrac{ \sqrt{3} }{4} \times 21 \times 21$

⠀⠀⠀⠀⠀⠀$\implies$$\sf \dfrac{ \sqrt{3} }{4} \times 441$

⠀⠀⠀⠀⠀⠀$\implies$ $\sf \dfrac{441 \sqrt{3} }{4} \: cm²$

$\underline\text{From the figure, we can say that,}$

$\color{green}\tt {Area \: of \: segment \: = \: Area \: of \: sector \: - \: Area \: of \: triangle}$

$\underline\text{Putting the suitable values :-}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀$\text{Area of the segment}$ = $\sf\red{231 - \frac{441}{4} \sqrt{3}\: cm²}$