MCQ 11 Mark
If the difference between the circumference and the radius of a circle is $37 \ cm ,$ then using $\pi=\frac{22}{7}$, the circumference $($in $cm )$ of the circle is:
AnswerLet the radius of the circle be $r$
Then the circumference of the circle $=2 \pi r$
According to the question,
$2 \pi r-r=37$
$\Rightarrow r(2 \pi-1)=37$
$\Rightarrow r\left(2 \times \frac{22}{7}-1\right)=37$
$\Rightarrow r\left(\frac{37}{7}\right)=37 $
$\Rightarrow r=7$
$\Rightarrow \text { Circumference }=2 \pi r=2 \times \frac{22}{7} \times 7=44$
View full question & answer→MCQ 21 Mark
The circumference of a circle is $22 \ cm .$ The area of its quadrant $($in $cm ^2 )$ is
- A
$\frac{77}{2}$
- B
$\frac{77}{4}$
- ✓
$\frac{77}{8}$
- D
$\frac{77}{16}$
AnswerCorrect option: C. $\frac{77}{8}$
Circumference of a circle $=2 \pi r$
$\Rightarrow 22=2 \pi r $
$\Rightarrow 11=\pi r$
$\Rightarrow \frac{22}{7} \times r=11$
$\Rightarrow r=\frac{7}{2} \ cm$
Now, area of a quadrant of a circle $=\frac{1}{4} \pi r^2$
$=\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$=\frac{77}{8} \ cm^2$
View full question & answer→MCQ 31 Mark
If the perimeter of a circle is half to that of a square, then the ratio of the area of the circle to the area of the square is
- A
$22: 7$
- B
$11: 7$
- C
$7: 11$
- ✓
$7: 22$
AnswerCorrect option: D. $7: 22$
Let radius of the circle be $r$ side of the square is $a \ cm$.
According to the question, perimeter of $t$ is half of perimeter of the square.
$\Rightarrow 2 \pi r=\frac{1}{2}(4 a)$
$\Rightarrow r=\frac{2 a}{2 \pi} \text { or } \frac{r}{a}=\frac{1}{\pi}$
$\frac{\text { Area of the circle }}{\text { Area of the square }} =\frac{\pi r^2}{a^2}$
$ =\pi \times \frac{1}{\pi^2}=\frac{1}{\pi} \text { or } \frac{7}{22}$
View full question & answer→MCQ 41 Mark
The area of a square that can be inscribed in a circle of area $\frac{1408}{7} cm^2$ is
- A
$321 cm^2$
- B
$642 cm^2$
- ✓
$128 cm^2$
- D
$256 cm^2$
AnswerCorrect option: C. $128 cm^2$
(c)
Area of circle $=\frac{1408}{7} cm^2$
or $\pi r^2=\frac{1408}{7}$
or $\frac{22}{7} \times r ^2=\frac{1408}{7}$
or $r ^2=\frac{1408}{7} \times \frac{7}{22}$
or $r=\sqrt{64}=8 cm$
Diameter of circle $=2 r=16 cm$
As square is inscribed in the circle, diameter of circle $=$ diagonal of square $=16 cm$
$
\begin{aligned}
\text { Area of square } & =\frac{(\text { diagonal of square })^2}{2} \\
& =\frac{16^2}{2}=\frac{256}{2}=128 cm^2
\end{aligned}
$ View full question & answer→MCQ 51 Mark
The minute hand of a clock is $84 \ cm$ long. The distance covered by the tip of minute hand from $10: 10 \ am$ to $10: 25 \ am$ is
- A
$44 \ cm$
- B
$88 \ cm$
- ✓
$132 \ cm$
- D
$176 \ cm$
AnswerCorrect option: C. $132 \ cm$
Length of minute hand $=$ Radius of the quadrant$/$sector so formed $=84 \ cm$.
In $1$ minute, minute hand makes an angle of $6^{\circ}$.
