Question

Calculate the area of a sector with radius 7m and sector angle 90 degrees.
take pie as 22 over 7

Answer 1

GivEn:

• Radius of Circle = 7 cm
• Sector angle = 90°

To find:

• Area of sector

SoluTion:

⋆ Reference of image is shown in diagram

$\setlength{\unitlength}{1.2mm}\begin{picture}(50,55)\thicklines\qbezier(25.000,10.000)(33.284,10.000)(39.142,15.858)\qbezier(39.142,15.858)(45.000,21.716)(45.000,30.000)\qbezier(45.000,30.000)(45.000,38.284)(39.142,44.142)\qbezier(39.142,44.142)(33.284,50.000)(25.000,50.000)\qbezier(25.000,50.000)(16.716,50.000)(10.858,44.142)\qbezier(10.858,44.142)( 5.000,38.284)( 5.000,30.000)\qbezier( 5.000,30.000)( 5.000,21.716)(10.858,15.858)\qbezier(10.858,15.858)(16.716,10.000)(25.000,10.000)\put(25,30){\line(5,0){20}}\put(25,30){\circle*{1}}\put(30,32){\sf\large{7 cm}}\put(25,30){\line(0, - 5){20}}\put(16,20){\sf\large{7 cm}}\put(28,30){\line(0, - 5){3}}\put(25,27){\line(5,0){3}}\put(30,25){\sf\large{ \sf 90^\circ }}\end{picture}$

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We know that,

$\star\;{\boxed{\sf{\purple{Area_{\;(sector)} = \dfrac{ \theta}{360^\circ} \times \pi r^2}}}}$

Now, Putting given values in formula,

$:\implies\sf \cancel{ \dfrac{90^\circ}{360^\circ}} \times \dfrac{22}{ \cancel{7}} \times \cancel{7} \times 7$

$:\implies\sf 4 \times 22 \times 7$

$:\implies{\boxed{\frak{\pink{616\;cm^2}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;Area\;of\;sector\;is\; \bf{616\;cm^2}.}}}$

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$\quad\boxed{\underline{\underline{\purple{\bigstar \: \bf\:Formula\:Related\:to\:Circle\:\bigstar}}}}$

$\;\;\bullet\;\;\sf Area_{\;(cricle)} = \pi r^2$

$\;\;\bullet\;\;\sf Area_{\;(Perimeter)} = 2 \pi r$

$\;\;\bullet\;\;\sf Area_{\;(semi - cricle)} = \dfrac{ \pi r^2}{2}$

$\;\;\bullet\;\;\sf Peri - meter_{\;(semi - cricle)} = \pi r + 2r$

$\;\;\bullet\;\;\sf Area_{\;(quadrant)} = \dfrac{ \pi r^2}{4}$

$\;\;\bullet\;\;\sf Area_{\;(sector)} = \dfrac{ \theta}{360^\circ} \times \pi r^2$

$\;\;\bullet\;\;\sf Peri - meter_{\;(sector)} = \dfrac{ \theta}{360^\circ} \times 2 \pi r + 2r$

$\;\;\bullet\;\;\sf Area_{\;(segment)} = \dfrac{ \theta}{360^\circ} \pi r^2 - \dfrac{1}{2} r^2 sin \theta$

$\;\;\bullet\;\;\sf Peri - meter_{\;(segment)} = \dfrac{\pi r \theta}{180^\circ} + 2r \dfrac{ \theta}{2}$