Question 12 Marks
Find the circumference of a circle whose area is $301.84\ cm^2.$
AnswerLet r cm be the radius of the circle. Then Area of a circle is
$\text{A}=\pi\text{r}^2\text{cm}^2$
$301.84\text{cm}^2=\pi\times\text{r}^2$
$301.84\text{cm}^2=\frac{22}{7}\times\text{r}^2$
$\text{r}^2=96.04\text{cm}^2$
$\text{r}=9.8\text{cm}$
We know that the Circumference of circle of radius r is
$\text{C}=2\pi\text{r}\text{ cm}$
So substituting the value of r in above formula
$\text{C}=2\times\frac{22}{7}\times9.8\text{cm}$
$=61.6\text{cm}$
View full question & answer→Question 22 Marks
A sector is cut-off from a circle of radius 21cm. The angle of the sector is 120°. Find the length of its arc and the area.
AnswerWe know that the arc length l and area A of a sector of circle at an angle $\theta$ of radius r is.
given by $\text{l}=\frac{\theta}{360^\circ}\times2\pi\text{r}$ and angle $\text{A}=\frac{\theta}{360^\circ}\pi\text{r}^2$
Let OAB is the given sector.
It is given that OA = 21m and angle $\angle\text{AOB}=120^\circ$
Now using the value of r and θ, we will find the value of l and A,
Arc length
$\text{l}=\frac{120^\circ}{360^\circ}\times2\times\frac{22}{7}\times21\text{cm}$
$=44\text{cm}$
Area of sector,
$\text{A}=\frac{120^\circ}{360^\circ}\times\frac{22}{7}\times21\times21$
$=462\text{cm}^2$ View full question & answer→Question 32 Marks
AB is a chord of a circle with centre O and radius 4cm. AB is of length 4cm. Find the area of the sector of the circle formed by chord AB.
AnswerRadius of the circle with centre O (r) = 4cm Length of chord AB = 4cm
$\therefore$ AOB is an equilateral triangle $\therefore\angle\text{AOB}=60^\circ$ Now area of the sector $\text{AOB}=\pi\text{r}^2\times\frac{\theta}{360^\circ}$ $=\pi(4)^2\times\frac{60^\circ}{360^\circ}\text{cm}^2$ $=16\pi\times\frac{1}{6}=\frac{8\pi}{3}\text{cm}^2$ View full question & answer→Question 42 Marks
The minute hand of a clock is $\sqrt{21}\text{cm}$ long. Find the area described by the minute hand on the face of the clock between 7:00 AM and 7:05 AM.
AnswerLength of minute hand of a clock (r) $=\sqrt{21}\text{cm}$
Period = 7 A:M. to 7.05 A:M. 5 minutes
$\therefore$ Central angle $(\theta)=\frac{5}{60}\times360^\circ=30^\circ$
Area of the sector $=\pi\text{r}^2\times\frac{\theta}{360^\circ}$
$=\frac{22}{7}\times(\sqrt{21})^2\times\frac{30^\circ}{360^\circ}$
$=\frac{22}{7}\times21\times\frac{1}{12}$
$=\frac{11}{2}=5.5\text{cm}^2$
View full question & answer→Question 52 Marks
What is the area of a square inscribed in a circle of diameter $p\ cm$?
AnswerDiameter $AC$ of the circle is $p$. Also AC is diagonal of square $ABCD$. Each angle of square is of $90^\circ$
Let x cm be the side $AB$ and $BC$ of square $ABCD.$
$AC^2 = AB^2+ BC^2 $
$\therefore$$P^2 = x^2 + x^2 P^2 = x^2 + x^2 2x^2 = P^2 x^2$ = $\frac{\text{P}^2}{2}$
Area of square $= x^2$ $=\frac{\text{P}^2}{2}\text{cm}^2$ View full question & answer→Question 62 Marks
How many revolutions a circular wheel of radius r metres makes in covering a distance of s metres?
AnswerRadius of circular of wheel (r) = r m
Circumference of a circular wheel = $2\pi\text{r}$
Distance to be covered = S m
Numbar of revolutions of circular wheel $=\frac{\text{Distance covered}}{\text{circumference of circular wheel}}=\frac{\text{S}}{2\pi\text{r}}$
View full question & answer→Question 72 Marks
A cow is tied with a rope of length 14m at the corner of a rectangular field of dimensions 20m x 16m, find the area of the field in which the cow can graze.
AnswerLet ABCD be a rectangular field of dimensions 20m x 16m. Suppose, a cow is tied at a point A.Let length of rope AE = 14m = r (say).