Therefore, in $15$ minutes it makes and angle of $15 \times 6^{\circ}=90^{\circ}$
Distance covered by the tip of the minute hand
$=\text { Length of arc }=\frac{\theta}{360} \times 2 \pi r$
$=\frac{90^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 84$
$=132 \ cm .$
View full question & answer→MCQ 61 Mark
The area of a quadrant of a circle where the circumference of circle is 176 m , is
- A
$2464 m^2$
- B
$1232 m^2$
- ✓
$616 m^2$
- D
$308 m^2$
AnswerCorrect option: C. $616 m^2$
(c)
$\begin{aligned} \text { Area of quadrant } & =\frac{\theta}{360^{\circ}} \times \pi r ^2 \\ & =\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 28 \times 28 \\ & =616 cm^2\end{aligned}$
View full question & answer→MCQ 71 Mark
Two concentric circles are centred at $O$. The area of shaded region, if outer and inner radii are $14 \ cm$ and $7 \ cm$ respectively, is
- ✓
$462 \ cm^2$
- B
$154 \ cm^2$
- C
$231 \ cm^2$
- D
$308 \ cm^2$
AnswerCorrect option: A. $462 \ cm^2$
Area of shaded region $=$ Area of outer circle $-$ Area of inner circle
$($Let $R =$ radius of outer circle and $r =$ radius of inner circle$)$
$=\pi R^2-\pi r^2=\pi\left(R^2-r^2\right)$
$=\frac{22}{7}\left\{(14)^2-(7)^2\right\}$
$=\frac{22}{7}(196-49)$
$=\frac{22}{7} \times 147$
$=22 \times 21$
$=462$
$\therefore$ Area of shaded region $=462 \ cm^2$
View full question & answer→MCQ 81 Mark
The perimeter of the sector of a circle of radius 14 cm and central angle $45^{\circ}$ is
Answer(d)
$\begin{aligned} & \text { Perimeter of sector of circle } \\ = & \text { length of arc }+2 \times \text { radius } \\ = & \frac{\theta}{360^{\circ}} \times 2 \pi r +2 \times 14 \\ = & \frac{45}{360^{\circ}} \times 2 \times \frac{22}{7} \times 14+28 \\ = & 11+28 \\ = & 39 cm\end{aligned}$
View full question & answer→MCQ 91 Mark
In the given figure, a circle is touching a semicircle at C and its diameter AB at O . If $AB =28$ cm , what is the radius of the inner circle?
Answer(c)
Here $A B$ is the diameter of the semicircle,
So, $AB =28 cm$
OA is the radius; so, $OA = OC =14 cm$
But OC is the diameter of the circle and we know that diameter $=2 \times$ radius
$
\begin{array}{rlrl}
\therefore & O C & =2 \times \text { radius } \\
\Rightarrow & 14 & =2 \times \text { radius } \\
\therefore & & \text { radius } & =\frac{14}{2}=7 cm
\end{array}
$
View full question & answer→MCQ 101 Mark
A circular arc of length $22 \ cm$ subtends an angle $\theta$ at the centre of the circle of radius $21 \ cm .$ The value of $\theta$ is
- A
$90^{\circ}$
- B
$50^{\circ}$
- ✓
$60^{\circ}$
- D
$30^{\circ}$
AnswerCorrect option: C. $60^{\circ}$
Length of arc$=\frac{\theta}{360^{\circ}} \times 2 \pi r$
$\Rightarrow 22=\frac{\theta}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21$
$\Rightarrow \theta=\frac{22 \times 360^{\circ} \times 7}{2 \times 21 \times 22}$
$\therefore \theta=60^{\circ}$
View full question & answer→MCQ 111 Mark
If the radius of a semi-circular protractor is 7 cm , then its perimeter is:
Answer(c)
Perimeter of a semicircle $=\pi r=22 cm$
View full question & answer→MCQ 121 Mark
The length of the arc of a circle of radius 14 cm which subtends an angle of $60^{\circ}$ at the centre of the circle is:
- ✓
$\frac{44}{3} cm$
- B
$\frac{88}{3} cm$
- C
$\frac{308}{3} cm$
- D
$\frac{616}{3} cm$
AnswerCorrect option: A. $\frac{44}{3} cm$
(a)
$l=r \times \theta=\frac{44}{3}$
View full question & answer→MCQ 131 Mark
In a circle of radius $14 \ cm ,$ an arc subtends an angle of $120^{\circ}$ at the centre. If $\sqrt{3}=1.73$ then the area of the segment of the circle is
- ✓
$120.56 \ cm^2$
- B
$124.63 \ cm^2$
- C
$118.24 \ cm^2$
- D
$130.57 \ cm^2$.
