Question 13 Marks
The circumference of a circle exceeds its diameter by $45\ cm.$ find the circumference of the circle. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerLet $r$ be the radius of the circle.
$⇒$ Diameter of a cirde$ = 2r$
And, circumference of a circle $=2\pi\text{r}$
It is given that,
Circumference of a circle $-$ Diameter of a circle $= 45\ cm$
$\Rightarrow2\pi\text{r}-2\text{r}=45$
$\Rightarrow2\text{r}(\pi-1)=45$
$\Rightarrow2\text{r}\Big(\frac{22}{7}-1\Big)=45$
$\Rightarrow\text{r}\Big(\frac{22-7}{7}\Big)=\frac{45}{2}$
$\Rightarrow\text{r}\times\frac{15}{7}=\frac{45}{2}$
$\Rightarrow\text{r}=\frac{45\times7}{15\times2}$
$\Rightarrow\text{r}=10.5\text{cm}$
$\therefore$ Circumference of a circle $=2\times\frac{22}{7}\times10.5=66\text{cm}$
View full question & answer→Question 23 Marks
Find the lengths of the arcs cut off from a circle of radius $12\ cm$ by a chord $12\ cm$ long. Also, find the area of the minor gment. $\big[\text{Take }\pi=3.14\text{ and }\sqrt{3}=1.73.\big]$
Answer$\triangle\text{OAB}$ is equilateral.
So, $\angle\text{AOB}=60$
Arc $\text{ACB}=\Big(2\pi\times12\times\frac{60}{360}\Big)\text{cm}$
$=4\pi\text{cm}$
$=(4\times3.14)\text{cm}$
$=12.56\text{cm}$
Length of arc $\text{BDA}=(2\pi12-\text{Arc}\text{ACB})\text{cm}$
$=24\pi-4\pi\text{cm}=(20\pi)\text{cm}$
$=(20\ 3.14)\text{cm}=62.8\text{cm}$
Area of the minor segment ACBA
$=\Big[\pi\times(12)^2\times\frac{60}{360}-\frac{\sqrt{3}}{4}\times(12)^2\Big]\text{cm}^2$
$=\Big(3.14\times12\times12\times\frac{60}{360}-\frac{1.73}{4}\times12\times12\Big)\text{cm}^2$
$=(75.36-62.28)\text{cm}^2=13.08\text{cm}^2$ View full question & answer→Question 33 Marks
In the given figure, find the area of the shaded region, if $ABCD$ is a square of side $14\ cm$ and $APO$ and $BPC$ are semicircles.
AnswerSide of a square $= 14\ cm$
$⇒$ Diameter of a semicircle $= 14\ cm$
$⇒$ Radius of a semicircle $= 7\ cm$
Now,
Area of a shaded region $=$ Area of a square $-$ Area of two semicircles
$=\Big[(14\times14)-\Big(\frac{22}{7}\times7\times7\Big)\Big]\text{cm}^2$
$=(196 - 154)\text{cm}^2$
$=42\text{cm}^2$
View full question & answer→Question 43 Marks
Find the area of the sector of a circle having radius $6\ cm$ and of angle $30^\circ . \big[\text{Take }\pi=3.14\big]$
AnswerRadius of a circle $= r = 6\ cm$
Central angle $=\theta=30^\circ$
$\therefore$ Area of the sector $=\frac{\pi\text{r}^2\theta}{360}$
$=\Big(\frac{3.14\times6\times6\times30^\circ}{360^\circ}\Big)\text{cm}^2$
$= 9.42\ cm^2$
View full question & answer→Question 53 Marks
The wheels of a car make $2500$ revolutions in covering a distance of $4.95\ km.$ Find the diameter of a wheel.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerDistance covered by the wheel in $1$ revolution
$=\Big(\frac{4.95\times1000\times100}{2500}\Big)\text{cm}=198\text{cm}$
