Area Related to Circles Test - 23

$$ Let\quad the\quad given\quad perimeter\quad be\quad P.\\ We\quad calculate\quad the\quad areas\quad of\quad the\quad given\quad figures\quad \\ and\quad compare.\\ Option\quad A\longrightarrow circle.\\ The\quad perimeter(circumference)=P\\ \Longrightarrow 2\pi r=P\quad (when\quad r=radius\quad of\quad the\quad circle.)\\ \Longrightarrow r=\frac { P }{ 2\pi  } \\ \therefore \quad ar.circle=\pi { r }^{ 2 }=\pi \times { \left( \frac { P }{ 2\pi  }  \right)  }^{ 2 }=\frac { { P }^{ 2 } }{ 4\times 3.14 } =\frac { { P }^{ 2 } }{ 12.56 } .\\ Option\quad B\longrightarrow Equilateral\quad triangle.\\ One\quad side=\frac { P }{ 3 } .\\ \therefore \quad area=\frac { \sqrt { 3 }  }{ 4 } { side }^{ 2 }=\frac { \sqrt { 3 }  }{ 4 } \times { \left( \frac { P }{ 3 }  \right)  }^{ 2 }=\frac { { P }^{ 2 } }{ 20.77 } .\\ Option\quad C\longrightarrow Square\\ side=\frac { P }{ 4 } \\ \therefore \quad area=\frac { P }{ 4 } \times \frac { P }{ 4 } =\frac { { P }^{ 2 } }{ 16 } .\\ Option\quad D\longrightarrow Regular\quad pentagon.\\ side=\frac { P }{ 5 } .\\ \therefore \quad ar.pentagon=\frac { 1 }{ 4 } \sqrt { 5\left( 5+2\sqrt { 5 }  \right)  } \times { side }^{ 2 }\\ =\frac { 1 }{ 4 } \sqrt { 5\left( 5+2\sqrt { 5 }  \right)  } \times { \left( \frac { P }{ 5 }  \right)  }^{ 2 }=\frac { { P }^{ 2 } }{ 19.95 } .\\ \therefore \quad comparing\quad the\quad areas\quad we\quad get\\ the\quad ar.circle\quad is\quad the\quad greatest.\\ Ans-\quad Option\quad A\\  $$ 

$$\Rightarrow \displaystyle \frac {\theta}{360}\, \times\, \pi r^2\, =\, 9\, \displaystyle \frac {3}{7}$$
$$\Rightarrow \displaystyle \frac {120}{360}\, \times\, \displaystyle \frac {22}{7}\, \times\, r^2\, =\, \displaystyle \frac {66}{7}$$
$$\Rightarrow r^2\, =\, \displaystyle \frac {66}{7}\, \times\, \displaystyle \frac {360}{120}\, \times\, \displaystyle \frac {7}{22}\, =\, 9$$
$$\Rightarrow r\, =\, \sqrt 9\, =\, 3$$ cm

Area of shaded region $$=$$ Area of trapezium $$-2$$ $$\times $$ Area of circle

Here $$r=12cm$$, $$\theta=60^o$$
Area of the corresponding segment  $$ABXA$$
$$=\displaystyle\left[\frac{\pi r^2\theta}{360}-\frac{1}{2}r^2\sin\theta\right]$$
$$=\displaystyle\left[\frac{3.14\times 12\times 12\times 60^o}{360^o}-\frac{1}{2}\times 12\times \sin 60^o\right]$$
$$=\displaystyle\left[\frac{314}{100}\times 12\times 2-6\times 12\times \displaystyle\frac{\sqrt 3}{2}\right]cm^2$$.    $$\displaystyle\left[\because \sin 60^o=\frac{\sqrt 3}{2}\right]$$
$$=\displaystyle\left[\frac{7536}{100}-36\times 1.73\right]cm^2$$
$$=[75.36-62.28]cm^2=13.08cm^2$$

Required area
= Area of sheet - 60 $$\displaystyle \times $$ area of 1 button
$$\displaystyle=\left ( 50\times 25-60\times \dfrac{22}{7}\times 1.4\times 1.4 \right )cm^{2}$$
$$\displaystyle =\left ( 1250-369.6 \right )cm^{2}$$
$$\displaystyle =880.4cm^{2}$$

We have,
$$OA=R=21m$$ and $$OC=r=14m$$
$$\therefore$$ Area of the flower bed
=Area of a quadrant of a circle of radius R - Area of a quadrant of a circle of radius r
$$=\displaystyle\frac{1}{4}\pi R^2-\frac{1}{4}\pi r^2$$
$$=\displaystyle\frac{\pi}{4}(R^2-r^2)$$
$$=\displaystyle\frac{1}{4}\times \frac{22}{7}(21^2-14^2)cm^2$$     $$[\therefore R=12m$$ and $$r=14m]$$
$$=\displaystyle\left(\frac{1}{4}\times \frac{22}{7}\times(21+14)(21-14)\right)m^2$$
$$=\displaystyle\left(\frac{1}{4}\times \frac{22}{7}\times 35\times 7\right)m^2=192.5m^2$$

Let ABCD be a square with side $$58$$ metres
Let its diagonals intersect at O

Then $$OA=OB=OC=OD$$ and

$$\displaystyle \angle AOD=\angle BOC=90^{\circ}$$

With radius$$= \displaystyle \dfrac{1}{2}AC\text{ arcs have been drawn namely } AED \ and \ CBF$$

$$\displaystyle \therefore \text{Radius of each arc}=\dfrac{1}{2}\times diagonal$$

$$\displaystyle =\dfrac{1}{2}\sqrt{2}\times 58$$

$$=1.41 \times 29 =40.89 metres.$$

$$\displaystyle 58\times 58+2\times \left [ \dfrac{22}{7}\times \left ( 40.89 \right )^{2}\times \dfrac{90}{360}-\dfrac{1}{2}\times \left ( 40.89 \right )^{2}90^{\circ} \right ]$$

$$\displaystyle 33.64+\dfrac{11\times \left ( 40.89 \right )^{2}}{7}-\left ( 40.89 \right )^{2}$$

$$=(3364+2627.4-1671.9)$$ sq metres

$$=4319.5$$ sq metres

Area of equilateral $$\Delta ABC=49\sqrt { 3 } { cm }^{ 2 }$$