Question 13 Marks
The diameter of a wheel of a car is $98\ cm$. How many revolutions will it make to travel $6160$ metres.
AnswerWe have: Diameter of the wheel of the car $=98\text{cm}$
$\therefore$ Circumference of the wheel of the car $=\pi\text{d}=\Big(\frac{22}{7}\times98\Big)\text{cm}=308\text{cm}$ Note that, in one revolution of the wheel, the car travels a distance equal to the circumference of the wheel.
$\therefore$ The distance travelled by the car in one revolution of the wheel $=308\text{cm}.$
Total distance travelled by the car $=6160\text{m}=616000\text{cm}.$
$\therefore$ Number of revolutions $=\frac{616000}{308}=2000.$
View full question & answer→Question 23 Marks
A wire in the form of a rectangle $18.7\ cm$ long and $14.3\ cm$ wide is reshaped and bent into the form of a circle. Find the radius of the circle so formed.
AnswerLength of the wire = Perimeter of the rectangle
$= 2(l + b ) = 2 × (18.7 + 14.3)$
$= 66\ cm$
Let the wire be bent in the form of a circle of radius $r\ cm$. Then,
circumference $= 66\ cm$
$\Rightarrow2\pi\text{r}=66\text{cm}$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=66\text{cm}$
$\Rightarrow\text{r}=\frac{66\times7}{2\times22}=10.5\text{cm}$
View full question & answer→Question 33 Marks
The ratio of the radii of two circles is $3 : 2$. What is the ratio of their circumferences?
AnswerWe have, the ratio of the radii $=3: 2$
So, let the radii of the two circles be $3 r$ and $2 r$ respectively.
Let $C_1$ and $C_2$ be the circumferences of the two circles of radii $3 r$ and $2 r$, respectively.
Then,$\mathrm{C}_1=2 \pi \times 3 \mathrm{r}=6 \pi \mathrm{r}, \text { and } \mathrm{C}_2=2 \pi \times 2 \mathrm{r}=4 \pi \mathrm{r}$
$\therefore \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{6 \pi \mathrm{r}}{4 \pi \mathrm{r}}=\frac{6}{4}=\frac{3}{2}$
$\mathrm{C}_1: \mathrm{C}_2=3: 2$
View full question & answer→Question 43 Marks
If the area of a circle is $50.24m^2$, find its circumference.
AnswerWe have:
Area of the circle $(A)$ $=\pi\text{r}^2=50.24\text{m}^2$
$=50.24\text{m}^2=\frac{22}{7}\times\text{r}^2$
$=\text{r}^2=\frac{50.24\times7}{22}=\frac{351.68}{22}=15.985\text{m}^2$
$=\text{r}=3.998\text{m}$
Circumference of circle $(C)$ $=2\pi\text{r}$
$=\text{C}=2\times\frac{22}{7}\times3.998\text{m}$
$=\text{C}=25.12\text{m}$
View full question & answer→Question 53 Marks
The perimeter of a circle is $4\pi\text{r cm}.$ What is the area of the circle?
AnswerWe have: Given perimeter of the circle $=4\pi\text{r cm}=2\pi(2\text{r)cm}$
We know that, the perimeter of a circle $=2\pi\text{r}$
$\therefore$ Radius of the circle $=2\text{r cm}$
Area of the circle $=\pi\text{r}^2=\pi(2\text{r})^2=4\pi\text{r}^2$
View full question & answer→Question 63 Marks
The area of a circle is $154 \mathrm{~cm}^2$. Find the radius of the circle.
AnswerLet the radius of the circle be $r\ cm$ .
$\text { Area of the circle }(A)=154 \mathrm{~cm}^2$
$\Rightarrow 154=\frac{22}{7} \times \mathrm{r}^2$
$\Rightarrow \mathrm{r}^2=\frac{154 \times 7}{22}$
$\Rightarrow \mathrm{r}^2=49$
$\Rightarrow \mathrm{r}=7 \mathrm{~cm}$
Hence, the radius of the circle is $7\ cm$ .
View full question & answer→Question 73 Marks
A rhombus has the same perimeter as the circumference of a circle. If the side of the rhombus is $2.2m$, find the radius of the circle.
AnswerWe have: The side of a rhombus $= 2.2m$
Let $C$ be the circumference of a circle having a radius $r\ cm$.
