MCQ 511 Mark
Area of a sector of angle $p \ ($in degrees$)$ of a circle with radius $R$ is :
- ✓
$\frac{\text{p}}{720}\times2\pi\text{R}^2$
- B
$\frac{\text{p}}{180}\times2\pi\text{R}$
- C
$\frac{\text{p}}{360}\times2\pi\text{R}$
- D
$\frac{\text{p}}{180}\times\pi\text{R}^2$
AnswerCorrect option: A. $\frac{\text{p}}{720}\times2\pi\text{R}^2$
Area of the sector of angle $p$ of a circle with radius $R$
$=\frac{\theta}{360}\times\pi\text{r}^2=\frac{\text{p}}{360}\times\pi\text{R}^2$
$=\frac{\text{p}}{2(360)}\times2\pi\text{R}^2$
$=\frac{\text{p}}{720}\times2\pi\text{R}^2$
View full question & answer→MCQ 521 Mark
The perimeter of a protractor is :
- A
$\pi\text{r}$
- B
$\pi+2\text{r}$
- C
$\pi+\text{r}$
- ✓
$\pi\text{r}+2\text{r}$
AnswerCorrect option: D. $\pi\text{r}+2\text{r}$
Let the radius of the protractor be $r$
$\therefore$ The perimeter of protractor $=$ Perimeter of semicircle $+$ Diameter of the semicircle
$\Rightarrow $The perimeter of protractor $=\pi\text{r}+2\text{r}$
View full question & answer→MCQ 531 Mark
The part of the circular region enclosed by two radii and the corresponding arc of a circle is called :
Answer
The part of the circular region enclosed by two radii and the corresponding arc of a circle is called a sector.
View full question & answer→MCQ 541 Mark
If the area of a sector of a circle bounded by an arc of length $5\pi\text{ cm}$ is equal to $20\pi\text{ cm}^2,$ then the radius of the circle :
- A
$12\ cm$
- B
$16\ cm$
- ✓
$8\ cm$
- D
$10\ cm$
AnswerCorrect option: C. $8\ cm$
Length of arc $=5\pi\text{ cm}$
area of sector $20\pi\text{ cm}^2$
Let the angle at the centre be $\theta$
then, $2\pi\text{r}\times\frac{\theta}{360^\circ}=5\pi$
$\text{r}\times\frac{\theta}{360^\circ}=\frac{5\pi}{2\pi}=\frac{5}{2}\ \dots(\text{i})$
Area $=\pi\text{r}\frac{\theta}{360^\circ}$
$\therefore\pi\text{r}^2\frac{\theta}{360^\circ}=20\pi$
$\pi\text{r}.\text{r}\frac{\theta}{360^\circ}=20\pi$
$\pi\text{r}\times\frac{5}{2}=20\pi$
$\Rightarrow\text{r}=\frac{20\pi\times2}{5\pi}$
$\Rightarrow\text{r}=8$
$\therefore$ Radius of the circle $= 8\ cm \ (c)$
View full question & answer→MCQ 551 Mark
The circumference of a circle is equal to the sum of the circumference 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. $28\ cm$
Let rem be the radius of the new circle.
We know :
Circumference of the new circle $=$ Circumference of the circle with diameter $36\ cm\ +$ Circumference of the circle with diameter $20\ cm$
Thus, we have :
$2\pi\text{r}=2\pi\text{r}_1+2\pi\text{r}_2$
$\Rightarrow2\pi\text{r}=(2\pi\times18)+(2\pi\times10)$
$\Rightarrow2\pi\text{r}=2\pi\times(18+10)$
$\Rightarrow2\pi\text{r}=(2\pi\times28)$
$\Rightarrow2\pi\text{r}=\Big(2\times\frac{22}{7}\times28\Big)$
$\Rightarrow2\pi\text{r}=176$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=176$
$\Rightarrow\text{r}=\Big(176\times\frac{7}{44}\Big)$
$\Rightarrow\text{r}=28\text{ cm}$
View full question & answer→MCQ 561 Mark
Tick the correct answer in the following : Area of a sector of angle $p\ ($in degrees$)$ of a circle with radius $R$ is,
- A
$\frac{\text{p}}{180}\times2\pi\text{R}$
- B
$\frac{\text{p}}{180}\times2\pi\text{R}^2$
- C
$\frac{\text{p}}{360}\times2\pi\text{R}$
- ✓
$\frac{\text{p}}{720}\times2\pi\text{R}^2$
AnswerCorrect option: D. $\frac{\text{p}}{720}\times2\pi\text{R}^2$
$\frac{\text{p}}{360^\circ}\times\pi\text{R}^2$
$=\frac{\text{p}}{720}\times2\pi\text{R}^2$
View full question & answer→MCQ 571 Mark
The length of the wire is $66m$. The number of circles of circumference $13.2\ cm$ can be made from the wire is :
AnswerGiven : Length of wire $= 66m = 6600\ cm$ and Circumference $= 13.2\ cm$
Number of circles $=\frac{\text{Length of wire}}{\text{Circumference of one circle}}$
$=\frac{6600}{13.5}=500$
