(2)     $\frac{77}{8}c{m}^2$

(4)

Since, radius of wheel $\left(r\right)=0.25m$

Total distance covered by a circular wheel = 11 km = 11000 m

$\Rightarrow \text{No. of revolutions}\times 2\pi r=11000$

$\Rightarrow \text{No. of revolutions}=\frac{11000\times 7}{2\times 22\times 0.25}=7000$

(2)      21 cm

(1)      $9\pi$ ${\mathrm{cm}}^2$

(2)

According to the question,

$$\begin{gathered} \pi R_1^2+\pi R_2^2=\pi R^2 \\ \Rightarrow \pi \left(R_1^2+R_2^2\right)=\pi R^2 \\ \Rightarrow R_1^2+R_2^2=R^2 \end{gathered}$$

(4)

Let a circle with centre at O, having radius r, is inscribed in a square of side 6 cm as shown in Fig.

$$\begin{gathered} Diameter of the circle=6cm\Rightarrow 2r=6\Rightarrow r=3 \mathrm{cm} \\ \mathrm{Area} \mathrm{of} \mathrm{circle}={\mathrm{\pi r}}^2=\mathrm{\pi }\times (3{)}^2=9\mathrm{\pi }{ \mathrm{cm}}^2 \end{gathered}$$

(2)

Let r be the radius of the circle and x be the length of each side of the square.

According to the question, Perimeter of circle = Perimeter of square

$$\begin{gathered} 2\pi r=4x\Rightarrow \pi r=2x\Rightarrow r=\frac{2x}{\pi } \left(i\right) \\ \Rightarrow \frac{\mathrm{Area} \mathrm{of} \mathrm{circle}}{\mathrm{Area} \mathrm{of} \mathrm{square}}=\frac{\pi r^2}{x^2}=\frac{\pi \times \frac{4x^2}{{\pi }^2}}{x^2}=\frac{4}{\pi } \\ \Rightarrow \frac{\mathrm{Area} \mathrm{of} \mathrm{circle}}{\mathrm{Area} \mathrm{of} \mathrm{square}}=\frac{4}{\frac{22}{7}}=\frac{4}{\frac{22}{7}}\times 7=\frac{28}{22}=\frac{14}{11} \\ Required ratio is 14:11 \end{gathered}$$

Let r be the radius of the circle. Then,

$$\begin{gathered} \pi r^2=64\pi \Rightarrow r^2=64\Rightarrow r=8 \mathrm{cm} \\ \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle} \\ =2\mathrm{\pi r}=2\mathrm{\pi }\times 8 \mathrm{cm}=16\mathrm{\pi } \mathrm{cm} \end{gathered}$$

(1)

Let the radii of two circles be $r_1 and r_2$. It is given that

$$\begin{gathered} \pi r_1^2:\pi r_2^2=4:9 \\ \Rightarrow r_1^2:r_2^2=4:9\Rightarrow r_1:r_2=2:3 \\ Let P_1 and P_2 be the perimeters of the two semi-circles.Then, \\ {P}_1=\pi r_1+2r_1 \\ {P}_2=\pi r_2+2r_2 \\ \Rightarrow P_1=(\pi +2)r_1\mathrm{and}{P}_2=(\pi +2)r_2 \\ \Rightarrow P_1:P_2=r_1:r_2=2:3 \end{gathered}$$

(1)

$$\begin{gathered} Area ofcircle ofradius R=\pi R^2 \\ \therefore Statement\left(i\right) is correct. \\ Circumference of circle=2\pi R \\ \therefore Statement \left(ii\right) is incorrect. \\ Area of circle=\pi R^2 \\ \frac{\mathrm{Area} \mathrm{of} \mathrm{circle}}{\mathrm{Circumference} \mathrm{of} \mathrm{circle}}=\frac{\pi R^2}{2\pi R}=\frac{R}{2} \\ \therefore Statement \left(iii\right) is correct \\ Circumference of circle=2\pi R \\ Let’stakeR=\frac{1}{2}, 2, 3 \\ WhenR=\frac{1}{2} \\ Area=\frac{\pi }{4} \\ Perimeter=\pi \\ \therefore Area < Perimeter \\ When R=2 \\ Area=4\pi Perimeter=4\pi \\ \therefore Area = Perimeter When R=3 \\ Area=9\pi Perimeter=6\pi \therefore Area>Perimeter \\ So, relation between area and perimeter of circle depends upon its radius. \\ Statement \left(iv\right) is incorrect. \end{gathered}$$

(3)

Let two circles touch internally (see Fig.)