Then, the perimeter of the rhombus $= 4 \times$ side $= 4 \times 2.2 = 8.8m.$
We know: Perimeter of the rhombus = Circumference of the circle
$\Rightarrow8.8\text{m}=2\pi\text{r}\Rightarrow\text{r}=\frac{8.8}{2\pi}$
$\Rightarrow\text{r}=\frac{8.8\times7}{2\times22}=1.4\text{m}$
The radius of the circle is $1.4m$.
View full question & answer→Question 83 Marks
A road which is $7m$ wide surrounds a circular park whose circumference is $352m$. Find the area of of road.
AnswerCircumference of the circular park $=2\pi\text{r}=352\text{m}$
$=2\pi\text{r}=352$
$=2\times\frac{22}{7}\times\text{r}=352$
$=\text{r}=56\text{m}$
Radius of the path including the $7m$ wide road $=(\text{r}+7)$
$=56+7=63\text{m}$
$\therefore$ Area of the road: $=\pi\times(63)^2-\pi\times(56)^2$
$=\frac{22}{7}\times63\times63-\frac{22}{7}\times56\times56$
$=22[9\times63-8\times56]$
$=22[567-448]$
$=2618\text{m}^2$
$\therefore$ Area of the road $=2618\text{m}^2$
View full question & answer→Question 93 Marks
A bicycle wheel makes $5000$ revolutions in moving $11\ km$. Find the diameter of the wheel.
AnswerWe have:
Total distance covered in $5000$ revolutions $= 11\ km = 11,000m$
$\therefore$ Distance covered in $1$ revolution $=\frac{11000}{5000}=\Big(\frac{11}{5}\Big)\text{m}.$
Distance covered in $1$ revolution = Circumference of the wheel
$\Big(\frac{11}{5}\Big)=\pi\text{d}$
$\Rightarrow\text{d}=\Big(\frac{11}{5\times\pi}\Big)=\Big(\frac{11\times7}{5\times22}\Big)\text{m}$
$\Rightarrow\text{d}=\Big(\frac{7}{10}\Big)=0.7\text{m}$
Thus, the diameter of the wheel is $0.7m = 70\ cm.$
View full question & answer→Question 103 Marks
A steel wire when bent in the form of a square encloses an area of $121cm^2$. If the same wire is bent in the form of a circle, find the area of the circle.
AnswerWe have:
$\text { Area of the square }=121 \mathrm{~cm}^2$
$\Rightarrow(\text { side })^2=(11)^2 \mathrm{~cm}^2$
$\Rightarrow \text { side }=11 \mathrm{~cm}.$
So, the perimeter of the square $=4$ (side)$=(4 \times 11) \mathrm{cm}=44 \mathrm{~cm}$.
Let $r$ be the radius of the circle.
Then,
Circumference of the circle $=$ Perimeter of the square
$=2 \pi \mathrm{r}=44$
$=2 \times \frac{22}{7} \times \mathrm{r}=44$
$=\mathrm{r}=7 \mathrm{~cm}$
$\therefore$ Area of the circle $=\pi \mathrm{r}^2=\frac{22}{7} \times 7 \times 7=154 \mathrm{~cm}^2$
View full question & answer→Question 113 Marks
How long will John take to make a round of a circular field of radius $21m$ cycling at the speed of 8km/hr?
AnswerWe have:
The radius of the circular field $= 21m$
$\therefore$ Circumference of the circular field $=2\pi\text{r}=2\times\frac{22}{7}\times21=132\text{m}$
If John cycles at the speed of $8$ km/hr (In $1$ hour John covers $8\ km = 8000m$), then,
John covers $8000 m$ in $1$ hour.
$\therefore$ Time required to cover $132 m$ $=\frac{132}{8000}=0.0165\text{ hours}$
$1$ hour $= 3600$ seconds
$\therefore$ $0.0615$ hours $= 0.0615 \times 3600 = 59.4$ seconds.
View full question & answer→Question 123 Marks
A wire of $5024m$ length is in the form of a square. It is cut and made a circle. Find the ratio of the area of the square to that of the circle.
AnswerWe have:
Perimeter of the square $=5024 \mathrm{~m}=$ Circumference of the circle
$\Rightarrow 4 \times$ Side of the square $=5024$
$\therefore$ Side of the square $=\frac{5024}{4}=1256 \mathrm{~m}$
Let the area of the square be $\mathrm{A}_1$ and the area of the circle be $\mathrm{A}_2$.
