Question 12 Marks
In the given figure, an equilateral triangle has been inscribed in a circle of radius 4cm. Find the area of the shaded region. $\big[\text{Take }\pi=3.14\text{ and }\sqrt{3}=1.73.\big]$
Answer
Drew
$\text{OD}\bot\text{BC}$
Because $\triangle\text{ABC}$ is equilateral, $\angle\text{A}=\angle\text{B}=\angle\text{C}=60^\circ.$
Thus, we have:
$\angle\text{OBD}=30^\circ$
$\Rightarrow\frac{\text{OD}}{\text{OB}}=\sin30^\circ$
$\Rightarrow\frac{\text{OD}}{\text{OB}}=\frac{1}{2}$
$\Rightarrow\text{OD}=\Big(\frac{1}{2}\times4\Big)\text{cm}[\because\text{OB}=\text{redius}]$
$\Rightarrow\text{OD}=2\text{cm}$
$\text{BD}^2=\big(\text{OB}^2-\text{OD}^2\big)$ [By Pythagoras' theorem]
$\Rightarrow\text{BD}^2=\big(4^2-2^2\big)\text{cm}^2$
$\Rightarrow\text{BD}^2=(16-4)\text{cm}^2$
$\Rightarrow\text{BD}^2=12\text{cm}^2$
$\Rightarrow\text{BD}=2\sqrt{3}\text{cm}$
Also,
$\text{BC}=2\times\text{BD}$
$=\big(2\times2\sqrt{3}\big)\text{cm}$
$\therefore$ Area of the shaded region = (Area of the circle) - (Area of $\triangle$ ABC)
$=\Big|(3.14\times4\times4)-\Big(\frac{\sqrt{3}}{4}\times4\sqrt{3}\times4\sqrt{3}\Big)\Big|\text{cm}^2$
$=\big|50.24-(12\times1.73)\big|\text{cm}^2$
$=(50.24-20.76)\text{cm}^2$
$=29.48\text{cm}^2$ View full question & answer→Question 22 Marks
The difference between the circumference and radius of a circle is 37cm. Using $\pi=\frac{22}{7},$ find the circumference of the circle.
AnswerLet r be the radius of a circle.
Then, circumference of a circle $=2\pi\text{r}$
We have,
Circumference - Radius = 37cm
$\Rightarrow2\pi\text{r}-\text{r}=37$
$\Rightarrow\text{r}(2\pi-1)=37$
$\Rightarrow\text{r}\Big(\frac{44}{7}-1\Big)=37$
$\Rightarrow\text{r}\Big(\frac{37}{7}\Big)=37$
$\Rightarrow\text{r}=\frac{37\times7}{37}$
$\Rightarrow\text{r}=7\text{cm}$
$\therefore$ Circumference of the circle $=2\pi\text{r}=\Big(2\times\frac{22}{7}\times7\Big)\text{cm}=44\text{cm}$
View full question & answer→Question 32 Marks
The radii of two circles are 19cm and 9cm Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
AnswerLet R be the radius of the circle.
Then, we have
Circumference of a circle of radius R = Circumference of a circle of radius 19cm + Circumference of a circle of radius 9cm
$\Rightarrow2\pi\text{R}=2\pi\times19+2\pi\times9$
⇒ R = 19 + 9
⇒ R = 28cm.
View full question & answer→Question 42 Marks
The perimeter of a sector of a circle of radius 5.6cm is 27.2cm. Find the area of the sector.
AnswerLet Obe the centre of the circle with radius 5.6cm and OACB be its sector with perimeter 27.2cm.
Thus, we have:
⇒ 5.6 + 5.6 + arc AB = 27.2
⇒ arc AB = 16cm
Now,
Area of the sector OACBO $=\Big(\frac{1}{2}\times\text{Radius}\times\text{l}\Big)\text{Square units}$
$=\Big(\frac{1}{2}5.6\times16\Big)\text{cm}^2$
$=44.8\text{cm}^2$
View full question & answer→Question 52 Marks
The area of the sector of a circle of radius $10.5 \ cm$, is $69.3 cm^2$. Find the central angle of the sector.