$\therefore$ The circumference of the wheel $= 198\ cm$
Let the diameter of the wheel be $d\ cm$
Then, $\pi\text{d}=198\Rightarrow\frac{22}{7}\times\text{d}=198$
$\Rightarrow\text{d}=\frac{198\times7}{22}=63\text{cm}$
Hence diameter of the wheel is $63\ cm$
View full question & answer→Question 63 Marks
A chord $PQ$ of a circle of radius $10\ cm$ subtends an angle of $60^\circ $ at the centre of the circle. Find the area of major and minor segments of the circle. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerRadius of a circle $= r = 10.5\ cm$
Area of a sector $= 69.3\ cm^2$
Now, area of the sector $=\frac{\pi\text{r}^2\theta}{360}$
$\Rightarrow69.3=\frac{\frac{22}{7}\times10.5\times10.5\times\theta}{360^\circ}$
$\Rightarrow69.3=\frac{11\times1.5\times10.5\times\theta}{180}$
$\Rightarrow\theta=\frac{69.3\times180}{11\times1.5\times10.5}$
$\Rightarrow\theta=72^\circ$
View full question & answer→Question 73 Marks
In a circle of radius $7\ cm, $ a square $ABCD$ is inscribed. Find the area of the circle which is outside the square.
Answer
Radius of a circle $= 7\ cm$
$⇒ $ Diagonal of the square$ = 2 × 7 = 14\ cm$
Now,
Area of the square $=\frac{1}{2}\times(\text{diagonal})^2=\Big(\frac{1}{2}\times14\times14\Big)\text{cm}^2=98\text{cm}^2$
Area of the circle $=\Big(\frac{22}{7}\times7\times7\Big)\text{cm}^2=154\text{cm}^2$
$\therefore$ Required area $=$ Area of the circle $-$ Area of the square
$ =(154-98) \mathrm{cm}^2$
$ =56 \mathrm{~cm}^2$ View full question & answer→Question 83 Marks
Find the radius of a circle whose perimeterter and are are numerically equal.
Answerlet $r$ be the radius of a circle.
Then, area of a circle $=\pi\text{r}^2$
Perimeter of a circle $=2\pi\text{r}$
It is given that,
Area of a circle $=$ Perimeter of a circle
$\Rightarrow\pi\text{r}^2=2\pi\text{r}$
$\Rightarrow \text{r}= 2 \text{ units}$
View full question & answer→Question 93 Marks
A park is in the form of a rectangle $120m$ by $90m.$ At the centre of the park, there is a circular lawn as shown in the figure. The area of the park excluding the lawn is $2950m^2$ Find the radius of the circular lawn. $\big[\text{Given }\pi = 3.14.\big]$
Answer$ \text { Area of rectangle }=(120 \times 90) \mathrm{m}^2$
$=10800 \mathrm{~m}^2$
$\text { Area of circular lawn }=[\text { Area of rectangle - Area of park excluding circular lawn }]$
$=[10800-2950] \mathrm{m}^2=7850 \mathrm{~m}^2$
$\text { Area of circular lawn }=7850 \Rightarrow \pi \mathrm{r}^2=7850 \mathrm{~m}^2$
$3.14 \times \mathrm{r}^2=7850 \mathrm{~m}^2$
$\text{r}^2=\Big(\frac{7850}{3.14}\Big)\text{m}^2$
$=2500\text{m}^2$
$\text{r}=\sqrt{2500}\text{m}$
or $\text{r}=50\text{m}$
Hence, radius of the circular lawn $= 50m$
View full question & answer→Question 103 Marks
In the given figure, $PQSR$ represents a flower bed. If $OP = 21\ m$ and $OR = 14\ m,$ find the area of the flower bed.