Area of the square $\left(A_1\right)=$ side $\times$ side $=\frac{5024}{4} \times \frac{5024}{4} \mathrm{~m}^2$
Circumference of the circle $=5024 \mathrm{~m}$
$=2 \pi \mathrm{r}=5024 \mathrm{~m}$
$=2 \times \frac{22}{7} \times \mathrm{r}=5024 \mathrm{~m}$
$=\mathrm{r}=\frac{5024 \times 7}{22 \times 2}$
Area of the circle $\left(\mathrm{A}_2\right)=\pi \mathrm{r}^2=\frac{22}{7} \times \frac{5024 \times 7}{22 \times 2} \times \frac{5024 \times 7}{22 \times 2}$
$=\frac{5024 \times 5024 \times 7}{22 \times 2 \times 2} \mathrm{~m}^2$
$\therefore \mathrm{~A}_1: \mathrm{A}_2=\frac{5024}{4} \times \frac{5024}{4}: \frac{5024 \times 5024 \times 7}{22 \times 2 \times 2}$
$\frac{\mathrm{~A}_1}{\mathrm{~A}_2}=\frac{5024}{4} \times \frac{5024}{4} \div \frac{5024 \times 5024 \times 7}{22 \times 2 \times 2}$
$\frac{\mathrm{~A}_1}{\mathrm{~A}_2}=\frac{\frac{5024 \times 5024}{4 \times 6}}{\frac{5024 \times 5024 \times 7}{222 \times 2 \times 2}}$
$\frac{\mathrm{~A}_1}{\mathrm{~A}_2}=\frac{11}{14}$
$\therefore \mathrm{~A}_1: \mathrm{A}_2=11: 14$
View full question & answer→Question 133 Marks
An ox in a kolhu (an oil processing apparatus) is tethered to a rope $3m$ long. How much distance does it cover in $14$ rounds?
AnswerWe have: Radius of the circular path traced by the ox in a kolhu $= 3m $
Distance covered by the ox in $1$ round = Circumference of the circular path
$=2\pi\text{r}=\Big(2\times\frac{22}{7}\times3\Big)\text{m}.$
$\therefore$ Distance covered in $14$ rounds
$=\Big(2\times\frac{22}{7}\times3\Big)\text{m}\times14=22\times12=264\text{m}.$
View full question & answer→Question 143 Marks
The diameter of the driving wheel of a bus is $140\ cm.$ How many revolutions per minute must the wheel make in order to keep a speed of $66\ km$ per hour?
AnswerWe have:
Diameter of the wheel $= 140\ cm$, desired speed of the bus $=66\text{km/hr}$
$\therefore$ Distance covered in $1$ revolution = The circumference of the wheel $=\pi\text{d}=\Big(\frac{22}{7}\times140\Big)\text{cm}=440\text{cm}.$
Now, the desired speed of the bus $=66\text{km/hr}=\frac{66\times1000\times100}{60}=110000\text{cm/min.}$
$\therefore$ Number of revolutions per minute $=\frac{110000}{440}=\frac{1000}{4}=250.$
Thus, the bus must make $250$ revolutions per minute to keep the speed at $66\ km/hr.$
View full question & answer→Question 153 Marks
The circumference of a circle is $3.14m$, find its area.
AnswerWe have:
Circumference of the circle $=3.14\text{m}=2\pi\text{r}$
$=3.14\text{m}=2\times\frac{22}{7}\times\text{r m}$
$=\text{r}=\frac{3.14\times7}{22\times2}\text{m}$
$=\text{r}=0.5\text{m}$
Area of the circle $(A)$ $=\pi\text{r}^2$
$=\text{A}=\frac{22}{7}\times0.5^2\text{m}^2=0.785\text{m}^2$
View full question & answer→Question 163 Marks
Find the radius of a circle, if its area is: $4\pi\text{cm}^2$
AnswerLet the radius of the circle be $r\ cm$.
$\therefore$ Area of the circle $(A)$ $=4\pi\text{cm}^2$
$\Rightarrow4\pi=\pi\times(\text{r})^2\text{cm}^2$
$\Rightarrow\text{r}^2=\Big(\frac{4\pi}{\pi}\Big)=4$
$\Rightarrow\text{r}=2\text{cm}$
View full question & answer→Question 173 Marks
Prove that the area of a circular path of uniform width $h$ surrounding a circular region of radius $r$ is $\pi\text{h}(2\text{r + h}).$
AnswerRadius of the circular region $= r$
Radius of the circular path of uniform width $h$ surrounding the circular region of radius $r = (r + h)$.
