Solution

Let $OACB$ be a sector of the circle making $60^{\circ}$ angle at centre $O$ of the circle.

Area of sector of angle $\theta=\dfrac{\theta}{360^{\circ}} \times \pi r^{2}$

Area of sector $OACB=\dfrac{60^{\circ}}{360^{\circ}} \times \dfrac{22}{7} \times(6)^{2}$

$=\dfrac{1}{6} \times \dfrac{22}{7} \times 6 \times 6=\dfrac{132}{7} cm^{2}$

Solution

Let the radius of the circle be $r$.

Circumference $=22 cm$

$2 \pi r=22$

$r=\dfrac{22}{2 \pi}=\dfrac{11}{\pi}$

Quadrant of circle will subtend $90^{\circ}$ angle at the centre of the circle.

Area of such quadrant of the circle $=\dfrac{90^{\circ}}{360^{\circ}} \times \pi \times r^{2}$

$=\dfrac{1}{4 \pi} \times \pi \times(\dfrac{11}{\pi})^{2}$

$=\dfrac{121}{4 \pi}=\dfrac{121 \times 7}{4 \times 22}$

$=\dfrac{77}{8} cm^{2}$

Solution

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates $360^{\circ}$.

In 5 minutes, minute hand will rotate $=\dfrac{360^{\circ}}{60} \times 5=30^{\circ}$

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of $30^{\circ}$ in a circle of $14 cm$ radius.

Area of sector of angle $\theta=\dfrac{\theta}{360^{\circ}} \times \pi r^{2}$

Area of sector of $30^{\circ}=\dfrac{30^{\circ}}{360^{\circ}} \times \dfrac{22}{7} \times 14 \times 14$

$=\dfrac{22}{12} \times 2 \times 14$

$=\dfrac{11 \times 14}{3}$

$=\dfrac{154}{3} cm^{2}$

Therefore, the area swept by the minute hand in 5 minutes is $\dfrac{154}{3} cm^{2}$.

Solution

Let $A B$ be the chord of the circle subtending $90^{\circ}$ angle at centre $O$ of the circle.

Area of major sector $OADB=(\dfrac{360^{\circ}-90^{\circ}}{360^{\circ}}) \times \pi r^{2}=(\dfrac{270^{\circ}}{360^{\circ}}) \pi r^{2}$ $=\dfrac{3}{4} \times 3.14 \times 10 \times 10$

$=235.5 cm^{2}$

Area of minor sector $OACB=\dfrac{90^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\dfrac{1}{4} \times 3.14 \times 10 \times 10$

$=78.5 cm^{2}$

Area of $\triangle OAB=\dfrac{1}{2} \times OA \times OB=\dfrac{1}{2} \times 10 \times 10$

$=50 cm^{2}$

Area of minor segment $A C B=$ Area of minor sector $OACB$ -

Area of $\triangle OAB=78.5-50=28.5 cm^{2}$

Solution

Radius $(r)$ of circle $=21 cm$

Angle subtended by the given arc $=60^{\circ}$

Length of an arc of a sector of angle $\theta=\dfrac{\theta}{360^{\circ}} \times 2 \pi r$

Length of $arc A C B=\dfrac{60^{\circ}}{360^{\circ}} \times 2 \times \dfrac{22}{7} \times 21$

$=\dfrac{1}{6} \times 2 \times 22 \times 3$

$=22 cm$

Area of sector $OACB=\dfrac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\dfrac{1}{6} \times \dfrac{22}{7} \times 21 \times 21$

$=231 cm^{2}$

In $\triangle OAB$,

$\angle OAB=\angle OBA(As OA=OB)$

$\angle OAB+\angle AOB+\angle OBA=180^{\circ}$

$2 \angle OAB+60^{\circ}=180^{\circ}$

$\angle OAB=60^{\circ}$

Therefore, $\triangle OAB$ is an equilateral triangle.