Answer
Area of flower bed $= ($area of quadrant $OPQ) - ($area of the quadrant $ORS)$
$=\Big[\frac{1}{4}\pi\text{r}^2_1-\frac{1}{4}\pi\text{r}^2_2\Big]$
$=\Big[\frac{1}{4}\times\frac{22}{7}\times21\times21-\frac{1}{4}\times\frac{22}{7}\times14\times14\Big]\text{m}^2$
$=[346.5-154]\text{m}^2=192.5\text{m}^2$ View full question & answer→Question 113 Marks
A circular disc of radius $6\ cm$ is divided into three sectors with central angles $90^\circ , 120^\circ $ and $150^\circ .$ what part of the whole circle is the secto with central angle $150^\circ ?$ Also, calculate the ratio of the areas of the three sectors.
Answer$\frac{\text{Area of sector with} \ \theta \ = \ 150^\circ}{\text{Area of the circle}}=\frac{\pi\times(6)^2\times\frac{150}{360}}{\pi\times(6)^2}$
$=\frac{150}{360}=\frac{5}{12}$
Required ratio $=\Big(36\pi\times\frac{90}{360}\Big):\Big(36\pi\times\frac{120}{360}\Big):\Big(36\pi\times\frac{150}{360}\Big)$
$=\frac{1}{4}:\frac{1}{3}:\frac{5}{12}=3:4:5$
View full question & answer→Question 123 Marks
The length of a chain used as the bounary of a semicircular park is $108\ m$. Find the area of the park $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerThe length of a chain used as the bounary of a semicircular park is $108\ m.$ Find the area of the park.
Let $r$ be the radius of the semicircular park.
Now, perimeter of a semicircular park $= 108\ m$
$\Rightarrow\pi\text{r}+2\text{r}=108$
$\Rightarrow\Big(\frac{22}{7}+2\Big)\text{r}=108$
$\Rightarrow\Big(\frac{22+14}{7}\Big)\text{r}=108$
$\Rightarrow\frac{36}{7}\text{r}=108$
$\Rightarrow\text{r}=\frac{108\times7}{36}=21\text{cm}$
$\therefore$ Area of the park $=\frac{1}{2}\times\frac{22}{7}\times21\times21=693\text{m}^2$
View full question & answer→Question 133 Marks
Two circular pieces of equal radii and maximim area, touching each other are cut out from a rectangular cardboard of dimensions $14\ cm × 7\ cm.$ Find the area of the $r$ maining cardboard.
AnswerSince the dimensions of a rectangular cardboard are $14\ cm × 7\ cm,$
the diameter of each circle is $7\ cm,$
Now,
Area of the rectangular cardboard $= 14 × 7 = 98\ cm^2$
Area of two circle $=2\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}=77\text{cm}^2$
$\therefore$ Area of the reamaining cardboard
$=$ Area of the rectangular cardboard $-$ Area of two circles
$= (98 - 77)\ cm^2$
$= 21\ cm^2$
View full question & answer→Question 143 Marks
In a circle of radius $10.5\ cm,$ the minor arc is one-fifth of the major arc. find the area of the sector corresponding to the major are. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerLet the major arc be $x \ cm$ long
Then, length of the minor arc $=\frac{1}{5}\times\text{cm}$
Circumference $=\Big(\text{x}+\frac{1}{5}\text{x}\Big)\text{cm}=\frac{6\text{x}}{5}\text{cm}$
$\frac{6\text{x}}{5}=2\times\frac{22}{7}\times\frac{21}{2}\Rightarrow\text{x}=55\text{cm}$
Required area $=\Big(\frac{1}{2}\times55\times\frac{21}{2}\Big)\text{cm}^2$
$\Big[\text{Area}=\frac{1}{2}\text{rl}\Big]$
$=2988.75\text{cm}^2$
View full question & answer→Question 153 Marks
In the given figure, find the area of the shaded region, where $ABCD$ is a square of side $14\ cm$ and all circles are of the same diameter.