∴ Statement (iii) is correct.Let two circles touch externally (see Fig. 12.7).From the figure,AC=AB+BC=r1+r2=distance between centresStatement (iv) is incorrect." />

$$\begin{gathered} From the figure, AC=Rand BC=r \\ AB=AC-BC=R-r \therefore Statement \left(i\right) is correct. \\ Dis\tan ce moved by a rotating wheel in one revolution \\ =2\pi R=\mathrm{circumference} \mathrm{of} \mathrm{circle} (\mathrm{see} \mathrm{Fig}. 12.5) \\ \therefore Statement \left(ii\right) is incorrect. \\ From Fig. 12.6, area enclosed by two concentric circles of radius Rand r(R>r): \\ =\pi R^2-\pi r^2=\pi (R^2-r^2) \end{gathered}$$

$$\begin{gathered} \therefore Statement \left(iii\right) is correct. \\ Let two circles touch externally (see Fig. 12.7). \\ From the figure, \\ AC=AB+BC=r_1+r_2=\mathrm{distance} \mathrm{between} \mathrm{centres} \\ \mathrm{Statement} \left(\mathrm{iv}\right) \mathrm{is} \mathrm{incorrect}. \end{gathered}$$

(3)

Let OAB be the given sector as shown in the figure.

Length of the corresponding arc, 

$$\begin{gathered} AB=\frac{60°}{360°}\times 2\pi r=\frac{1}{6}\times 2\times \frac{22}{7}\times 21=22 \mathrm{cm} \\ \Rightarrow \mathrm{AB}=22 \mathrm{cm} \\ \therefore \mathrm{Perimeter} \mathrm{of} \mathrm{the} \mathrm{sector}=\mathrm{OA}+\mathrm{AB}+\mathrm{BO}=21+22+21=64 \mathrm{cm} \end{gathered}$$

(1)

$$\begin{gathered} We have r=14 \mathrm{cm} and \theta =90° \\ \therefore \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{arc}=\frac{\theta }{360°}\times 2\pi r \\ =\frac{90°}{360°}\times 2\times \frac{22}{7}\times 14=\frac{1}{4}\times 4\times 22=22 \mathrm{cm} \end{gathered}$$

(4)