AnswerArea of the sector of circle $=\frac{\pi\text{r}^2\theta}{360}=69.3$
Radius $= 10.5cm$
$\Rightarrow\frac{\pi\times(10.5)^2\times\theta}{360}=69.3$
$\Rightarrow\theta=\frac{69.3\times360\times7}{10.5\times10.5\times22}=72^\circ$
View full question & answer→Question 62 Marks
OACB is a quadrant of a circle with centre O and its radius is 3.5cm. If OD = 2cm. find the area of $\Big[\text{Take }\pi=\frac{22}{7}.\Big]$
- Quadrant OACB
- The shaded region.
Answer
- Area of the quadrant OACB $=\Big(\frac{1}{4}\times\frac{22}{7}\times3.5\times3.5\Big)\text{cm}^2$
$=\Big(\frac{1}{4}\times\frac{22}{7}\times\frac{35}{10}\times\frac{35}{10}\Big)\text{cm}^2$
$=\frac{77}{8}\text{cm}^2$
$=9.625\text{cm}^2$
- Area of the shaded region= Area of the quadrant OACB - Area of $\triangle\text{AOD}$
$=\Big|\Big(\frac{77}{8}\Big)=\Big(\frac{1}{2}\times3.5\times2\Big)\Big|\text{cm}^2$
$=\Big(\frac{77}{8}-\frac{35}{10}\Big)\text{cm}^2$
$=\frac{49}{8}\text{cm}^2$
$=6.125\text{cm}^2$ View full question & answer→Question 72 Marks
The areas of two circles are in th ratio $4 : 9$. What is the ratio between their circumferences?
AnswerLet$ R_1$ and $R_2$_ be the radii of two circles respectively.
Then, we have $\frac{\pi\text{R}^2_1}{\pi\text{R}^2_2}=\frac49$
$\Rightarrow\frac{\text{R}^2_1}{\text{R}^2_2}=\frac49$
$\Rightarrow\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2=\Big(\frac23\Big)^2$
$\Rightarrow\frac{\text{R}_1}{\text{R}_2}=\frac23$
Now, $\frac{2\pi\text{R}_1}{2\pi\text{R}_1}=\frac{\text{R}_1}{\text{R}_2}=\frac23$
$\therefore$ Ratio of their circumference is 2 : 3.
View full question & answer→Question 82 Marks
The radius of the wheel of a vehicle is 42cm. How many revolutions will it complete in a 19.8-km-long journey? $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerRadius of the wheel = 42cm
Circumference of wheel $=2\pi\text{r}=\Big(2\times\frac{22}{7}\times42\Big)\text{cm}=264\text{cm}$
Distance travelled = 19.8km = 1980000cm
Number of revolutions $=\Big(\frac{1980000}{264}\Big)=7500$
View full question & answer→Question 92 Marks
A wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}\text{cm}^2.$ The same wire is bent to form a circle. Find the area enclosed by the circle. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerArea of equilateral $=\frac{\sqrt{3}\text{a}^2}{4}=121\sqrt{3}$
$\text{a}^2=121\times\frac{\sqrt{3}}{\sqrt{3}}\times4$
$\text{a}^2=484\Rightarrow\text{a}=\sqrt{484}$
Perimeter of equilaterral triangle = 3a = (322)cm
= 66cm
Circumference of circle = perimeter of circle
$2\pi\text{r}=66\Rightarrow\text{r}\times\frac{7}{22\times2}=10.5\text{cm}$
Area of circle $=\pi\text{r}^2=\Big(\frac{22}{7}\times10.5\times10.5\Big)\text{cm}^2$
$=346.5\text{cm}^2$
View full question & answer→Question 102 Marks
Find the area of a ring whose outer and inner radii are respectively 23cm and 12cm.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerArea of outer circle $=\pi_1^2=\Big(\frac{22}{7}\times23\times23\Big)\text{cm}^2$
$=1662.5\text{cm}^2$
Area of inner cirdee $=\pi\text{r}_2^2=\Big(\frac{22}{7}\times12\times12\Big)\text{cm}^2$
$=452.2\text{cm}^2$
Area of ring = outer area - inner area
$=(1662.5- 452.5)\text{cm}^2=1210\text{cm}^2$
View full question & answer→Question 112 Marks
The minute hand of a clock is 12cm long. Find the area swept by in it 35 minutes.