View full question & answer→MCQ 581 Mark
The perimeter of a circular field is $242m$. The area of the field is :
- A
$9317 \mathrm{~m}^2$
- B
$18634 \mathrm{~m}^2$
- ✓
$658.5 \mathrm{~m}^2$
- D
AnswerCorrect option: C. $658.5 \mathrm{~m}^2$
Let the radius be $r \ cm.$
We know,
Cirumference of the circle $=2\pi\text{r}$
Thus, we have :
$2\pi\text{r}^2=242$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=242$
$\Rightarrow\frac{44}{7}\times\text{r}=242$
$\Rightarrow\text{r}=\Big(242\times\frac{7}{44}\Big)$
$\Rightarrow\text{r}=\frac{77}{2}$
$\therefore$ Area of the circle $=\pi\text{r}^2$
$=\Big(\frac{22}{7}\times\frac{77}{2}\times\frac{77}{2}\Big)\text{m}^2$
$=4658.5\text{m}^2$
View full question & answer→MCQ 591 Mark
The area of the square that can be inscribed in a circle of radius $12\ cm$ is :
- A
$576 \text{ sq.cm}$
- B
$500 \text{ sq.cm}$
- C
$144 \text{ sq.cm}$
- ✓
$288 \text{ sq.cm}$
AnswerCorrect option: D. $288 \text{ sq.cm}$
According to question, Diameter $(BD) = 2 \ \times $ radius $= 2 \times 12 = 24\ cm$
$\therefore\text{Area of square}=\frac{1}{2}\times\text{d}^2$
$=\frac{1}{2}\times24\times24=288\text{ sq.cm}$ View full question & answer→MCQ 601 Mark
Area of the largest triangle that can be inscribed in a semi $-$ circle of radius $r$ units is :
- ✓
$\text{r}^2\text{ sq. units}$
- B
$\frac{1}{2}\text{r}^2\text{ sq. units}$
- C
$2\text{r}^2\text{ sq. units}$
- D
$\sqrt{2}\text{r}^2\text{ sq. units}$
AnswerCorrect option: A. $\text{r}^2\text{ sq. units}$
Take a point $C$ on the circumference of the semi $-$ circle and join it by the end points of diameter $A$ and $B.$
$\therefore\angle\text{C}=90^\circ\ [$by property of circle$]$
$[$angle in a semi $-$ circle are right angle$]$
So, $\triangle\text{ABC}$ is right angled triangle.
$\therefore$ Area of largest $\triangle\text{ABC}=\frac{1}{2}\times\text{AB}\times\text{CD}$
$=\frac{1}{2}\times2\text{r}\times\text{r}$
$\text{r}^2\text{ sq. units}\ \text{(a)}$ View full question & answer→MCQ 611 Mark
Tick the correct answer in the following and justify your choice : If the perimeter and thearea of a circle are numerically equal, then the radius of the circle is :
- ✓
$2$ units
- B
$\pi$ units
- C
$4$ units
- D
$7$ units
AnswerCorrect option: A. $2$ units
Circumference of circle $= $ Area of circle
$\Rightarrow\ 2\pi\text{r}-\pi\text{r}^2$
$\Rightarrow\ \text{r}=2\text{ units}$
View full question & answer→MCQ 621 Mark
The area of a square is the same as the area of a circle. 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 a be side of the square.
We know
Area of a square $= a^2$
Let $r$ be the radius of the circle.
We know :
Area of a circle $=\pi\text{r}^2$
Because the area of the square is the same as the area of the circle, we have:
$\text{a}^32=\pi\text{r}^2$
$\Rightarrow\frac{\text{r}^2}{\text{a}^2}=\frac{1}{\pi}$
$\Rightarrow\frac{\text{r}}{\text{a}}=\frac{1}{\sqrt{\pi}}$
$\therefore$ Ratio of their perimeters
$=\frac{2\pi\text{r}}{4\text{a}}$
$\big[$because perimeter o the circle is $2\pi\text{r}$ and perimeter of the square. is $4a\big]$
$=\frac{\pi\text{r}}{2\text{a}}$
$=\frac{\pi}{2}\times\frac{\text{r}}{\text{a}}$
$=\frac{\pi}{2}\times\frac{1}{\sqrt{\pi}}\Big[\text{Since}\frac{\text{r}}{\text{a}}=\frac{1}{\sqrt{\pi}}\Big]$
$=\frac{\sqrt{\pi}}{2}$
$=\sqrt{\pi}:2$
View full question & answer→MCQ 631 Mark
The circumference of a circle is $100m$. The side of a square inscribed in the circle is :
AnswerCorrect option: D. $\frac{100\sqrt{2}}{\pi}$
We have given the circumference of the circle that is $100\ cm$.
If $d$ is the diameter of the circle, then
its circumference will be $\pi\text{d}.$
$\therefore\pi\text{d}=100$
$\therefore\text{d}=\frac{100}{\pi}$
We obtained diameter of the circle.