$$\begin{gathered} \Delta AOB is right-angled at O \\ Length of chord AB: \\ AB=\sqrt{A{O}^2+B{O}^2}=\sqrt{r^2+r^2}=\sqrt{2}r=\sqrt{2}\times \frac{5}{\sqrt{2}}=5 \mathrm{cm} \\ \left(\mathrm{iii}\right) \mathrm{is} \mathrm{correct}. \\ \mathrm{Area} \mathrm{o} \mathrm{f}\mathbf{\Delta }\mathrm{AOB}: \\ \mathrm{Area}=\frac{1}{2}\times \mathrm{AO}\times \mathrm{BO}=\frac{1}{2}{\mathrm{r}}^2=\frac{1}{2}{\left(\frac{5}{\sqrt{2}}\right)}^2=\frac{25}{4} {\mathrm{cm}}^2 \\ \mathrm{Thus}\left(\mathrm{ii}\right) \mathrm{is} \mathrm{incorrect}(\mathrm{because} \mathrm{statement}(\mathrm{ii})\mathrm{says}\frac{25}{2},\mathrm{not}\frac{25}{4}). \\ \mathrm{Length} \mathrm{of} \mathrm{arc} \mathrm{ACB}: \\ \mathrm{Arc} \mathrm{length}=\frac{\mathrm{\theta }}{360°}\times \mathrm{circumference} \\ =\frac{90°}{360°}\times 2\mathrm{\pi r}=\frac{1}{4}\times 2\mathrm{\pi }\times \frac{5}{\sqrt{2}}=\frac{5\sqrt{2}\mathrm{\pi }}{4} \mathrm{cm} \\ \mathrm{But} \mathrm{statement}\left(\mathrm{iv}\right) \mathrm{says}\frac{3\sqrt{2}\mathrm{\pi }}{4},\mathrm{so}\left(\mathrm{iv}\right) \mathrm{is} \mathrm{incorrect}. \\ \mathrm{Finding} \mathrm{areas} \mathrm{of} \mathrm{segments} \\ \mathrm{Let} {\mathrm{A}}_1=\mathrm{area} \mathrm{of} \mathrm{minor} \mathrm{segment} \mathrm{ACB} \\ \mathrm{Let} {\mathrm{A}}_2=\mathrm{area} \mathrm{of} \mathrm{major} \mathrm{segment} \mathrm{ADB} \\ \mathrm{Area} \mathrm{of} \mathrm{a} \mathrm{segment}: \frac{\mathrm{\theta }}{360°}{\mathrm{\pi r}}^2-\frac{1}{2}{\mathrm{r}}^2 \\ \mathrm{For\theta }=90°: \\ {\mathrm{A}}_1=\left[\frac{\mathrm{\pi }}{360}\times 90-\frac{1}{2}\right]{\mathrm{r}}^2=\left[\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right]{\left(\frac{5}{\sqrt{2}}\right)}^2 \\ {\mathrm{A}}_1=\left(\frac{25\mathrm{\pi }}{8}-\frac{25}{4}\right){\mathrm{cm}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{entire} \mathrm{circle}: \\ {\mathrm{\pi r}}^2=\mathrm{\pi }\times {\left(\frac{5}{\sqrt{2}}\right)}^2=\frac{25\mathrm{\pi }}{2} \\ {\mathrm{A}}_2=\frac{25\mathrm{\pi }}{2}-{\mathrm{A}}_1 \\ {\mathrm{A}}_2=\frac{25\mathrm{\pi }}{2}-\left(\frac{25\mathrm{\pi }}{8}-\frac{25}{4}\right)=\frac{75\mathrm{\pi }}{8}+\frac{25}{4} \\ \mathrm{Required} \mathrm{difference}: \\ {\mathrm{A}}_2-{\mathrm{A}}_1=\left(\frac{75\mathrm{\pi }}{8}+\frac{25}{4}\right)-\left(\frac{25\mathrm{\pi }}{8}-\frac{25}{4}\right) \\ =\frac{25}{4}\left(\mathrm{\pi }+2\right) {\mathrm{cm}}^2 \\ \therefore \left(\mathrm{i}\right) \mathrm{is} \mathrm{correct} \end{gathered}$$

(2)