AnswerAngle described by the minute hand in 60 minutes
$=360^\circ$
Angle described by the minute hand in 35 minutes $=\Big(\frac{360}{60}\times35\Big)^\circ$
$=210^\circ$
Now,
$\text{r}=12\text{cm}$ and $\theta=210^\circ$
$\therefore$ Required area swept by the minute hand in 35 minutes = Area of the sector with
$\text{r}=12\text{cm}$ and $\theta=210^\circ$
$=\frac{\pi\text{r}^2\theta}{360}$
$=\Big(\frac{22}{7}\times12\times12\times\frac{210}{360}\Big)\text{cm}^2$
$=264\text{cm}^2$
View full question & answer→Question 122 Marks
Find the length of the arc of a circle of diameter 42cm which subtends an angle of 60° at the centre.
AnswerDiameter of a circle = 42cm
⇒ Radius of a circle $=\text{r}=\frac{42}{2}=21\text{cm}$
Central angle $=\theta=60^\circ$
$\therefore$ Length of the arc $=\frac{2\pi\text{r}\theta}{360}$
$=\Bigg(\frac{2\times\frac{22}{7}\times21\times60^\circ}{360^\circ}\Bigg)\text{cm}$
$= 22\text{cm.}$
View full question & answer→Question 132 Marks
In the give figure, the sectors of two concentric circles of radii 7cm and 3.5cm are shown. Find the area of the shaded region.
AnswerArea of shaded region
$=\pi\big[\text{R}^2-\text{r}^2\big]\frac{\theta}{360^\circ}$
$=\frac{22}{7}\big[7^2-(3.5)^2\big]\frac{30^\circ}{360^\circ}$
$=\frac{22}{7}(7+3.5)(7-3.5)\times\frac{1}{12}$
$=\frac{22}{7}\times10.5\times3.5\times\frac{1}{12}$
$=22\times\frac{5}{10}\times\frac{35}{10}\times\frac{1}{4}=\frac{77}{8}=9.62\text{cm}^2.$
View full question & answer→Question 142 Marks
The circumforence of a circle is 39.6cm. Find its area. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerCircumference of circle $=2\pi\text{r}=39.6\text{cm}$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=39.6$
$\text{r}=\Big(39.6\times\frac{7}{44}\Big)\text{cm}=6.3$
$\text{r}=6.3\text{cm}$
Area of cride $=\pi\text{r}^2=\Big(\frac{22}{7}\times6.3\times6.3\Big)\text{cm}^2$
$=124.74\text{cm}^2$
View full question & answer→Question 152 Marks
The circumference of a circle is 22cm. find the area of its quadrant.
AnswerLet r be the radius of a circle
Circumference of a circle = 22cm
$\Rightarrow2\pi\text{r}=22$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=22$
$\Rightarrow\text{r}=\frac72\text{cm}$
Area of quadrant of circle $=\frac14\times\pi\text{r}^2=\Big(\frac{1}{4}\times\frac{22}{7}\times\frac72\times\frac72\Big)\text{cm}^2=\frac{77}{8}\text{cm}^2$
View full question & answer→Question 162 Marks
A path of $8\ m$ width runs around the outside of a circular park whose radius is$ 17\ m$. Find the area of the path.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerInner radius of the circular park $= 17\ m$
Width of the path $= 8m$
Outer radius of the circular park = (17 + 8)m = 25m
Area of path $=\pi\Big[(25)^2-(17)^2\Big]\text{cm}^2$
$=\pi(25+17)(25-17)\text{m}^2$
$=\Big[\frac{22}{7}\times42\times8\Big]\text{m}^2$
$\therefore$ Area $= 1056m^2$
View full question & answer→Question 172 Marks
The circumferences of two circles are in the ratio $2 : 3$ what is the ratio between their areas?
AnswerLet $R_1$ and $R_2$ be the radii of two circles respectively.