AnswerCorrect option: A. $120.56 \ cm^2$
$=\left(\frac{\pi r^2 \theta}{360}-r^2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\right)$
$=\left(\frac{22}{7} \times 14 \times 14 \times \frac{120}{360}\right)-\left(14 \times 14 \times \sin 60^{\circ} \cos 60^{\circ}\right)$
$=\frac{616}{3}-\frac{\sqrt{3}}{2} \times \frac{1}{2} \times 14 \times 14$
$=205.33-49 \times 1.73$
$=205.33-84.77$
$=120.56 \ cm^2$
View full question & answer→MCQ 141 Mark
If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is
- A
$4: \pi$
- ✓
$\pi: 4$
- C
$\pi: 7$
- D
$7: \pi$
AnswerCorrect option: B. $\pi: 4$
According to the question,
$4 a=2 \pi r$
$\Rightarrow \frac{a}{r}=\frac{\pi}{2}$
$=\frac{\text { Area of square }}{\text { Area of circle }}$
$=\frac{a^2}{\pi r^2}=\frac{1}{\pi} \times\left(\frac{\pi}{2}\right)^2$
$=\frac{\pi^2}{4 \pi}$
$=\frac{\pi}{4}$
Hence, required ratio $=\pi: 4$
View full question & answer→MCQ 151 Mark
The area of a circle is equal to the sum of the areas of two circles of radii $24 \ cm$ and $7 \ cm .$ The diameter of the new circle is
- ✓
$25 \ cm$
- B
$31 \ cm$
- C
$50 \ cm$
- D
$62 \ cm$
AnswerCorrect option: A. $25 \ cm$
Area of the new circle
$=\left\{\pi \times(24)^2+\pi(7)^2\right\} \ cm^2$
$=\{\pi \times(576+49)\} \ cm^2$
$=625 \pi \ cm^2$
$\therefore \pi R^2=625 \pi$
$\Rightarrow R^2=625$
$\Rightarrow R=\sqrt{625}=25 \ cm$
$\therefore$ Radius of the new circle $=25 \ cm$
View full question & answer→MCQ 161 Mark
The circumference of a circle is equal to the sum of the circumferences of two circles having diameters $36 \ cm$ and $20 \ cm .$ The radius of the new circle is
- A
$16 \ cm$
- ✓
$28 \ cm$
- C
$42 \ cm$
- D
$56 \ cm$
AnswerCorrect option: B. $28 \ cm$
circumference of new circle
$=(2 \pi \times 18+2 \pi \times 10)$
$\Rightarrow 2 \pi R$
$=2 \pi(18+10)$
$=2 \pi \times 28 \ cm$
View full question & answer→MCQ 171 Mark
The areas of two concentric circles are $1386 \ cm^2$ and $962.5 \ cm^2$. The width of the ring is
- A
$2.8 \ cm$
- ✓
$3.5 \ cm$
- C
$4.2 \ cm$
- D
$3.8 \ cm$
AnswerCorrect option: B. $3.5 \ cm$
According to the question,
$\pi R^2=1386$
$\Rightarrow R^2=1386 \times \frac{7}{22}$
$\Rightarrow R^2=441$
$\Rightarrow R=\sqrt{441}=21 \ cm$
$\pi r^2=962.5$
$\Rightarrow r^2=\frac{9625}{10} \times \frac{7}{22}=\frac{49 \times 25}{4}$
$\Rightarrow r=\frac{7 \times 5}{2}=\frac{35}{2} \ cm$
Width of the ring
$(R-r)=21-\frac{35}{2}=\frac{7}{2}=3.5 \ cm$
View full question & answer→MCQ 181 Mark
The circumference of two circles are in the ratio $3: 4$. The ratio of their areas is
- A
$3: 4$
- B
$4: 3$
- ✓
$9: 16$
- D
$16: 9$
AnswerCorrect option: C. $9: 16$
According to the question,
$\frac{2 \pi R_1}{2 \pi R_2}=\frac{3}{4}$
$\Rightarrow \frac{R_1}{R_2}=\frac{3}{4}$
$\Rightarrow\left(\frac{R_1}{R_2}\right)^2=\left(\frac{3}{4}\right)^2$
$\Rightarrow \frac{R_1{ }^2}{R_2{ }^2}=\frac{9}{16}$
$\Rightarrow \frac{\pi R_1{ }^2}{\pi R_2{ }^2}$
$=\frac{9}{16}$
Hence, required ratio $=9: 16$
View full question & answer→MCQ 191 Mark
The areas of two circles are in the ratio $9: 4$. The ratio of their circumference is
- ✓
$3: 2$
- B
$4: 9$
- C
$2: 3$
- D
$81: 16$
AnswerCorrect option: A. $3: 2$
According to the questions,
$\frac{\pi R_1{ }^2}{\pi R_2{ }^2}=\frac{9}{4}$
$\Rightarrow \frac{R_1{ }^2}{R_2{ }^2}=\frac{9}{4}$
$\Rightarrow \frac{R_1}{R_2}=\sqrt{\frac{9}{4}}$
$\Rightarrow \frac{R_1}{R_2}=\frac{3}{2}$
$\Rightarrow \frac{2 \pi R_1}{2 \pi R_2}$
$=\frac{3}{2}$
Hence, required ratio $=3: 2$
View full question & answer→MCQ 201 Mark
The radius of a wheel is 0.25 m . How many revolutions will it make in covering 11 km .