$$\begin{gathered} Area of sector =\frac{\theta }{360}\times \mathrm{area} \mathrm{of} \mathrm{circle} \\ \mathrm{Area} \mathrm{of} \mathrm{sector} \mathrm{at} \mathrm{P} = \frac{\angle \mathrm{P}}{360}\times {\mathrm{\pi r}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{sector} \mathrm{at} \mathrm{Q} = \frac{\angle \mathrm{Q}}{360}\times {\mathrm{\pi r}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{sector} \mathrm{at} \mathrm{R} =\frac{\angle \mathrm{R}}{360}\times {\mathrm{\pi r}}^2 \\ \mathrm{Total} \mathrm{area} \mathrm{of} \mathrm{shaded} \mathrm{region} =\left(\frac{\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}}{360}\right)\times {\mathrm{\pi r}}^2 \\ \mathrm{We} \mathrm{know} \\ (\mathrm{Sum} \mathrm{of} \mathrm{interior} \mathrm{angles} \mathrm{of} \mathrm{triangle}) \mathrm{Area} \mathrm{of} \mathrm{shaded} \mathrm{region} \\ =\frac{180}{360}{\mathrm{\pi r}}^2=\frac{1}{2}{\mathrm{\pi r}}^2=\frac{1}{2}\times \frac{22}{7}\times 14\times 14=308 {\mathrm{cm}}^2 \\ \therefore \left(\mathrm{ii}\right) \mathrm{is} \mathrm{correct}. \\ \mathrm{Hence}, \mathrm{we} \mathrm{don}’\mathrm{t} \mathrm{need} \mathrm{angles} \mathrm{to} \mathrm{find} \mathrm{area} \mathrm{of} \mathrm{shaded} \mathrm{region}, \mathrm{so} \left(\mathrm{i}\right) \mathrm{is} \mathrm{incorrect}. \\ \mathrm{Arc} \mathrm{length} = \mathrm{Radius} \times \mathrm{Angle} \mathrm{in} \mathrm{radians} \therefore \left(\mathrm{iv}\right) \mathrm{is} \mathrm{correct}. \\ \mathrm{But} \mathrm{we} \mathrm{need} \mathrm{angles} \mathrm{to} \mathrm{find} \mathrm{respective} \mathrm{arc} \mathrm{length} \mathrm{and} \mathrm{areas} \mathrm{of} \mathrm{respective} \mathrm{sectors}. \therefore \left(\mathrm{iii}\right) \mathrm{is} \mathrm{correct} \\ \mathrm{Hence} \left(\mathrm{ii}\right), \left(\mathrm{iii}\right) \mathrm{and} \left(\mathrm{iv}\right) \mathrm{are} \mathrm{correct}. \end{gathered}$$

(1)

$$\begin{gathered} SinceAB\parallel CD, \\ \angle BAD+\angle ADC=180° \\ (Co-interior angles) \\ \Rightarrow \angle BAD+60°=180° \\ \Rightarrow \angle BAD=120° \\ \Rightarrow \theta =120° \mathrm{and} r=3 \mathrm{cm} \\ \therefore \mathrm{Area} \mathrm{of} \mathrm{shaded} \mathrm{region}=\frac{\mathrm{\theta }}{360°}\times {\mathrm{\pi r}}^2 \\ =\frac{120°}{360°}\times \mathrm{\pi }\times (3{)}^2=\frac{1}{3}\times 9\mathrm{\pi }=3\mathrm{\pi } {\mathrm{cm}}^2 \end{gathered}$$

(2)

We have length of the square board = 2a

⇒ Diameter of circle = 2a

⇒ Radius of circle = a

$\therefore Area of square=(2a{)}^2=4a^2$

$And,area of circle inscribed=\pi a^2$

Area of shaded region = Area of square − Area of circle

$=4a^2-\pi a^2=(4-\pi )a^2$

(1)

AGE and FEB are quarter circles with radius $\frac{14}{2}=7 \mathrm{cm}$

Therefore, Area of quarter circles AGE + Area of quarter circle FEB

$$\begin{gathered} =\frac{\pi (7{)}^2}{4}+\frac{\pi (7{)}^2}{4}=\frac{\pi (7{)}^2}{2} \\ =\frac{22}{7}\times \frac{(7{)}^2}{2}=77 c{m}^2 \therefore \left(ii\right) \mathrm{is} \mathrm{correct} \\ \mathrm{Area} \mathrm{of} \mathrm{upper} \mathrm{semi}-\mathrm{circle} (\mathrm{diameter} \mathrm{DC}) \\ =\frac{1}{2}\times \frac{\mathrm{\pi }(\mathrm{DC}{)}^2}{4} \\ =\frac{22}{7}\times \frac{14\times 14}{4}\times \frac{1}{2}=77 {\mathrm{cm}}^2 \\ \mathrm{Area} \mathrm{of} \mathrm{blue} \mathrm{shaded} \mathrm{region} \\ =\mathrm{Area} \mathrm{of} \mathrm{rectangle} \mathrm{ABCD}-\mathrm{Area} \mathrm{of} \mathrm{region} (\mathrm{AEG} + \mathrm{FEB})-\mathrm{Area} \mathrm{of} \mathrm{semicircle} \mathrm{with} \mathrm{diameter} \mathrm{DC} \\ =(50\times 14)-77-77 \\ =546 \mathrm{cm}2 \\ \therefore \left(\mathrm{i}\right) \mathrm{is} \mathrm{correct} \end{gathered}$$

