Question

find the area of sector of circle of radius 7cm and central angle 60​

Answer 1

Answer:

sinx

The integral

(d) nonesinx

The integral

Soa

dx

converges

(a) absolutely

(b) uniformly

(c) conditionally

(d) none

Answer 2

${\large{\underbrace{ \underline{\sf{Understanding \: the \: Question}}}}}$

Here in this question we have to find area of sector of circle with given radius and central angle as 7cm and 60° respectively. We can find the area of sector by just applying the formula for area of sector.

So let's start!

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${ \star} \boxed{ \sf{Area\: of\: sector=\dfrac{\theta}{360 \degree}\pi {r}^{2}}}$

$\sf{\implies Area\: of\: sector=\dfrac{60 \degree}{360 \degree} \times \dfrac{22}{7} \times {7}^{2}}$

$\sf{\implies Area\: of\: sector=\dfrac{60 \degree}{360 \degree} \times \dfrac{22}{7} \times 49}$

$\sf{\implies Area\: of\: sector= \cancel{\dfrac{60 \degree}{360 \degree} \times \dfrac{22}{7} \times 49}}$

$\sf{ \implies Area\: of\: sector=25.66}$

Hence the required area of sector is 25.66cm².

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ADDITIONAL INFORMATION:

What is the difference between minor sector and major sector?

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

FORMULAE RELATED TO AREA:

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

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NOTE-Kindly see this answer on web for better understanding.