Area of $\triangle OAB=\dfrac{\sqrt{3}}{4} \times(\text{ Side })^{2}$

$=\dfrac{\sqrt{3}}{4} \times(21)^{2}=\dfrac{441 \sqrt{3}}{4} cm^{2}$

Area of segment $A C B=$ Area of sector $O A C B$ - Area of $\triangle O A B$

$=(231-\dfrac{441 \sqrt{3}}{4}) cm^{2}$

Solution

Radius ( $r$ ) of circle $=15 cm$

Area of sector OPRQ $=\dfrac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\dfrac{1}{6} \times 3.14 \times(15)^{2}$

$=117.75 cm^{2}$

In $\triangle OPQ$,

$\angle OPQ=\angle OQP($ As OP $=OQ)$

$\angle OPQ+\angle OQP+\angle POQ=180^{\circ}$

$2 \angle OPQ=120^{\circ}$

$\angle OPQ=60^{\circ}$

$\triangle OPQ$ is an equilateral triangle.

Area of $\triangle OPQ=\dfrac{\sqrt{3}}{4} \times(\text{ side })^{2}$

$=\dfrac{\sqrt{3}}{4} \times(15)^{2}=\dfrac{225 \sqrt{3}}{4} cm^{2}$

$=56.25 \sqrt{3}$

$=97.3125 cm^{2}$

Area of segment PRQ = Area of sector OPRQ - Area of $\triangle OPQ$

$=117.75$ - 97.3125

$=20.4375 cm^{2}$

Area of major segment PSQ = Area of circle - Area of segment PRQ

$ \begin{aligned} & =\pi(15)^{2}-20.4375 \\ & =3.14 \times 225-20.4375 \\ & =706.5-20.4375 \\ & =\quad 686.0625 cm^{2} \end{aligned} $

Solution

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.

$SV=VT$

In $\triangle OVS$,

$ \begin{aligned} & \dfrac{O V}{O S}=\cos 60^{\circ} \\ & \dfrac{O V}{12}=\dfrac{1}{2} \\ & O V=6 cm \end{aligned} $

$\dfrac{SV}{SO}=\sin 60^{\circ}=\dfrac{\sqrt{3}}{2}$

$\dfrac{SV}{12}=\dfrac{\sqrt{3}}{2}$

$SV=6 \sqrt{3} cm$

$ST=2 SV=2 \times 6 \sqrt{3}=12 \sqrt{3} cm$

Area of $\triangle OST=\dfrac{1}{2} \times ST \times OV$

$=\dfrac{1}{2} \times 12 \sqrt{3} \times 6$

$=36 \sqrt{3}=36 \times 1.73=62.28 cm^{2}$

Area of sector OSUT $=\dfrac{120^{\circ}}{360^{\circ}} \times \pi(12)^{2}$

$=\dfrac{1}{3} \times 3.14 \times 144=150.72 cm^{2}$

Area of segment SUT = Area of sector OSUT - Area of $\triangle OST$

$=150.72-62.28$

$=88.44 cm^{2}$

Solution

From the figure, it can be observed that the horse can graze a sector of $90^{\circ}$ in a circle of $5 m$ radius.

Area that can be grazed by horse $=$ Area of sector $OACB$

$=\dfrac{90^{\circ}}{360^{\circ}} \pi r^{2}$

$=\dfrac{1}{4} \times 3.14 \times(5)^{2}$

$=19.625 m^{2}$

Area that can be grazed by the horse when length of rope is $10 m$ long $=\dfrac{90^{\circ}}{360^{\circ}} \times \pi \times(10)^{2}$

$=\dfrac{1}{4} \times 3.14 \times 100$

$=78.5 m^{2}$

Increase in grazing area $=(78.5-19.625) m^{2}$

$=58.875 m^{2}$

Solution

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Radius of circle $=\dfrac{35}{2} mm$

Circumference of brooch $=2 \pi r$

$=2 \times \dfrac{22}{7} \times(\dfrac{35}{2})$

$=110 mm$

Length of wire required $=110+5 \times 35$

$=110+175=285 mm$

It can be observed from the figure that each of 10 sectors of the circle is subtending $36^{\circ}$ at the centre of the circle.