Answer
Side of the square $ABCD = 14\ cm$
Area of square $ABCD = 14 14 = 196\ cm^2$
Radius of each circle $=\frac{14}{4}=3.5\text{cm}$
Area of the circles $= 4$ area of one circle
$=4\times\pi(3.5)^2$
$=4\times\frac{22}{7}\times3.5\times3.5$
$=154\text{cm}^2$
Area of shaded region $=$ Area of square $-$ area of $4$ circles
$= 196 - 154 = 42\ cm^2$ View full question & answer→Question 163 Marks
If the area of a circle is numerically equal to twice its circumference then what is the diameter of the circle?
AnswerLet r be the radius of the circle.
It is given that,
Area of a circle $= 2 ×$ Circumference of a circle
$\Rightarrow\pi\text{r}^2=2\times2\pi\text{r}$
$⇒ r^2= 4r$
$⇒ r =4$
$⇒$ Diameter $= 2r = 2 × 4 = 8cm$
View full question & answer→Question 173 Marks
The sum of the radii of two circles is $7\ cm,$ and the differece of their circumferences is $8\ cm$. Find the circumferences of the circleas.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerLet the radii of cricles be $x\ cm$ and $(7 - x)\ cm$
Then,
$2\pi\text{x}-[2\pi(7-\text{x})]=8$
$2\pi\text{x}-[14\pi-2\pi\text{x}]=8$
$2\pi\text{x}-14\pi+2\pi\text{x}=8$
$4\pi\text{x}-14\pi=8$
$2\pi\text{x}=4+7\pi$
$2\pi\text{x}=4+22$
$2\pi\text{x}=26$
Substitute the value of $2\pi\text{x}$ in $2\pi(7-\text{x})$
$=14\pi-2\pi\text{x}=14\times\frac{22}{7}-26$
$=44-26=18\text{cm}$
Circumference of the circles are $26\ cm$ and $18\ cm$
View full question & answer→Question 183 Marks
A racetrack is in the form of a ring whose inner circumfarence is $352m$ and outer circumference is $396m$. Find the width and the area of the track. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerLet $r\ m$ and $R\ m$ be the radii of inner circle and outer boundaries respectively.
Then, $2\pi\text{r}=352$ and $2\pi\text{R}=396$
$\text{r}=\frac{352}{2\pi},\ \text{R}=\frac{396}{2\pi}$
Width of the track $= (R - r)m$
$=\Big(\frac{396}{2\pi}-\frac{352}{2\pi}\Big)\text{m}=\Big(\frac{44}{2\pi}\Big)\text{m}$
$=\Big(\frac{44}{2}\times\frac{7}{22}\Big)\text{m}=7\text{m}$
Area the track $=\pi(\text{R}^2-\text{r}^2)=\pi(\text{R}+\text{r})(\text{R}-\text{r})$
$=\Big[\pi\Big(\frac{352}{2\pi}+\frac{396}{2\pi}\Big)\times7\Big]\text{m}^2$
$=\Big[\Big(\pi\times\frac{748}{2\pi}\Big)\times7\Big]\text{m}^2=(374\times7)\text{m}^2$
$=2618\text{m}^2$
View full question & answer→Question 193 Marks
The radius of a circle with centre $O$ Is $7\ cm$. Two radii $OA$ and $OB$ are drawn at right angles to each other find the area of minor and major segments. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerRadius of the circle $= r = 7\ cm$
Central angle $=\theta=90^\circ$
$=\frac{22}{7}\times7\times7\times\frac{90}{360}-\frac{1}{2}\times7\times7\times\sin90^\circ$
$=\frac{77}{2}-\frac{49}{2}$
$=\frac{7-49}{2}$
$=\frac{28}{2}$
$=14\text{cm}^2$
Area of a cirde $=\pi^2=\frac{22}{7}\times7\times7=154\text{cm}^2$
Area of the major segment $=$ Area of a circle $-$ Area of the monor segment
$ =(154-14) \mathrm{cm}^2$
$ =140 \mathrm{~cm}^2$
View full question & answer→Question 203 Marks
The perimeter of a certain sector of a circle of radius $6.5\ cm$ is $31\ cm$. Find the area of the sector.