$\therefore$ Area of the path $=\pi(\text{r + h})^2-\pi\text{r}^2$
$=\pi\text{r}^2+\pi\text{h}^2+2\pi\text{rh}-\pi\text{r}^2$
$=\pi\text{h}(2\text{r + h})$
View full question & answer→Question 183 Marks
A piece of wire is bent in the shape of an equilateral triangle of each side $6.6\
cm$. It is re-bent to form a circular ring. What is the diameter of the ring?
AnswerWe have: Length of the wire = The perimeter of the equilateral triangle $= 3 \times side $
$= 3 \times 6.6 = 19.8\ cm.$
Let the wire be bent to form a circular ring of radius $r\ cm.$
Then, Circumference $= 19.8cm$
$\Rightarrow2\pi\text{r}=19.8\text{cm}$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=19.8\text{cm}$
$\Rightarrow\text{r}=\frac{19.8\times\text{7}}{2\times22}=3.15\text{cm}$
So, the diameter of the ring $= 2 \times 3.15 = 6.30\ cm$.
View full question & answer→Question 193 Marks
The diameter of a wheel of a car is $63\ cm$. Find the distance travelled by the car during the period, the wheel makes $1000$ revolutions.
AnswerIt may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
Now, the diameter of the wheel $= 63\ cm$
$\therefore$ Circumference of the wheel $=\pi\text{d}=\Big(\frac{22}{7}\times63\Big)\text{cm}=198\text{cm}$
Thus, the cycle covers $198\ cm$ in one revolution.
$\therefore$ The distance covered by the cycle in $1000$ revolutions $= (198 \times 1000) = 198000\ cm = 1980m.$
View full question & answer→Question 203 Marks
Find the radius of a circle, if its area is: $55.44\text{m}^2$
AnswerLet the radius of the circle be $r\ cm$.
$\therefore$ Area of the circle $(A)$ $=55.44\text{m}^2$
$=55.44=\pi\text{r}^2\text{m}^2$
$=\text{r}^2=\frac{55.44\times7}{22}$
$=\text{r}^2=17.64$
$=\text{r}=4.2\text{m}$
View full question & answer→Question 213 Marks
A boy is cycling such that the wheels of the cycle are making $140$ revolutions per minute. If the diameter of the wheel is $60\ cm$, calculate the speed per hour with which the boy is cycling.
AnswerWe have: The diameter of the wheel $= 60\ cm$
Distance covered by the wheel in $1$ revolution = Circumference of the wheel
$\therefore$ Distance covered by the wheel in $1$ revolution $=\pi\text{d}=\Big(\frac{22}{7}\times60\Big)\text{cm}.$
$\therefore$ Distance covered in $140$ revolutions $=\Big(\frac{22}{7}\times60\times140\Big)\text{cm}$
$=\Big(\frac{184800}{7}\Big)\text{cm}=26400\text{cm}$
Thus, the wheel covers $26400\ cm$ in $1$ minute.
Then, Speed $=\Big(\frac{26400}{100}\times60\Big)\text{m/hr}=(264\times60)\text{m/hr}$ Speed $=\Big(\frac{264\times60}{1000}\Big)\text{km/h}=15.84\text{km/hr}$
The speed with which the boy is cycling is $15.84\ km/hr.$
View full question & answer→Question 223 Marks
A wire is looped in the form of a circle of radius $28\ cm$. It is re-bent into a square form. Determine the length of the side of the square.
AnswerWe have:
The radius of the circle $= 28\ cm$
$\therefore$ Circumference of the circle $=2\pi\text{r}=2\times\frac{22}{7}\times28=176\text{cm}$
Let a cm be the side of the square.
Then,
The circumference of the circle = The perimeter of the square
$\Rightarrow176=4\times\text{a}$
$\Rightarrow\text{a}=\Big(\frac{176}{4}\Big)\text{cm}=44\text{cm}$
The side of the square is $44\ cm.$
View full question & answer→Question 233 Marks
Find the radius of a circle, if its area is: $1.54\text{km}^2$
AnswerLet the radius of the circle be $r\ cm.$
$\therefore$ Area of the circle $(A)$ $=1.54\text{km}^2$
$=1.54=\pi\text{r}^2\text{km}^2$
$=\text{r}^2=\frac{1.54\times7}{22}$
$=\text{r}^2=0.49$
$=\text{r}=0.7\text{km}=700\text{m}$
View full question & answer→