Question

8. The length of the minute hand of a clock is 14 cm. The area swept by the
minute hand in 5 minutes is
a. 153.9 cm^2
b. 102.6 cm^2
c. 51.3 cm^2
d. 205.2 cm^2​

Answer 1

Answer :-

$\:\\\large{\boxed{\underline{\sf{Question's\;\;Analysis\;\;:-}}}}$

Here the concept of area of sector has been used. We are given the length of the minute hand of clock. So that will be the radius of the circular clock. Now we know that the angle at centre will be 360° of clock. So when we will divide this 360° by 12, we can get the angle formed at centre by minute hand in five minutes. And then we can apply the values in the formula of Area of Sector and find the values.

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★ Formula Used :-

$\:\\\large{\boxed{\sf{\odot\;\;Angle\;formed\;by\;minute\;hand\;at\;centre\;in\;5\;minutes\;=\;\bf{\dfrac{360^{\circ}}{12}}}}}$

$\:\\\large{\boxed{\sf{\odot\;\;Area\;of\;Sector\;=\;\bf{\dfrac{\pi r^{2} \theta}{360^{\circ}}}}}}$

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★ Question :-

The length of the minute hand of a clock is 14 cm. The area swept by the minute hand in 5 minutes is ?

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★ Solution :-

Given,

» Length of minute hand = r = 14 cm

» Time taken for the minute hand = 5 minutes

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~ For angle (θ) formed by minute hand in 5 minutes :-

$\:\\\qquad\large{\sf{:\longrightarrow\:\;\;Angle\;formed\;by\;minute\;hand\;at\;centre\;in\;5\;minutes,\;\bf{\theta}\;=\;\bf{\dfrac{360^{\circ}}{12}\;\;=\;\;\underline{\underline{30^{\circ}}}}}}$

$\:\\\large{\boxed{\boxed{\tt{Angle\;formed\;at\;centre\;by\;minute\;hand\;in\;5\;minutes_{(\theta)}\;=\;\bf{30^{\circ}}}}}}$

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~ For the Area swept by minute hand in 5 minutes :-

$\:\\\qquad\large{\sf{:\Longrightarrow\:\;\;Area\;of\;Sector\;=\;\bf{\dfrac{\pi r^{2} \theta}{360^{\circ}}}}}$

$\:\\\qquad\large{\sf{:\Longrightarrow\:\;\;Area\;of\;Sector\;=\;\bf{\dfrac{\dfrac{22}{7}\:\times\:(14)^{2}\:\times\:30^{\circ}}{360^{\circ}}}}}$

$\:\\\qquad\large{\sf{:\Longrightarrow\:\;\;Area\;of\;Sector\;=\;\bf{\dfrac{22\:\times\:(14)^{2}\:\times\:30^{\circ}}{7\:\times\:360^{\circ}}}}}$

$\:\\\qquad\large{\sf{:\Longrightarrow\:\;\;Area\;of\;Sector\;=\;\bf{\dfrac{22\:\times\:(14)^{2}\:\times\:30^{\circ}}{7\:\times\:360^{\circ}}\;\;=\;\;\underline{\underline{51.3\;\;cm^{2}}}}}}$

Hence, the correct option is

Option C.) 51.3 cm²

$\:\\\large{\underline{\underline{\rm{Thus,\;area\;swept\;by\;minute\;hand\;in\;5\;minutes\;is\;\;\boxed{\bf{51.3\;\;cm^{2}}}}}}}$

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★ More Formulas to know :-

$\:\\\sf{\leadsto\;\;Length\;of\;arc\;=\;\dfrac{2\pi r \theta}{360^{\circ}}}$

$\:\\\sf{\leadsto\;\;Area\;subtended\;by\;cord\;=\;\dfrac{\pi r^{2} \theta}{360^{\circ}}\;-\;\dfrac{1}{2}\:\times\:r^{2}\:\sin \theta}$

$\:\\\sf{\leadsto\;\;Area\;of\;Quadrant\;=\;\dfrac{\pi r^{2}}{4}}$

$\:\\\sf{\leadsto\;\;Area\;of\;Circle\;=\;\pi r^{2}}$

$\:\\\sf{\leadsto\;\;Perimeter\;of\;Circle\;=\; 2 \pi r}$

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★ Figure for the Question :-

$\setlength{\unitlength}{1.2mm}\begin{picture}\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,30){\line(5, - 4){16}}\put(25,30){\circle*{1}}\put(24,32){\sf\large{O}}\put(15,40){\sf\large{Major Sector}}\put(5,14){\sf\large{A}}\put(25,30){\line(- 5, -4){16}}\put(43,14){\sf\large{B}}\put(14,16){\sf\large{Minor Sector}}\end{picture}$

This is the figure to show the difference between the Major Sector and Minor Sector. The minor sector is the area swept by minute hand in 5 minutes and the major sector is the left area.

* Note :- Kindly view this answer from web to see the displayed image here. Also use desktop site to see it.

Answer 2

Given,

The length of the minute hand of a clock is 14 cm.

To find :

The area swept by the  minute hand in 5 minutes .

Solution :

Here, angle created by the clock in 1 minute = 360/60

∴ Angle created by the clock in 5 minutes , θ = 6 * 5 = 30°

Now, radius of clock, r = 14 cm

Now we know,

⇒ Area swept = θ/360° × πr²

⇒ Area swept = 30°/360° * 3.14 * 14²

⇒ Area swept = 1/12 * 3.14 * 196

⇒ Area swept = 1/12 * 615.44

⇒ Area swept = 51.3 cm²

Therefore,

Area swept by the minute hand = 51.3 cm²        [Option : c]