Then, we have
$\frac{2\pi\text{R}_1}{2\pi\text{R}_2}=\frac23$
$\Rightarrow\frac{\text{R}_1}{\text{R}_2}=\frac23$
Now, $\frac{\pi\text{R}^2_1}{\pi\text{R}^2_1}=\frac{\text{R}^2_1}{\text{R}^2_2}=\Big(\frac{\text{R}_1}{\text{R}_2}\Big)^2=\Big(\frac{2}{3}\Big)^2=\frac49$
$\therefore$ Ratio of their areas is 4 : 9.
View full question & answer→Question 182 Marks
A pendulum swings through an angl of 30° and describes an arc 8.8cm in length. Find the length of the pendulum.
AnswerLength of the pendulum = radius of sector = r cm
Arc length = 8.8
$\Rightarrow2\times\frac{22}{7}\times\text{r}\times\frac{30}{360}=8.8$
$\Rightarrow\text{r}=\frac{8.8\times7\times360}{2\times22\times30}=16.8\text{cm}$
View full question & answer→Question 192 Marks
What is the diameter of a circle whose area is equal to the sum of the areas of two circles of diameters 10cm and 24cm?
AnswerLet the radius of the large circle be R.
Then, we have
Area of large circle of radius R
= Area of a circle of radius 5cm + Area of a circle of radius 12cm
$\Rightarrow\pi\text{R}^2=\Big(\pi\times5^2+\pi\times12^2\Big)$
$\Rightarrow\pi\text{R}^2=\big(25\pi+144\pi\big)$
$\Rightarrow\pi\text{R}^2=169\pi$
$\Rightarrow\text{R}^2=169$
$\Rightarrow\text{R}=13\text{cm}$
⇒ Diameter = 2R = 26cm
View full question & answer→Question 202 Marks
The radii of two circles are $8\ cm$ and 6cm find the radius of the circle having area equal to the sum of the areas of the two circles.
AnswerLet R be the radius of the circle.
Then, we have
Area of a circle of radius R = Area of a circle of radius $8\ cm$ + Area of a circle of radius $6\ cm$
$\Rightarrow\pi\text{R}^2=\pi\times(8)^2+\pi\times(6)^2$
$\Rightarrow R^2 = 64 + 36$
$\Rightarrow R^2= 100$
$\Rightarrow R = 10cm.$
View full question & answer→Question 212 Marks
The minute hand of a clock is 7.5cm long. find the area of the face of the clock described by the minute hand in 56 minutes.
AnswerAngle described by the minuts hand in 60 minuts = 360°
Angle described by the minuts hand in 35 minuts $=\Big(\frac{360}{60}\times35\Big)^\circ$
$=210^\circ$
now,
r = 12cm and $\theta$ = 210°
$\therefore$ Reqiuired area described by the minuts hand oin 35 minuts = area of the sector where
r = 12cm and $\theta $ = 210°
$=\frac{\pi\text{r}^2\theta}{360}$
$=\Big(\frac{22}{7}\times12\times12\times\frac{210}{360}\Big)\text{cm}^2$
$=264\text{cm}^2$
View full question & answer→Question 222 Marks
A chord of a circle of radius 14cm a makes a right angle at the centre. Find the area of the sector.
AnswerHere, r = 14cm
and $\angle \text{AOB}=90^\circ$
$\therefore$ Area of monor sector
$=\frac{90^\circ}{360^\circ}\times\pi\text{r}^2$
$=\frac{1}{4}\times\frac{22}{7}\times14\times14$
$=154\text{cm}^2$
View full question & answer→Question 232 Marks
The wheels of the locomotive of a train are 2.1min radius. They make 75 revolutions in one minute. Find the speed of the train in km per hour.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerRadius of wheel = 2.1m
Circumference of wheel $=(2\pi\text{r})\text{m}\Big(2\times\frac{22}{7}\times2.1\Big)\text{m}=13.2\text{m}$
Distance covered in one revolution = 13.2m
Distance covered in 75 revolution = (13.275)m = 990m
$=\frac{990}{1000}\text{km}$
Distance a covered in 1 minute $=\frac{99}{100}\text{km}$
Distance a covered in 1 hour $=\Big(\frac{99}{100}\times60\Big)\text{km}=59.4\text{km}$
View full question & answer→Question 242 Marks
In a circle of radius 21cm, an arc subtends an angle of 60° at the centre Find the length of the arc.