Look at the figure, diameter of the circle is also the diagonal of the square $\text{ABCD}.$
We know that if we have diagonal of the circle we can calculate the side of the square, using the formula given below,
$\text{side=}\sqrt{2}\times\text{diagonal}$
Substituting the value of diagonal we get,
$\text{side}=\sqrt{2}\times\frac{100}{\pi}$
Therefore, side of the inscribed square is $\frac{100\sqrt{2}}{\pi}\text{cm}$
Hence, the correct answer is option $(d).$ View full question & answer→MCQ 641 Mark
The perimeter of the sector $\text{OAB}$ shown in the following figure, is :
- ✓
$\frac{64}{3}\text{ cm}$
- B
$26\text{ cm}$
- C
$\frac{64}{5}\text{ cm}$
- D
$19\text{ cm}$
AnswerCorrect option: A. $\frac{64}{3}\text{ cm}$
Radius of sector of $60^\circ = 7\ cm$
$\therefore$ perimeter $=$ arc $AB + 2r$
$=2\pi\text{r}\times\frac{60}{360}+2\times7$
$=2\times\frac{22}{7}\times7\times\frac{1}{6}+14$
$=\frac{22}{3}+14=\frac{64}{3}\text{ cm}\ \text{(a)}$
View full question & answer→MCQ 651 Mark
The perimeter of a semicircular protractor whose radius is $7\ cm$ is :
- A
$72\ cm$
- B
$27\ cm$
- ✓
$36\ cm$
- D
$18\ cm$
AnswerCorrect option: C. $36\ cm$
The perimeter of circular protractor
$\pi\text{r}+\text{2r}=\frac{22}{7}\times7+2\times7=36\text{ cm}$
View full question & answer→MCQ 661 Mark
The area of the largest triangle that can be inscribed in a semi $-$ circle of radius $r,$ is :
- ✓
$r^2 $
- B
$ 2 r^2 $
- C
$ r^3 $
- D
$ 2 r^3 $
AnswerCorrect option: A. $r^2 $
The largest triangle inscribed in a semi $-$ circle of radius $r,$ can be $\triangle\text{ABC}$ as shown in the figure,
whose base $= AB = 2r$
and altitude $OC = r$
Area of triangle $=\frac{1}{2}\text{base}\times\text{altitude}$
$=\frac{1}{2}\times2\text{r}\times\text{r}=\text{r}^2\ (a)$ View full question & answer→MCQ 671 Mark
Choose the correct answer from the given four options : If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius $R,$ then
- ✓
$\mathrm{R}_1+\mathrm{R}_2= R.$
- B
$\mathrm{R}_1+\mathrm{R}_2 > R.$
- C
$\mathrm{R}_1+\mathrm{R}_2 < R.$
- D
Nothing definite can be said about the relation among $R_1,R_2$ and $R$.
AnswerCorrect option: A. $\mathrm{R}_1+\mathrm{R}_2= R.$
According to the given condition,
Circumference of circle $=$ Circumference of first $+$ Circumference of second circle
$\therefore\ \ 2\pi\text{R}=2\pi\text{R}_1+2\pi\text{R}_2$
$\Rightarrow\ \ \text{R}=\text{R}_1+\text{R}_2$
View full question & answer→MCQ 681 Mark
If the circumference of a circle and the perimeter of a square are equal, then :
- A
Area of the circle $ < $ area of the square
- B
Area of the circle $= \frac{1}{2}$ Area of the square
- C
Area of the circle $=$ area of the square
- ✓
Area of the circle $ > $ area of the square
AnswerCorrect option: D. Area of the circle $ > $ area of the square
Let the radius of the circle be $r$ and side of the square be a.
Then, according to the question,
$2\pi\text{r}=4\text{a}$
$\Rightarrow\text{a}=\frac{2\pi\text{r}}{4}=\frac{\pi\text{r}}{2}...\text{(i)} $
Now, the ratio of their areas,
$\pi\text{r}^2$ and $\text{a}^2$
$\Rightarrow\pi\text{r}^2$ and $\Big(\frac{\pi\text{r}}{2}\Big)^2\ [$From eq.$(i)]$
$\Rightarrow\pi\text{r }$ and $\frac{\pi^2\text{r}^2}{4}$
Therefore, Area of the circle $ > $ Area of the square
View full question & answer→MCQ 691 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. $154\ cm^2$
Let the radius be $r \ cm$.
We know,
cirumference of the circle $=2\pi\text{r}$
Thus, we have :
$2\pi\text{r}^2-\text{r}=37$
$\Rightarrow\text{r}(2\pi-1)=37$
$\Rightarrow\text{r}\Big(2\times\frac{22}{7}-1\Big)=37$
$\Rightarrow\text{r}\Big(\frac{37}{7}\Big)=37$
$\Rightarrow\text{r}=\Big(37\times\frac{7}{37}\Big)$
$\Rightarrow\text{r}=7\text{ cm}$
Now,
Circumference of the circle $=\pi\text{r}^2$
$=\Big(\frac{22}{7}\times7\times7\Big)\text{ cm}^2$
$=154\text{ cm}^2$
View full question & answer→MCQ 701 Mark
The area of a circle is $2464 \text{ sq.cm},$ then its diameter is given by :
- ✓
$28\ cm$
- B
$7\ cm$
- C
$56\ cm$
- D
$14\ cm$
AnswerCorrect option: A. $28\ cm$
Given : Area of circle $= 2464 \text{ sq.cm}$
$\Rightarrow\pi\text{r}^2=2464$
$\Rightarrow\text{r}^2=\frac{2464}{22}\times 7=784$
$\Rightarrow\text{r}=28\text{ cm}$
View full question & answer→MCQ 711 Mark
If a chord of a circle of radius $14\ cm$ subtends a right angle at the centre of the circle, then the area of the sector is :
- A
$128 \text{ sq.cm}$
- ✓
$154 \text{ sq.cm}$
- C
$156 \text{ sq.cm}$
- D
$142 \text{ sq.cm}$
AnswerCorrect option: B. $154 \text{ sq.cm}$
Area of the sector $=\frac{\theta}{360^\circ}\times\pi\text{r}^2$
$\Rightarrow $ Area of the sector $=\frac{90^\circ}{360^\circ}\times\frac{22}{7}\times(14)^2$
$\Rightarrow $ Area of the sector $=\frac{1}{4}\times\frac{22}{7}\times14\times14$
$\Rightarrow $ Area of the sector $=154\text{ sq.cm}$ View full question & answer→MCQ 721 Mark
The area of a circle is $220\ cm^2$. The area of ta square inscribed in it is
- A
$49\ cm^2$
- B
$70\ cm^2$
- ✓
$140\ cm^2$
- D
$150\ cm^2$
AnswerCorrect option: C. $140\ cm^2$
Area of a circle $= 220\ cm^2$
$\therefore$ Radius $(r) =\sqrt{\frac{\text{A}}{\pi}}=\sqrt{\frac{220\times7}{22}}$
$=\sqrt{70}\text{ cm}$
Diagonal of square $=$ diameter of the circle
$=2\times\text{r}=2\times\sqrt{70}\text{ cm}$
$\therefore$ Area of square $= \Big(\frac{\text{Diagonal}}{\sqrt{2}}\Big)^2=\Big(\frac{2\sqrt{70}}{\sqrt{2}}\Big)^2$
$=\frac{4\times70}{2}=140\text{ cm}^2 \ \text{(c)}$
View full question & answer→MCQ 731 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. $120.56\ cm^2$
Radius of the circle $, r = 14\ cm$
Draw a perpendicular $OD$ to chord $AB$. It will bisect $AB$.