AnswerLet sector of circle is $OAB$
Perimeter of a sector of circle $= 31\ cm$
$OA + OB + $length of arc $AB = 31\ cm$
$6.5 + 6.5 + arc\ AB = 31\ cm$
arc $AB = 31 - 13$
$= 18\ cm$
Area of circle $=\frac{1}{2} lr$
$=\frac{1}{2}\times18\times6.5=58.5\text{ cm}^2$ View full question & answer→Question 213 Marks
Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii $4\ cm$ and $3\ cm.$
AnswerLet the r radius of the large circle be $R.$
Then, We have
Area of large circle of radius $R$
$=$ Area of a circle of radius $4\ cm$ $+$ Area of a circle of radius $3\ cm$
$\Rightarrow\pi\text{R}^2=\Big(\pi\times4^2+\pi\times3^2\Big)$
$\Rightarrow\pi\text{R}^2=(16\pi+9\pi)$
$\Rightarrow\text{R}^2=25$
$\Rightarrow\text{R}=5\text{cm}$
$\Rightarrow\text{Diameter}=2\text{R}=10\text{cm}$
View full question & answer→Question 223 Marks
A sector is cut from a circle of radius $21\ cm.$ The angle of the sector is $150^\circ .$ Find the length of the arc and the area of the sector. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerLength of the are $=\frac{2\pi\text{r}\theta}{360},\ \text{r}=21\text{cm},\theta=150^\circ$
$=\Big(\frac{2\pi\times21\times150}{360}\Big)\text{cm}=(17.5\pi\text{cm})$
Length of are $=\Big(17.5\times\frac{22}{7}\Big)\text{cm}=55\text{cm}$
Area of the sector $=\frac{\pi^2\theta}{360}=\Big(\frac{\pi\times21\times21\times150}{360}\Big)\text{cm}^2$
$=\Big(\frac{22}{7}\times183.75\Big)\text{cm}^2=577.5\text{cm}^2$
View full question & answer→Question 233 Marks
A horse is tethered to one comer of a field which is in the shape of an equilateral triangle of side $12m$ lf the lrngth of the rope is $7m,$ find the area of the field which the horse cannot geaze. $\text{Take }\sqrt{3}=1.732$ the answer correct to $2$ places of decimal $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerEach angle of equilateral triangle is $60$
Area which cannot be grazed $= ($area of equilateral $\triangle\text{ABC})$
$= ($area of the sector with $r = 7m, \theta = 60^\circ )$
$=\Big[\frac{\sqrt{3}}{4}\times(12)^2-\frac{22}{7}\times(7)^2\times\frac{60}{360}\Big]\text{m}^2$
$=\Big[(\sqrt{3}\times12\times3)-\frac{(22\times7)}{6}\Big]$
$=62.35-25.66\text{m}^2$
$=36.68\text{m}^2$
Area that the horse connot graze is $36.68m^2$ View full question & answer→Question 243 Marks
From a rectangular sheet of paper $ABCD$ with $AB = 40\ cm$ and $AD = 28\ cm,$ a semicircular portion with $BC$ as diameter is cut off. Find the area of the remaining paper.