AnswerRadius of a circle = r = 21cm
Central angle $=\theta=60^\circ$
$\therefore$ Length of the arc $=\frac{2\pi\text{r}\theta}{360}$
$=\Big(2\times\frac{22}{7}\times21\times\frac{60}{360}\Big)\text{cm}$
$=22\text{cm}$
View full question & answer→Question 252 Marks
The minute hand of a dock is 15cm long. Calculate the area kwept by it in 20 minutes. $\big[\text{Take }\pi=3.14\big]$
AnswerAngle described by the ninute hand in 60 minutes = 360°
Angle described by minute hand in 20 minutes $=\Big(\frac{360}{60}\times20\Big)=120^\circ$
Required area swept by the minute hand in 20 minutes
= Area of the sector$($with r = 15cm and $\theta=120^\circ\big)$
$=\Big(\frac{\pi\text{r}^2\theta}{360^\circ}\Big)\text{cm}^2$
$=\Big(3.14\times15\times15\times\frac{120^\circ}{360^\circ}\Big)$
$=235.5\text{cm}^2$
View full question & answer→Question 262 Marks
A square is inscribed in a circle. ind the ratio of the areas rf the circle and the square.
Answer
Let the radius of the circle be r cm.
Then, diagonal of the square = diameter of the circle = 2r cm
Area of the circle $=\pi\text{r}^2$ sq. units
Area of the square $=\frac{1}{2}\times\text{diagonal}^2=\frac12\times4\text{r}^2=2\text{r}^2$ sq. units
Now, $\frac{\text{Area of the circle}}{\text{Area of the square}}=\frac{\pi\text{r}^2}{2\text{r}^2}=\frac{\pi}{2}$
$\therefore$ Required Ratio is $\pi :2.$ View full question & answer→Question 272 Marks
Find the perimeter of a semicircular protractor whose diameter is 14 cm.
AnswerDiameter of a semicircular protractor = 14cm
⇒ Radius of a semicircular protractor = r = 7cm
$\therefore$ Perimeter of a semicircular protractor $=(\pi\text{r}+2\text{r})$
$= \text{r} (\pi+2)$
$=7\Big(\frac{22}{7}+2\Big)\text{cm}$
$=7\Big(\frac{22+14}{7}\Big)\text{cm}$
$=7\times\frac{36}{7}$
$=36\text{cm}$
View full question & answer→Question 282 Marks
The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84cm, find its speed in km per hour. $\Big[\text{Take }\pi=\frac{22}{7}\Big]$
AnswerDistance covered in 1 revolution $=\pi\times\text{d}$
$=\Big(\frac{22}{7}\times84\Big)\text{cm}$
$=264\text{cm}$
Distance covered in 1 second = (5 × 264)cm
$=1320\text{cm}$
Distance covered in 1 hour = (60 × 60 × 1320)cm
$=4752000\text{cm}$
$=\Big(\frac{4752000}{1000\times100}\Big)\text{km}$
$=47.52\text{km}$
View full question & answer→Question 292 Marks
Find the area of a circle whose circurnference is $8\pi.$
AnswerLet r be the radius of the circle.
Circumference of a circle $=8\pi$
$\Rightarrow2\pi\text{r}=8\pi$
$\Rightarrow\text{r}=\frac82$
$\Rightarrow\text{r}=4$
$\therefore$ Area of a circle $=\pi\text{r}^2=\pi\times4\times4=16\pi$
View full question & answer→Question 302 Marks
In a circle of radius 21cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.
Answerlet ACB be the given arc subtending at an angle of 60° at the centre.
Now, we have:
$\text{r}=21\text{cm}$ and $\theta=60^\circ$
$\therefore$ Length of the arc ACB $=\frac{2\pi\text{r}}{360}$
$=\Big(2\times\frac{22}{7}\times21\times\frac{60}{360}\Big)$
$=22\text{cm}$
View full question & answer→Question 312 Marks
The area of a circle is $98.56 cm^2$ find its circumference. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerLet r be the radius of the circle.