$\angle\text{A}=180^\circ-(90^\circ+60^\circ)=30^\circ$
$\cos30^\circ=\frac{\text{AD}}{\text{OA}}$
$\Rightarrow\frac{\sqrt{3}}{2}=\frac{\text{AD}}{14}$
$\Rightarrow\text{AD}=7\sqrt{3}$
$\Rightarrow\text{AB}=2\times\text{AD}=14\sqrt{3}\text{ cm}$
$\sin30^\circ=\frac{\text{OD}}{14}$
$\Rightarrow\text{OD}=7\text{ cm}$
Area of minor segment $=$ Area of sector $\text{OAPB}\ -$ Area of tringle $\text{AOB}$
$=\frac{\theta}{360^\circ}\pi(\text{OA})^2-\frac{1}{2}\times\text{OD}\times\text{AB}$
$=\frac{120^\circ}{360^\circ}\times\frac{22}{7}(14)^2-\frac{1}{2}\times7\times14\sqrt{3}$
$=205.33-84.77$
$=120.56\text{ cm}^2$
Hence, the correct answer is option $(a).$ View full question & answer→MCQ 741 Mark
One side of a rhombus is $20\ cm$ long and one of its diagonals measure $24\ cm.$ The area of the rhombus is :
- A
$192\ cm^2$
- B
$480\ cm^2$
- C
$240\ cm^2$
- ✓
$384\ cm^2$
AnswerCorrect option: D. $384\ cm^2$
We have,
$AB = BC = CD = DA = 20\ cm$ and $BD = 24\ cm$
Also, $\text{BO}=\frac{\text{BD}}{2}=\frac{24}{2}=12\text{ cm}$
In $\triangle\text{AOB},$
Using Pythagoras theorem
$\text{AO}^2=\text{AB}^2-\text{BO}^2$
$=20^2-12^2$
$=400-144$
$\Rightarrow\text{AO}^2=256$
$\Rightarrow\text{AO}=\sqrt{256}$
$\Rightarrow\text{AO}=16\text{ cm}$
$\Rightarrow\text{AC}=2\text{AO}=2\times16=32\text{ cm}$
Now, the area of the rhombus $\text{ABCD} =\frac{1}{2}\times\text{AC}\times\text{BD}$
$=\frac{1}{2}\times32\times24$
$=384\text{ cm}^2$ View full question & answer→MCQ 751 Mark
The ratio of the outer and inner perimeters of a circular path is $23 : 22$. If the path is $5$ metres wide, the diameter of the inner circle is :
- A
$55m$
- B
$110m$
- ✓
$220m$
- D
$230m$
AnswerCorrect option: C. $220m$
Ratio in the outer and inner perimeter of a circular path $= 23 : 22$
Width of path $= 5m$
Let $R$ and $r$ be the radii of outer and inner path then $R- r = 5m ….(i)$
and $\frac{2\pi\text{R}}{2\pi\text{r}}=\frac{23}{22}$
$\Rightarrow\frac{\text{R}}{\text{r}}=\frac{23}{22}$
$\Rightarrow22\text{R}=23\text{r}$
$\Rightarrow\text{R}=\frac{23}{22}\text{r}$
$\therefore$ From $(i)$
$\frac{23}{22}\text{r}-\text{r}=5$
$\Rightarrow\text{r}\Big(\frac{23}{22}-1\Big)=5$
$\Rightarrow\text{r}\Big(\frac{1}{22}\Big)=5$
$\Rightarrow\text{r}=5\times22=110\text{m}$
$\therefore$ Diameter of inner circle $= 2r = 2 \times 110$
$= 220m \ \ \ (c)$
View full question & answer→MCQ 761 Mark
The area of the sector of angle $\theta^\circ$ of a circle with radius $R$ is :
- A
$\frac{2\pi\text{R}\theta}{180}$
- B
$\frac{\pi\text{R}^2\theta}{180}$
- C
$\frac{2\pi\text{R}\theta}{360}$
- ✓
$\frac{\pi\text{R}^2\theta}{360}$
AnswerCorrect option: D. $\frac{\pi\text{R}^2\theta}{360}$
$\frac{\pi\text{R}^2\theta}{360}$
View full question & answer→MCQ 771 Mark
Choose the correct answer from the given four options : If the circumference of a circle and the perimeter of a square are equal, then :
- A
Area of the circle $ =$ Area of the square.