Answer(d)
Distance moved in 1 revolution $=2 \pi R$
$
=2 \times \frac{22}{7} \times \frac{25}{100}=\frac{11}{7} m
$
Total distance covered $=11 km=(11 \times 1000) m$ $=11000 m$
Number of revolutions $=11000 \times \frac{7}{11}=7000$
View full question & answer→MCQ 211 Mark
The diameter of a wheel is $40 \ cm .$ How many revolutions will it make in covering $176 m ?$
AnswerDistance moved in $1$ revolution
$=\pi d=\left(\frac{22}{7} \times \frac{40}{100}\right) m$
$=\frac{44}{35} m$
Total distance covered $=176 m$
$\therefore$ Number of revolutions
$=176 \times \frac{35}{44}$
$=140$
View full question & answer→MCQ 221 Mark
In making $1000$ revolutions, a wheel covers $88 \ km .$ The diameter of the wheel is
- A
$14 m$
- B
$24 m$
- ✓
$28 m$
- D
$40 m$
AnswerCorrect option: C. $28 m$
Distance moved in $1$ revolution
$=\frac{88000}{1000}=88 m$
$\pi d=88 $
$\Rightarrow \frac{22}{7} \times d=88$
$\Rightarrow d=88 \times \frac{7}{22}$
$=28 m$
View full question & answer→MCQ 231 Mark
The length of a arc of a sector of angle $\theta^{\circ}$ of a circle with radius $R$ is
- A
$\frac{2 \pi R \theta}{180}$
- B
$\frac{\pi R ^2 \theta}{180}$
- ✓
$\frac{2 \pi R \theta}{360}$
- D
$\frac{\pi R ^2 \theta}{360}$
AnswerCorrect option: C. $\frac{2 \pi R \theta}{360}$
(c)
The length of a arc of a sector of angle $\theta^{\circ}$ of a circle with Radius $R$ is $\frac{2 \pi R \theta}{360}$.
View full question & answer→MCQ 241 Mark
The area of a square is the same as the area of a circles. Their perimeters are in the ratio
- A
$1: 1$
- B
$2: \pi$
- C
$\pi: 2$
- ✓
$\sqrt{\pi}: 2$
AnswerCorrect option: D. $\sqrt{\pi}: 2$
Let the side of a square be $'a\ '$ unit and radius of circle be $R$ units.
$\therefore a^2=\pi R^2 $
$\Rightarrow \frac{R^2}{a^2}=\frac{1}{\pi}$
$\Rightarrow \frac{R}{a}=\frac{1}{\sqrt{\pi}}$
Ratio of their perimeters $=\frac{2 \pi R}{4 a }=\frac{\pi}{2} \times \frac{ R }{ a }$
$=\frac{\pi}{2} \times \frac{1}{\sqrt{\pi}}=\frac{\sqrt{\pi}}{2}$
$=\sqrt{\pi}: 2$
View full question & answer→MCQ 251 Mark
On decreasing the radius of a circle by $30 \%$, its area is decreased by
- A
$30 \%$
- B
$60 \%$
- C
$45 \%$
- ✓
$51 \%$
AnswerCorrect option: D. $51 \%$
(d)
Let the original radius be 100 units.
$\therefore$ Then, original area $=\pi R^2=\pi \times(100)^2$ $=10000 \pi$
New radius $=70$ units.