(1)

$$\begin{gathered} Since∆ABC is isosceles \\ AB=BC=21 \mathrm{cm} \\ \mathrm{Area} \mathrm{of} ∆\mathrm{ABC}: \\ =\frac{1}{2}\times \mathrm{AB}\times \mathrm{BC}=\frac{1}{2}\times (21{)}^2=220.5 {\mathrm{cm}}^2 \\ \therefore \mathrm{Statement} \left(\mathrm{iv}\right) \mathrm{is} \mathrm{correct}. \\ ∆\mathrm{ABC} \mathrm{is} \mathrm{isosceles} \mathrm{with}\angle \mathrm{B}=90° \\ \mathrm{So}, \mathrm{the} \mathrm{other} \mathrm{angles} \mathrm{each} \mathrm{measure} 45°. \\ \Rightarrow \mathrm{Angle} \mathrm{of} \mathrm{sector} \mathrm{CBD}=45° \\ \therefore \mathrm{Statement} \left(\mathrm{iii}\right) \mathrm{is} \mathrm{correct}. \\ \mathrm{Area} \mathrm{of} \mathrm{sector} \mathrm{CBD} \\ =\frac{\mathrm{\theta }}{360°}\times {\mathrm{\pi r}}^2 \\ =\frac{45}{360}\times 3.14\times 21\times 21 \\ =173.0925 {\mathrm{cm}}^2 \\ \therefore \mathrm{Statement} \left(\mathrm{ii}\right) \mathrm{is} \mathrm{incorrect}. \\ \mathrm{Area} \mathrm{of} \mathrm{triangle} \mathrm{outside} \mathrm{circular} \mathrm{region} \\ =\mathrm{Area} \mathrm{of} ∆\mathrm{ABC}-\mathrm{Area} \mathrm{of} \mathrm{sector} \mathrm{CBD} \\ =220.5-173.0925=47.4075 {\mathrm{cm}}^2 \\ \therefore \mathrm{Statement} \left(\mathrm{i}\right) \mathrm{is} \mathrm{correct}. \end{gathered}$$

(4)

Length of rectangle (AC), l = sum of diameter of all inner circles 

$$\begin{gathered} =3\times 20=60 \mathrm{cm} \\ \mathrm{Width} \mathrm{of} \mathrm{rectangle} \mathrm{BC}, \\ \mathrm{b}=\mathrm{diameter} \mathrm{of} \mathrm{smaller} \mathrm{circle}=20 \mathrm{cm} \\ \mathrm{Radius} \mathrm{of} \mathrm{bigger} \mathrm{circle} \left(\mathrm{OB}\right), \\ \mathrm{Diagonal} \mathrm{of} \mathrm{Rectangle} \\ \mathrm{R}=\frac{\sqrt{{\mathrm{AC}}^2+{\mathrm{BC}}^2}}{2}=\frac{\sqrt{(60{)}^2+(20{)}^2}}{2}=\frac{\sqrt{4000}}{2}=\frac{\sqrt{1000}}{1} \\ \mathrm{Area} \mathrm{of} \mathrm{rectangle} =\mathrm{l}\times \mathrm{b}=60\times 20=1200 {\mathrm{cm}}^2 \\ \therefore \mathrm{Statement} \left(\mathrm{iii}\right) \mathrm{is} \mathrm{correct}. \\ =2(\mathrm{l}+\mathrm{b})= \\ 2(60+20)=160 \mathrm{cm} \\ \therefore \mathrm{Statement} \left(\mathrm{iv}\right) \mathrm{is} \mathrm{incorrect}. \\ \mathrm{Area} \mathrm{of} \mathrm{bigger} \mathrm{circle} ={\mathrm{\pi R}}^2=\mathrm{\pi }\times (\sqrt{1000}{)}^2=1000\mathrm{\pi } {\mathrm{cm}}^2 \\ \therefore \mathrm{Statement} (\mathrm{i}) \mathrm{is} \mathrm{correct}. \\ \mathrm{Option} (\mathrm{d}) \mathrm{is} \mathrm{correct}. \end{gathered}$$