- ✓
Area of the circle $ > $ Area of the square.
- C
Area of the circle $ < $ Area of the square.
- D
Nothing definite can be said about the relation between the areas of the circle and square.
AnswerCorrect option: B. Area of the circle $ > $ Area of the square.
According to the given condition,
Circumference of a circle $=$ Perimeter of square
$2\pi\text{r}=4\text{a}$
$[$where $,r$ and $a$ are radius of circle and side of square respectively$]$
$\Rightarrow\ \frac{22}{7}\text{r}=2\text{a}$
$\Rightarrow11\text{r}=7\text{a}$
$\Rightarrow\ \text{a}=\frac{11}{7}\text{a}$
$\Rightarrow\text{r}=\frac{7\text{a}}{11}\ \dots(\text{i})$
Now, area of circle, $\text{A}_1=\pi\text{r}^2$
$=\pi\Big(\frac{7\text{a}}{11}\Big)^2=\frac{22}{7}\times\frac{49\text{a}^2}{121}\ \ [\text{from Eq. (i)}]$
$=\frac{14\text{a}^2}{11}\ \dots(\text{ii})$
and area of square, $A_2= (a)_2...(iii)$
From Eqs. $(ii)$ and $(iii), \text{A}_1=\frac{14}{11}\text{A}_2$
$\therefore\ \ \text{A}_1>\text{A}_2$
Hence, Area of the circle $ > $ Area of the square.
View full question & answer→MCQ 781 Mark
If he area of a sector of a circle is $\frac{7}{20}$ of the area of the circle, then the sector angle is equal to :
- A
$110^\circ$
- B
$130^\circ$
- C
$100^\circ$
- ✓
$126^\circ$
AnswerCorrect option: D. $126^\circ$
We have given that area of the sector is $\frac{7}{20}$ of the area of the circle.
Therefore, area of the sector $=\frac{7}{20}\text{x}$ area of the circle.
$\therefore\frac{\theta}{360}\times\pi\text{r}^2=\frac{7}{20}\times\pi\text{r}^2$
Now we will simplify the equation as below,
$\frac{\theta}{360}=\frac{7}{20}$
Now we will multiply both sides of the equation by $360,$
$\therefore\theta=\frac{7}{20}\times360$
$\therefore\theta=126$
Therefore, sector angle is $126^\circ .$
Hence, the correct answer is option $(d)$.
View full question & answer→MCQ 791 Mark
The area of incircle of an equilateral triangle is $154\ cm^2$. The perimeter of the triangle is :
- A
$71.5\ cm$
- B
$71.7\ cm$
- C
$72.3\ cm$
- ✓
$72.7\ cm$
AnswerCorrect option: D. $72.7\ cm$
Area of incircle of equilateral triangle is $154\ cm^2$
We have to find the perimeter of the triangle.
So we will use area to get,
Area of incircle $= 154$
$\pi\text{r}^2=154$
$\text{r}=\sqrt{\frac{154}{\pi}}\text{ cm}$
As triangle is equilateral so,
$\angle\text{OCM}=30^\circ$
So,
$\tan30^\circ=\frac{\text{r}}{\text{MC}}$
$\text{MC}=\sqrt{\frac{154(3)}{\pi}}\text{ cm}$
So,
$\text{AC}=2\text{(MC)}$
$=2\bigg(\sqrt{\frac{154(3)}{\pi}}\bigg)\text{cm}$
Therefore perimeter of the triangle is,
$=3\text{(AC)}$
$=6\Big(\sqrt{\frac{462}{3.14}}\Big)$
$=72.7\text{ cm}$
Therefore the answer is $(d)$. View full question & answer→MCQ 801 Mark
The perimeter of a sector of a circle with radius $'r’$ and with sector angle $\theta$ is given by :
- A
$2\text{r}+\frac{\theta}{360}3\pi\text{r}$
- B
$2\text{r}+\frac{\theta}{180}2\pi\text{r}$
- C
$\frac{2\pi\text{r}\theta}{360}$
- ✓
$2\text{r}+\frac{\theta}{360}2\pi\text{r}$
AnswerCorrect option: D. $2\text{r}+\frac{\theta}{360}2\pi\text{r}$
Perimeter of sector $= 2 \ \times $ Radius $+$ Length of arc
$\Rightarrow$ Perimeter of sector $=2\text{r}+\frac{0}{360^\circ}\times2\pi\text{r}$
View full question & answer→MCQ 811 Mark
If the difference between the circumference and radius of a circle is $37\ cm, $ then using $\pi=\frac{22}{7}$ the circumference $($in $\ cm)$ of the circle is :
AnswerWe know that circumference; $C$ of the circle with radius $r$ is equal to $2\pi\text{r}$
We have given difference between circumference and radius of the citcle that is $37\ cm,$
$\therefore\text{C}-\text{r}=2\pi\text{r}-\text{r}$
$\therefore(2\pi-1)\text{r}=37$
Substituting we $\pi=\frac{22}{7}$ get,