New area $=\pi \times 70^2=4900 \pi$
Decrease $\%=\left(\frac{10000 \pi-4900 \pi}{10000 \pi} \times 100\right) \%$
$
=\frac{5100}{10000} \times 100=51 \%
$
View full question & answer→MCQ 261 Mark
On increasing the diameter of a circle by $40 \%$, its area will be increased by
- A
$40 \%$
- B
$80 \%$
- ✓
$96 \%$
- D
$82 \%$
AnswerCorrect option: C. $96 \%$
(c)
Let original diameter be 100 units. Then, original radius $=50$
Original area $=\pi R ^2=\pi \times 50^2=2500 \pi$.
New diameter = 140
New radius $=70$
$\therefore$ New area $=\pi(70)^2=4900 \pi$.
Increase $\%=\left(\frac{2400}{2500} \times 100\right) \%=96 \%$
View full question & answer→MCQ 271 Mark
The area of a sector of angle $0^{\circ}$ of a circle with radius $R$ is
- A
$\frac{2 \pi R \theta}{360}$
- ✓
$\frac{\pi R ^2 \theta}{360}$
- C
$\frac{\pi R ^2 \theta}{180}$
- D
$\frac{2 \pi r \theta}{360}$
AnswerCorrect option: B. $\frac{\pi R ^2 \theta}{360}$
(b)
Area of a sector of angle $\theta^{\circ}$ of a circle with Radius
$
R=\frac{\pi R^2 \theta}{360}
$
View full question & answer→MCQ 281 Mark
In a circle of radius $21 \ cm$ , an arc subtends an angle of $60^{\circ}$ at the centre. The length of the are is
- A
$21 \ cm$
- ✓
$22 \ cm$
- C
$18.16 \ cm$
- D
$23.5 \ cm$
AnswerCorrect option: B. $22 \ cm$
$\text { Arc length }=\frac{2 \pi r \theta}{360}$
$=2 \times \frac{22}{7} \times 21 \times \frac{60}{360}$
$=22 \ cm$
View full question & answer→MCQ 291 Mark
The perimeter of a circular field is $242 \ m$ . The area of the field is
- A
$9317 m^2$
- B
$18634 m^2$
- ✓
$4658.5 m^8$
- D
$9000 m^3$.
AnswerCorrect option: C. $4658.5 m^8$
According to the question,
$2 \pi R=242$
$\Rightarrow 2 \frac{22}{7} \times R=242$
$\Rightarrow R=242 \times \frac{7}{44}=\frac{77}{2} m$
$\therefore \text { Area }=\pi R^2=\frac{22}{7} \times \frac{77}{2} \times \frac{77}{2}=4658.5 m^2$
View full question & answer→MCQ 301 Mark
The difference between the circumference and radius of a circle is $37 \ cm$ . The area of the circle is
- A
$111 \ cm^2$
- B
$184 \ cm^2$
- ✓
$154 \ cm^2$
- D
$259 \ cm^2$.
AnswerCorrect option: C. $154 \ cm^2$
According to the question,
$2 \pi R-R=37$
$\Rightarrow R\left(2 \times \frac{22}{7}-1\right)=37$
$\Rightarrow R=37 \times \frac{7}{37}=7 \ cm$
$\text { Area }=\pi R^2=\frac{22}{7} \times 7 \times 7=154 \ cm^2$
View full question & answer→MCQ 311 Mark
The area of a circle is $49 \pi^2$. Its circumference is
- A
$7 \pi \ cm$
- ✓
$14 \pi \ cm$
- C
$21 \pi \ cm$
- D
$28 \ cm$
AnswerCorrect option: B. $14 \pi \ cm$
$\because \text { Area of circle }=49 \pi$
$\pi R ^2=49 \pi \Rightarrow R =\sqrt{49}=7$
$\therefore \text { Circumference }=2 \pi R =2 \times \pi \times 7=14 \pi \ cm$
View full question & answer→MCQ 321 Mark
The area of a circle is $38.5 \ cm^2$. The circumference of the circle is
- A
$6.2 \ cm$
- B
$12.1 \ cm$
- C
$11 \ cm$
- ✓
$22 \ cm$
AnswerCorrect option: D. $22 \ cm$
$\because \text { Area of circle }=38.5 \ cm^2$
$\pi R ^2=38.5$
$\Rightarrow \frac{22}{7} \times R ^2=38.5$
$\Rightarrow R ^2=\frac{38.5 \times 7}{22}$
$\Rightarrow R ^2=\frac{49}{4} \Rightarrow R =\frac{7}{2} \ cm$
$\text { Circumference }=2 \pi R =2 \times \frac{22}{7} \times \frac{7}{2}=22 \ cm$
View full question & answer→