Therefore, area of each sector $=\dfrac{36^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\dfrac{1}{10} \times \dfrac{22}{7} \times(\dfrac{35}{2}) \times(\dfrac{35}{2})$

$=\dfrac{385}{4} mm^{2}$

Solution

There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending $\dfrac{360^{\circ}}{8}=45^{\circ}$ at the centre of the assumed flat circle.

Area between two consecutive ribs of circle $=\dfrac{45^{\circ}}{360^{\circ}} \times \pi r^{2}$

$ \begin{aligned} & =\dfrac{1}{8} \times \dfrac{22}{7} \times(45)^{2} \\ & =\dfrac{11}{28} \times 2025=\dfrac{22275}{28} cm^{2} \end{aligned} $

Solution

It can be observed from the figure that each blade of wiper will sweep an area of a sector of $115^{\circ}$ in a circle of $25 cm$ radius.

Area of such sector $=\dfrac{115^{\circ}}{360^{\circ}} \times \pi \times(25)^{2}$

$=\dfrac{23}{72} \times \dfrac{22}{7} \times 25 \times 25$

$=\dfrac{158125}{252} cm^{2}$

Area swept by 2 blades $=2 \times \dfrac{158125}{252}$

$=\dfrac{158125}{126} cm^{2}$

Solution

It can be observed from the figure that the lighthouse spreads light across a

sector of $80^{\circ}$ in a circle of $16.5 km$ radius.

Area of sector $OACB=\dfrac{80^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\dfrac{2}{9} \times 3.14 \times 16.5 \times 16.5$

$=189.97 km^{2}$

Solution

It can be observed that these designs are segments of the circle.

Consider segment APB. Chord AB is a side of the hexagon. Each chord will substitute $\dfrac{360^{\circ}}{6}=60^{\circ}$ at the centre of the circle.

In $\triangle OAB$,

$\angle OAB=\angle OBA(As OA=OB)$

$\angle A O B=60^{\circ}$

$\angle OAB+\angle OBA+\angle AOB=180^{\circ}$

$2 \angle OAB=180^{\circ}-60^{\circ}=120^{\circ}$

$\angle OAB=60^{\circ}$

Therefore, $\triangle OAB$ is an equilateral triangle.

Area of $\triangle OAB=\dfrac{\sqrt{3}}{4} \times(\text{ side })^{2}$

$\begin{aligned}=\dfrac{\sqrt{3}}{4} \times(28)^{2}=196 \sqrt{3} & =196 \times 1.7 \\ & =333.2 cm^{2}\end{aligned}$

Area of sector OAPB $=\dfrac{60^{\circ}}{360^{\circ}} \times \pi r^{2}$

$=\dfrac{1}{6} \times \dfrac{22}{7} \times 28 \times 28$

$=\dfrac{1232}{3} cm^{2}$

Area of segment APB = Area of sector OAPB - Area of $\triangle O A B$

$=(\dfrac{1232}{3}-333.2) cm^{2}$

Therefore, area of designs $=6 \times(\dfrac{1232}{3}-333.2) cm^{2}$

$ \begin{aligned} & =(2464-1999.2) cm^{2} \\ & =464.8 cm^{2} \end{aligned} $

Cost of making $1 cm^{2}$ designs $=$ Rs 0.35

Cost of making $464.76 cm^{2}$ designs $=464.8 \times 0.35=$ Rs 162.68

Therefore, the cost of making such designs is Rs 162.68 .

Solution

We know that area of sector of angle $\theta=\dfrac{\theta}{360^{\circ}} \times \pi R^{2}$

Area of sector of angle $P=\dfrac{p}{360^{\circ}}(\pi R^{2})$

$=(\dfrac{p}{720^{\circ}})(2 \pi R^{2})$

Hence, (D) is the correct answer.