$\therefore\Big(2\times\frac{22}{7}-1\Big)\text{r}=37$
$\therefore\Big(\frac{44-7}{7}\Big)\text{r}=37$
$\therefore\Big(\frac{37}{7}\Big)\text{r}=37$
Dividing both sides of the equation by $\frac{7}{37},$ we get,
$\therefore\text{r}=7$
Threfore, circumference of the circle will be
$2\pi\text{r}=2\times\frac{22}{7}\times7$
$=44\text{ cm}^2$
Hence, the correct choice is $(b).$
View full question & answer→MCQ 821 Mark
The area of a circle inscribed in a square of side $'a\ ’$ units is :
- A
$\pi\text{a}^2 \text { sq units}$
- B
$\frac{1}{3}\pi\text{a}^2\text{ sq units}$
- C
$\frac{1}{2}\pi\text{a}^2\text{ sq units}$
- ✓
$\frac{1}{4}\pi\text{a}^2\text{ sq units}$
AnswerCorrect option: D. $\frac{1}{4}\pi\text{a}^2\text{ sq units}$
According to the question,
Diameter of circle $=$ side of a square
$\Rightarrow\text{d}=\text{a}$
$\Rightarrow\text{r}=\frac{\text{a}}{2}$
Now, Area of the circle $=\pi\text{r}^2=\pi(\frac{\text{a}}{2})^2$
$\Rightarrow$ Area of the circle $=\frac{1}{4}\pi\text{a}^2\text{ sq.units}$
View full question & answer→MCQ 831 Mark
The area of a circle is $49\pi\text{ cm}^2.$ Its circumference is :
- A
$7\pi\text{ cm}$
- ✓
$14\pi\text{ cm}$
- C
$21\pi\text{ cm}$
- D
$28\pi\text{ cm}$
AnswerCorrect option: B. $14\pi\text{ cm}$
Let the radius be $r \ cm.$
We know,
Area of a circle
$=\pi\text{r}^2$
Thus, we have :
$\pi\text{r}^2=49$
$\Rightarrow\text{r}^2=49$
$\Rightarrow\text{r}^2=\sqrt{49}$
$\Rightarrow\text{r}=7$
Now,
Circumference of the circle $=2\pi\text{r}$
$=\Big(2\times\frac{22}{7}\times\frac{7}{2}\Big)\text{ cm}$
$=14\pi\text{ cm}$
View full question & answer→MCQ 841 Mark
Choose the correct answer from the given four options : It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters $16m$ and $12m$ in a locality. The radius of the new park would be :
AnswerArea of first circular park, whose diameter is $16m$
$=\pi\text{r}^2=\pi\Big(\frac{16}{2}\Big)^2=64\pi\text{m}^2$
$\Big[\because\text{r}=\frac{\text{d}}{2}=\frac{16}{2}=8\text{m}\Big]$
Area of second circular park, whose diameter is $12m$
$=\pi\Big(\frac{12}{2}\Big)^2=\pi(6)^2=36\pi\text{m}^2$
$\Big[\because\text{r}=\frac{\text{d}}{2}=\frac{12}{2}=6\text{m}\Big]$
According to the given condition,
Area of single circular park $=$ Area of first circular park $+$ Area of second circular park
$\pi\text{R}^2=64\pi+36\pi\ [\because R$ be the radius of single circular park$]$
$\Rightarrow\ \ \pi\text{R}^2=100\pi$
$\Rightarrow\text{R}^2=100$
$\therefore\ \ \text{R}=10\text{m}$
View full question & answer→MCQ 851 Mark
The area of a sector of a circle with a sector angle $\theta$ is given by :
- A
$\frac{2\pi\text{}r}{360}$
- B
$\frac{2\pi\text{}r^2\theta}{180}$
- C
$\frac{\pi\text{r}\theta}{360}$
- ✓
$\frac{2\pi\text{r}^2\theta}{360}$
AnswerCorrect option: D. $\frac{2\pi\text{r}^2\theta}{360}$
The area of a sector of a circle with sector angle $\theta$ is given by$\frac{2\pi\text{}r^2\theta}{360^\circ},$
where $r =$ radius of the circle.
View full question & answer→MCQ 861 Mark
If the circumference of a circle increases from $4\pi$ to $8\pi,$ then its area is :
AnswerLet the circumference $\text{C}=4\pi$
$\therefore2\pi\text{r}=4\pi$
$\therefore\text{r}=2$
Therefore, area of the circle when radius of the circle is $2$ can be calculated as below,
$\pi\text{r}^2=\frac{22}{7}\times4\ \dots(1)$
Now when circumference is $\text{C}=8\pi,$
then the radius of the circle is calculated as below,
$\therefore2\pi\text{r}=8\pi$
$\therefore\text{r}=4$
Therefore, area of the circle when radius of the circle is $4$ can be calculated
as below $, \pi\text{r}^2=\frac{22}{7}\times16$
$\therefore\pi\text{r}^2=4\Big(\frac{22}{7}\times4\Big) \dots(2)$
Therefore, from equation $(1)$ and $(2) $ we can say that its area is quadrupled.