Area of the cirde $= 98.56cm^2$
$\Rightarrow\pi\text{r}^2=98.56$
$\Rightarrow\frac{22}{7}\times\text{r}^2=98.56$
$\Rightarrow\text{r}^2=\frac{98.56\times7}{22}$
$\Rightarrow\text{r}^2=31.36$
$\Rightarrow\text{r}=5.6$
$\therefore$ Circumference of a circle $=2\times\frac{22}{7}\times5.6=35.2\text{cm}$
View full question & answer→Question 322 Marks
The wheel of a motorcycle is of radius 35cm. How many revolutions per minute must the wheel make so as to keep a speed of 66km/hr? $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerDistance covered by a wheel in 1 minute
$=\Big(\frac{72.6\times1000\times100}{60}\Big)\text{cm}=121000\text{cm}$
Circumference of a wheel $=\Big(2\times\frac{22}{7}\times70\Big)\text{cm}=440\text{cm}$
Number of revolution in 1min $=\Big(\frac{122000}{440}\Big)=275$
View full question & answer→Question 332 Marks
A boy is cycling in such a way that the wheels of his bicycle. are making 140 revolutions per minute. If the diameter of a wheel is 60cm calculate the speed (in km/hr) at which th boy is cycling.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerRadius of the wheel $=\text{r}=\frac{60}{2}=30\text{cm}$
Circumference of the wheel $=2\pi\text{r}\Big(2\times\frac{22}{7}\times30\Big)\text{cm}$
$=\frac{1320}{7}\text{cm}$
Distance covered in 140 revolution
$=\Big(\frac{1320}{7}\times140\Big)\text{cm}=(1320\times20)\text{cm}$
$=26400\text{cm}=\frac{26400}{100}\text{m}=264\text{m}=\frac{264}{1000}\text{km}$
Distance covered in one hour $=\Big(\frac{264}{1000}\times60\Big)\text{km}=15.84\text{km}$
View full question & answer→Question 342 Marks
In the given figure, ABCD is a square each of whose sides measures 28cm. Find the area of the shaded region. $\Big[\text{Take }\pi=\frac{22}{7}.\Big]$
AnswerLet rbe the radius of the circle.
Thus, we have:
$\text{r}=\frac{28}{2}\text{cm}$
$=14\text{cm}$
Now,
Area of the shaded region = (Area of the square ABCD) - 4(Area of the sector where r = 14cm and = $\theta$ 90°)
$=\Big|(28\times28)-4\Big(\frac{22}{7}\times14\times14\times\frac{90}{360}\Big)\Big|\text{cm}^2$
$=\big|784-4(154)\big|\text{cm}^2$
$=(784-616)\text{cm}^2$
$=168\text{cm}^2$
View full question & answer→Question 352 Marks
The radius of a circle is 17.5cm. Find fhe area of the sector e dosed by two radii and an arc 44cm in length.
AnswerLength of arc of circle = 44cm
Radius of circle = 17.5cm
Area of sector $=\frac{1}{2}\text{ lr }\Big(\frac{1}{2}\times44\times17.5\Big)\text{ cm}^2$
$=(22\times17.5)\text{cm}^2=385\text{cm}^2$
View full question & answer→Question 362 Marks
A park os of the shape of a circle of diameter 7m. it is surrounded by a path of width of 0.7m find the expenditure of cementing the path, if its cost is rs. 110 sq m.$\Big[\text{Use }\pi=\frac{22}{7}\Big]$
Answer
The Diameter of park = 7m
$\therefore$ Radius $=\frac{7}{2}=3.5\text{m}$
The width of path = 0.7m
$\therefore$ Radius of park with path
$=3.5+0.7=4.2\text{m}$
Area of the path $=\pi(4.2)^2-\pi(3.5)^2$
$=\frac{22}{7}(17.64-12.25)$
$=\frac{22}{7}\times5.39=22\times0.77$
$= 16.94\text{m}^2$
Cost of the cementing the path
= 16.94 × 110
= Rs. 1863.40 View full question & answer→Question 372 Marks
What is the perimeter of a square which circumscribes a circle of radius a cm?
