Question

29. The length of a minute hand of a clock is 5cm. Find the area and perimeter swept by the minute
hand in 3 days.
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Answer 1

Answer:

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Answer 2

Question :-

The length of a minute hand of a clock is 5cm. Find the area and perimeter swept by the minute hand in 3 days.

❥︎ Answer

❥︎ Given :-

The length of a minute hand of a clock is 5cm.

❥︎ To Find :-

The area and perimeter swept by the minute hand in 3 days.

❥︎ Concept used :-

${{ \boxed{\large{\bold\red{Area_{(Sector)}\:\ = \:\pi r^2 \dfrac{θ}{360} }}}}} \:$

${ { \boxed{\large{\bold\red{Length_{(Sector)}\:\ = \:2\pi r \dfrac{θ}{360} }}}}} \:$

❥︎ Solution :-

The length of minute hand of a clock is 5 cm.

Therefore, radius = 5 cm.

❥︎ We know,

In 1 day, number of hours = 24

In 3 days, number of hours = 24 × 3 = 72 hours.

❥︎ Angle subtended at the centre by minute hand in 1 hour

is 360°.

So, Angle subtended at the centre by minute hand in 72 hours = 72 × 360° = 25920°.

❥︎ Now,

${{ \boxed{\large{\bold\red{Area_{(Sector)}\:\ = \:\pi r^2 \dfrac{θ}{360} }}}}} \:$

On substituting the values r = 15 and θ = 25920°, we get

$\bf\implies \:Area = \dfrac{22}{7} \times 15 \times 15 \times \dfrac{25920}{360}$

$\bf\implies \:Area = \dfrac{356400}{7} {cm}^{2}$

❥︎ Now,

${ { \boxed{\large{\bold\red{Length_{(Sector)}\:\ = \:2\pi r \dfrac{θ}{360} }}}}} \:$

On substituting the values r = 15 and θ = 25920°, we get

$\bf\implies \:Length = 2 \times \dfrac{22}{7} \times 15 \times \dfrac{25920}{360}$

$\bf\implies \:Length = \dfrac{47520}{7} \: cm$