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 871 Mark
The area of a sector of a circle with radius $21\ cm$ and sector angle $120^\circ$ is :
- A
$156 \text{ sq.cm}$
- B
$426 \text{ sq.cm}$
- ✓
$462 \text{ sq.cm}$
- D
$288 \text{ sq.cm}$
AnswerCorrect option: C. $462 \text{ sq.cm}$
Area of the sector $=\frac{\theta}{360^\circ}\times\pi\text{r}^2$
$\Rightarrow$ Area of the sector $=\frac{120^\circ}{360^\circ}\times\frac{22}{7}\times21\times21$
$=462\text{ sq.cm}$
View full question & answer→MCQ 881 Mark
The perimeter of a triangle is $30\ cm$ and the circumference of its incircle is $88\ cm$. The area of the triangle is :
- A
$70\ cm^2$
- B
$140\ cm^2$
- ✓
$210\ cm^2$
- D
$420\ cm^2$
AnswerCorrect option: C. $210\ cm^2$
We have to find the area of the given triangle.
Perimeter of triangle is $30\ cm$.
Let the radius of the circle be $r.$
We have,
Circumference of incircle $= 88$
$2\pi\text{r}=88$
$\text{r}=14$
Therefore,
$\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{OAB})+\text{ar}(\triangle\text{OBC})+\text{ar}(\triangle\text{OAC})$
$=\frac{1}{2}\text{(r)}\text{(AB+BC+CA)}$
$=\frac{1}{2}(14)(30)\text{ cm}^2$
$=210\text{ cm}^2$
Therefore the answer is $(c).$ View full question & answer→MCQ 891 Mark
If the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle of radius $R,$ then :
- ✓
$\text{R}_1+\text{R}_2=\text{R}$
- B
$\text{R}_1+\text{R}_2 > \text{R}$
- C
$\text{R}_1+\text{R}_2 < \text{R}$
- D
$\text{None of these}$
AnswerCorrect option: A. $\text{R}_1+\text{R}_2=\text{R}$
Because the sum of the circumferences of two circles with radii $R_1$ and $R_2$ is equal to the circumference of a circle with radius $R.$ we have :
$2\pi\text{R}_1+2\pi\text{R}_2=2\pi\text{R}$
$\Rightarrow2\pi(\text{R}_1+\text{R}_2)=2\pi\text{R}$
$\Rightarrow\text{R}_1+\text{R}_2=\text{R}$
View full question & answer→MCQ 901 Mark
The diameter of a wheel is $40\ cm$. How many revolutions will it make in covering $176m?$
AnswerDistance covered by the wheel in $1$ revolution $=\pi\text{d}$
$=\Big(\frac{22}{7}\times40\Big)\text{ cm}$
$=\frac{880}{7}\text{ cm}$
$=\frac{880}{7\times100}\text{m}$
Number of revolutions required to cover $176m =\bigg(\frac{176}{\frac{880}{7\times100}}\bigg)$
$=\Big(176\times100\times\frac{7}{880}\Big)$
$=140$
View full question & answer→MCQ 911 Mark
A wire can be bent in the form of a circle of radius $56\ cm$. If it is bent in the form fo a square, then its area will be :
- A
$3520\ cm^2$
- B
$6400\ cm^2$
- ✓
$7744\ cm^2$
- D
$8800\ cm^2$
AnswerCorrect option: C. $7744\ cm^2$
Radius of circular wire $(r) = 56\ cm$
Circumference $=2\pi\text{r}=2\times\frac{22}{7}\times56\text{ cm}=352\text{ cm}$
Now perimeter of square $= 352\ cm$
$\therefore$ side of square $=\frac{352}{4}=88\text{ cm}$
and area of square $=\text{(side)}^2=(88)^2$
$=7744\text{ cm}^2\ (\text{c})$
View full question & answer→MCQ 921 Mark
If a line meets the circle in two distinct points, it is called :
AnswerA secant line, also simply called a secant, is a line meet two points in a circle.
View full question & answer→MCQ 931 Mark
If the sum of the circumference of two circles with radii $r_1$ and $r_2$ is equal to the circumference of a circle of radius $r,$ then :
- ✓
$r=r_1+r_2 $
- B
$ r_1+r_2>r $
- C
$ r_1+r_2$
- D
AnswerCorrect option: A. $r=r_1+r_2 $
The radius of the two circles $r_1$ and $r_2$
Now, according to the given condition
The circumference of the circle with radius $r =$ circumference of the circle with radius $r_1 \ +$ circumference of the circle with radius $r_2$
$2\pi\text{r}=2\pi\text{r}_1+2\pi\text{r}_2$
$2\pi\text{r}=2\pi(\text{r}_1+\text{r}_2)$
$\text{r}=\text{r}_1+\text{r}_2$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 941 Mark
Find the area of shaded portion, where the length of it is $14$ units and radius of the upper semicircle is $7$ units :
A AnswerRequried area of figure $ =$ Area of shaded portion.
If we substract area of semi circle from area of rectangle then add the area of upper semi circle,
we get the net area or alternatively, if we calculate the area of rectangle, we get the net area $= l \times b$
$= 14 \times 14$
$= 196$
View full question & answer→MCQ 951 Mark
If the sum of the areas of two circles with radii $r_1$ and $r_2$ is equal to the area of a circle of radius $r,$ then $\text{r}^2_{1}+\text{r}^2_{1}$
- A
$ > r^2$
- ✓
$= r^2$
- C
$ < r^2$
- D
AnswerCorrect option: B. $= r^2$
Sum of area of two circles with radii $r_1$ and $r_2$
$=\pi\text{r}^2_{1}+\pi\text{r}^2_{2}=\pi(\text{r}^2_{1}+\text{r}^2_{2})$
and area of a circle with radius $\text{r}=\pi\text{r}^2$
$\therefore\pi(\text{r}^2_{1}+\text{r}^2_{2})=\pi\text{r}^2$
$\Rightarrow\text{r}^2_{1}+\text{r}^2_{2}=\text{r}^2\ \ \text{(b)}$
View full question & answer→MCQ 961 Mark
The area of a circular path of uniform width $h$ surrounding a circular region of radius $r$ is :
- A
$\pi(2\text{r}+\text{h})\text{r}$
- ✓
$\pi(2\text{r}+\text{h})\text{h}$
- C
$\pi(\text{h}+\text{r})\text{r}$
- D
$\pi(\text{h}+\text{r})\text{h}$
AnswerCorrect option: B. $\pi(2\text{r}+\text{h})\text{h}$
We have
$OA = r$
$AB = h$
Therefore, radius of the outer circle will be $r + h$
Now we will find the area between the two circles.