AnswerLength of a rectangular sheet of paper $= AB = 40cm$
Breadth of a rectangular sheet of paper $= AD = 28cm$
$⇒$ Area of a rectangular sheet of paper $= AB × AD = 40 × 28 = 1120\ cm^2$
Diameter of a Semicircular portion $= AD = 28\ cm$
$⇒$ Radius $= 14cm$
$⇒$ Area of a Semicircular portion $=\frac{1}{2}\times\frac{22}{7}\times14\times14=308\text{cm}^2$
$\therefore$ Area of the remaining paper $=$ Area of a rectangular sheet of paper $-$ Area of a Semicircular portion
$= (1120 - 308)\ cm^2$
$= 812\ cm^2$
View full question & answer→Question 253 Marks
The short and long hands of a clock are $4\ cm$ and $6\ cm$ long respectively. find the sum of distances travelled by their tips in $2$ days. $[\text{Take }\pi\ =3.14]$
AnswerIn $2$ days, the shot hand will complete $4$ rounds
$\therefore$ Distance travelled by its tip in $2$ days
$= 4($circumference of the circle with $r = 4\ cm)$
$=(4\times2\pi\times4)\text{cm}=32\pi\text{cm}$
In $2$ days, the long hand will complete $48$ rounds
$\therefore$ length moved by its tip
$= 48($circumference of the circle eith $r = 6\ cm)$
$=(48\times2\pi\times6)\text{cm}=576\pi\text{cm}$
$\therefore$ Sum of the lengths moved
$=(32\pi+576\pi)=608\pi\text{cm}$
$=(608\times3.14)\text{cm}=1909.12\text{cm}$
View full question & answer→Question 263 Marks
In the given figure, the shape of the top of a table is that of a sector of a circle with centre $O$ and $\angle\text{AOB}=90^\circ.$ If $AO = OB = 42\ cm$ then find the perimeter of the top of the table.
Answer$\angle\text{AOB}=90^\circ$
$AO = OB = 42\ cm$
$⇒$ Radius of a circle $= 42\ cm$
$\therefore$ Required perimeter $=$ Circumference of a circle $-$ Length of arc $AB + (AO + OB)$
$=\Big\{\Big(2\times\frac{22}{7}\times42\Big)-\Big(2\times\frac{22}{7}\times42\times\frac{90}{360}\Big)+(42+42)\Big\}\text{cm}$
$=\Big\{\Big(2\times\frac{22}{7}\times42\Big)-\Big(2\times\frac{22}{7}\times42\times\frac{90}{360}\Big)+(42+42)\Big\}\text{cm}$
$=(264-66+84)\text{cm}$
$=282\text{cm}$
View full question & answer→Question 273 Marks
In the given figure, $APB$ and $CQD$ are semicircles of diameter $7\ cm$ each, while $ARC$ an $BSD$ are semicircles of diameter $14\ cm$ each. Find the,
- Perimeter,
- Area of the shad d region.
Answer
- Perimeter of the shaded region
$=$ Perimeter of semicircles $(ARC + BSD)\ +\ $Perimeter of semicircles $(APB + CQD)$
$=\Big\{2\Big(\frac{22}{7}\times7\Big)+2\Big(\frac{22}{7}\times\frac{7}{2}\Big)\Big\}\text{cm}$
$=(44+22)\text{cm}$
$=66\text{cm}$
- Area of the shaded region
$=$ Area of semicircles $(ARC + BSD)\ -\ $Area of semicircles $(APB + CQD)$
$=\Big\{2\Big(\frac{1}{2}\times\frac{22}{7}\times7\times7\Big)-2\Big(\frac{1}{2}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\Big)\Big\}\text{cm}^2$
$=\Big(2\times77-2\times\frac{77}{4}\Big)\text{cm}^2$
$=(154-38.5)\text{cm}^2$
$=115.5\text{cm}^2$ View full question & answer→Question 283 Marks
A horse is placed for grazing inside a reangular field $70\ m$ by $2\ m.$ It is tethered to one comer by a rope $21\ m$ long. On how much area an it graze$?$ How much area is left ungrazed? $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer
Area ehich the horse can graze $=$ Area of the quadrant of radius $21\ m$
$=\Big(\frac{1}{4}\times\frac{22}{7}\times21\times21\Big)\text{m}^2$
$=346.5\text{m}^2$
Area ungrazed $=[(70\times52)-346.5]\text{m}^2$
$=3293.5\text{m}^2$ View full question & answer→