Question

i) In the given figure , area of the sector is A cm2 and perimeter of the sector
is 50 cm. Prove that
a)
$theta \: = \frac{360}{\pi} ( \frac{25}{r} - 1)$
b)
$a = 25r - {r}^{2}$
​

Answer 1

Answer:

radius of circle = r cm

angle subtended at centre = ∅

a )      perimeter of sector = radius + radius + arc

                         50           = r + r + $\frac{theta}{360}$2$\pi$r

                              50      = 2r + 2r( $\frac{theta}{360}$ $\pi$)

                               50     = 2r ( 1 + $\frac{theta}{360}$ $\pi$ )

                            $\frac{50}{2r}$     =  1 + $\frac{theta}{360}$ $\pi$

                           $\frac{25}{r}$     =   1 + $\frac{theta}{360}$ $\pi$

                       $\frac{25}{r} -1$   =   $\frac{theta}{360}$ $\pi$

                      theta         =   $\frac{360}{\pi }$ ( $\frac{25}{r} -1$)                                           ........(1)

b)    area of sector =   $\frac{theta}{360}$ $\pi r^{2}$

           A     =  $\frac{\frac{360 (\frac{25}{r} -1)}{\pi }}{360}$ ×$\pi$ $r^{2}$                [(putting the value of theta from eq. ..(1)]    

           A    = $\frac{25}{r}$ × $r^{2}$ $-r^{2}$

           A   =  25 r $-r^{2}$

please mark it as brainliest answer.