AnswerSince square circumscribes a circle of radius a cm
We have,
Side of the square = 2 × radius of circle = 2a cm
Then, Perimeter of the square = (4 × 2a) = 8a cm
View full question & answer→Question 382 Marks
A copper wire when bent in the form of a square encloses an area of $484cm^2$ The same wire is now bent in the form of d circle. find the area enclosed by the circle. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
AnswerArea of surface = (side)$^2 = 484cm^2$
⇒ Side $=\sqrt{484}\text{cm}=22\text{cm}$
Perimeter of square = 4 side = 422 = 88cm
Ciecumference of circle = perimeter of square
$2\pi\text{r}=88\text{cm}\Rightarrow\text{r}=\frac{88\times7}{2\times22}=14\text{cm}$
Area of cirde $=\pi\text{r}^2=\Big(\frac{22}{7}\times14\times14\Big)\text{cm}^2=616\text{cm}^2$
View full question & answer→Question 392 Marks
A Wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}\text{cm}^2.$ If the same wire is bent into the form of a circle, what will be the area of the circle? $\Big[\text{Take }\pi=\frac{22}{7}.\Big]$
AnswerLet a cm be the side of the equilateral triangle.
Now,
Area of the equilateral triangle $=\frac{\sqrt{3}}{4}\text{a}^2$
We have
$\frac{\sqrt{3}}{4}\text{a}^2=121\sqrt{3}$
$\Rightarrow\frac{\text{a}^2}{4}=121$
$\Rightarrow\text{a}^2=484$
$\Rightarrow\text{a}=22$
Perimeter of the triangle = Circumference of the circle
Perimeter of the triangle = (22 + 22 + 22)cm
= 66cm
Now, let r cm be the radius of the circle.
We know:
Circumference of the circle $=2\pi\text{r}$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=66$
$\Rightarrow\text{r}=\Big(66\times\frac{7}{44}\Big)\text{cm}$
$\Rightarrow\text{r}=\frac{21}{2}\text{cm}$
Also,
Area of the circle $=\pi\text{r}^2$
$=\Big(\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2}\Big)\text{cm}^2$
$=\frac{693}{2}\text{cm}^2$
$=346.5\text{cm}^2$
View full question & answer→Question 402 Marks
The circumference of a circle is 8cm. Find the area of the sictor whose central angle is 72°.
AnswerLet r be the radius of a circle.
Circumference of a circle = 8cm
Central angle $=\theta=72^\circ$
Now,
Circmference of a circle $=2\pi\text{r}$
$\Rightarrow88=2\pi\text{r}$
$\Rightarrow88=2\times\frac{22}{7}\times\text{r}$
$\Rightarrow\text{r}=\frac{88\times7}{2\times22}$
$\Rightarrow\text{r}=14\text{cm}$
$\therefore$ Area of the sector $=\frac{\pi\text{r}^2\theta}{360}=\Big(\frac{22}{7}\times14\times14\times\frac{72}{360}\Big)\text{cm}^2$
$=123.2\text{cm}^2$
View full question & answer→Question 412 Marks
A sector of 56°, cut out from a circle, contains $17.6 cm^2$. Find the radius of the circle.
Answer$\theta=56^\circ$ and let radius is r cm
Area of sector $=\frac{\pi\text{r}^2\theta}{360^\circ}=17.6\text{cm}^2$
$\Rightarrow\frac{22}{7}\times\text{r}^2\times\frac{56^\circ}{360^\circ}=17.6$
$\text{r}^2=\Big(\frac{17.6\times360\times7}{22\times56}\Big)\text{cm}^2$
$\text{r}^2=36\text{cm}^2\Rightarrow\text{r}=\sqrt{36}\text{cm}=6\text{cm}$
Hence radius $= 6cm.$
View full question & answer→Question 422 Marks
The circumference of a circle is 22cm. Find its area. $\Big[\text{Take }\pi=\frac{22}{7}\Big]$
AnswerLet rem be the radius of the circle.
Now,
Circumference of the circle:
$2\pi\text{r}=22$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=22$
$\Rightarrow\text{r}=\Big(22\times\frac{7}{44}\Big)\text{cm}$
$\Rightarrow\text{r}=\frac{7}{2}$
Also,
Area of the circle $=\pi\text{r}^2$
$=\Big(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\Big)\text{cm}^2$
$=\frac{77}{2}\text{cm}^2$
$=38.5\text{cm}^2$
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