Area of the circular path $-$ area of the outer circle $-$ area of the inner cicrle.
$\therefore$ Area of the circular path $=\pi\Big[(\text{r}+\text{h})^2\Big]-\pi\text{r}^2$
$\therefore$ Area of the circular path $=\pi(\text{r}^2+2\text{rh}+\text{h}^2)-\pi\text{r}^2$
$\therefore$ Area of the circular path $=\pi(\text{r}^2+2\text{rh}+\text{h}^2-\text{r}^2)$
Cancelling $r^2$ we get,
Area of the circular path $=\pi(2\text{rh}+\text{h}^2)$
$\therefore$ Area of the circular path $=\pi(2\text{r}+\text{h})\text{h}.$
Hence, the correct answer is option $(b).$ View full question & answer→MCQ 971 Mark
If the area of a circle is equal to the sum of the areas of two circles of diameters $10\ cm$ and $24\ cm,$ then diameter of the large circle $($in $\ cm)$ is :
AnswerLet the diameter of the larger circle be $d$
Now, Area of larger circle $=$ Area of circle having diameter $10\ cm \ + $ Area of circle having diameter $24\ cm$
$\Rightarrow\pi\Big(\frac{\text{d}}{2}\Big)^2=\pi\Big(\frac{10}{2}\Big)^2+\pi\Big(\frac{24}{2}\Big)^2$
$\Rightarrow\Big(\frac{\text{d}}{2}\Big)^2=(5)^2+(12)^2$
$\Rightarrow\Big(\frac{\text{d}}{2}\Big)^2=25+144$
$\Rightarrow\Big(\frac{\text{d}}{2}\Big)^2=13^2$
$\Rightarrow\frac{\text{d}}{2}=13$
$\Rightarrow\text{d}=26\text{ cm}$
Hence, the correct answer is option $(b).$
View full question & answer→MCQ 981 Mark
The diameter of a circular park is $17.5m$. It is surrounded by a path of width $3.5m$. The area of the path is :
- ✓
$231 \text{ sq.m}$
- B
$175 \text{ sq.m}$
- C
$220 \text{ sq.m}$
- D
$255 \text{ sq.m}$
AnswerCorrect option: A. $231 \text{ sq.m}$
The diameter of the park $= 17.5m$
$\therefore$ The radius of the park $\text{r}=\frac{17.5}{2}\text{m}$
And Radius of the park including path $\text{R}=\frac{17.5}{2}+3.5=\frac{24.5}{2}\text{m}$
Now,Area of the path $=\pi\Big(\text{R}^2-\text{r}^2\Big)$
$=\frac{22}{7}\Bigg[\Big(\frac{24.5}{2}\Big)^2-\Big(\frac{17.5}{4}\Big)^2\Bigg]$
$=\frac{22}{7}\Big(\frac{24.5}{2}+\frac{17.5}{2}\Big) $
$\Big(\frac{24.5}{2}-\frac{17.5}{2}\Big) $
$=\frac{22}{7}\times\frac{7.0}{2}\times\frac{42}{2}=231\text{ sq.m}$ View full question & answer→MCQ 991 Mark
The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters $36\ cm$ and $20\ cm$ is :
- A
$14\ cm$
- ✓
$28\ cm$
- C
$56\ cm$
- D
$42\ cm$
AnswerCorrect option: B. $28\ cm$
Let required radius be $R$.
Then according to the question,
$2\pi\text{R}=2\pi\text{r}_1+2\pi\text{r}_2$
$=\pi\Big(\text{r}_1+\text{r}_2\Big)$
$\Rightarrow\text{R}=\text{r}_1+\text{r}_2$
$\Rightarrow\text{R}=\frac{36}{2}+\frac{20}{2}$
$=18+10=28\text{ cm}$
View full question & answer→MCQ 1001 Mark
Choose the correct answer from the given four options : If the sum of the areas of two circles with radii $R_1$ and $R_2$ is equal to the area of a circle of radius $R,$ then :
- A
$\text{R}_1+\text{R}_2=\text{R}$
- ✓
$\text{R}_1^2+\text{R}_2^2=\text{R}^2$
- C
$\text{R}_1+\text{R}_2<\text{R}$
- D
$\text{R}_1^2+\text{R}_2^2<\text{R}^2$
AnswerCorrect option: B. $\text{R}_1^2+\text{R}_2^2=\text{R}^2$
According to the given condition,
Area of circle $=$ Area of first circle $+$ Area of second circle
$\therefore\ \ \pi\text{R}^2=\pi\text{R}_1^2+\pi\text{R}^2_2$
$\Rightarrow\ \ \text{R}^2=\text{R}^2_1+\text{R}^2_2$
View full question & answer→