$\therefore$ Area of the field in which the cow graze = Area of sector $\text{AFEG}=\frac{\theta}{360^\circ}\times\pi\text{r}^2$ $=\frac{90}{360^\circ}\times\pi(14)^2$ [So, the angle between two adjacent sides of a rectangles is 90°] $=\frac{1}{4}\times\frac{22}{7}\times196=154\text{m}^2$ View full question & answer→Question 82 Marks
Find the angle subtended at the centre of a circle of radius 5cm by an arc of length $\frac{5\pi}{3}\text{cm}.$
AnswerWe know that the arc length l of a sector of an angle $θ$ in a circle of radius r is
$\text{l}=\frac{\theta}{360^\circ}\times2\pi\text{r}$
It is given that r = 5cm and length $\text{l}=\frac{5\pi}{3}\text{cm}.$ Substituting these value in above equation,
$\frac{5\pi}{3}=\frac{\theta}{360^\circ}\times2\pi\times5$
$5\pi\times360^\circ=\theta\times2\pi\times5\times3$
$\theta=60^\circ$
Hence, the angle subtended at the centre of circle is 60°.
View full question & answer→Question 92 Marks
In the following figure, ABCD is a rectangle, having $AB= 20\ cm$ and $BC = 14\ cm$. Two sectors of $180^\circ $ have been cut off.
Calculate: The length of the boundary of the shaded region
AnswerABCD is a rectangle whose Length $AB = 20\ cm$ and width $BC = 14\ cm$
$\therefore$ Area of the rectangle$ = l \times b = 20 \times 14 = 280\ cm^2$
Radius of each semicurcle (r) $=\frac{\text{BC}}{2}=\frac{14}{2}$ $=7$
$\therefore$ Area of two semicircles $=2\times\frac{1}{2}\pi\text{r}^2=\pi\text{r}^2$
$=\frac{22}{7}\times7\times7=154\text{cm}^2$
Length of boundary of the shaded region $=\text{AB}+\text{CD}+2\times$ circumference if semicircle
$=20+20+2\times\pi\text{r}$
$=40+2\times\frac{22}{7}\times7=40+44=84\text{cm}$
View full question & answer→Question 102 Marks
The radii of two circles are 8cm and 6cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
AnswerRadius of circles are 8cm and 6cmArea of circle with radius $8\text{cm}=\pi(8)^2=64\pi\text{cm}^2$
Area of circle with radius $6\text{cm}=\pi(6)^2=36\pi\text{cm}^2$Areas sum $=64\pi+36\pi=100\pi\text{cm}^2$
Radius of circle be x cm Area $=\pi\text{x}^2$ $\pi\text{x}^2=100\pi$ $\text{x}^2=100\Rightarrow\text{x}=\sqrt{100}=10\text{cm}$
View full question & answer→Question 112 Marks
A circular pond is $17.5\ m$ in diameter. It is surrounded by a $2\ m$ wide path. Find the cost of constructing the path at the rate of $₹ 25\ per\ m^2$.
AnswerDiameter of the pond =$ 17.5\ m$
Radius of the pond = $8.75\ m$
Radius of the pond with the path = $8.75 + 2 = 10.75\ m$
Area of the path = Area of the pond along with the path $(−)$ area of the pond
Area of the path = $\pi\text{R}^2-\pi\text{r}^2=\pi\big[(10.75)^2-(8.75)^2\big]$ $=\pi219.5=122.46\text{m}^2$
Cost of constructing the path $= 25 × 122.46 = Rs 3061.5$
View full question & answer→Question 122 Marks
In a circle of radius 6cm, a chord of length 10cm makes an angle of 110° at the centre of the circle. Find:
The area of the sector OAB.
AnswerArea of sector $=\frac{\theta}{360^\circ}\times\pi\text{r}^2$
$=\frac{110}{360}\times\frac{22}{7}\times6\times6$
$=\frac{242}{7}\text{cm}^2$
View full question & answer→Question 132 Marks
A steel wire when bent in the form of a square encloses an area of $121\text{cm}^2$. If the same wire is bent in the form of a circle, find the area of the circle.
AnswerArea of square formed by a wire $= 121\text{cm}^2$
$\therefore$ Side of square $(\text{a})=\sqrt{\text{Area}}=\sqrt{121}=11\text{cm}$
Perimeter of the square $= 4\times \text {side} = 4 \times 11 = 44\ cm$
$\therefore$ Circumference of the circle formed by the wire $= 44\ cm$
Let r be the radius $\therefore2\pi\text{r}=44\Rightarrow2\frac{22}{7}\text{r}=44$
$\Rightarrow\text{r}=\frac{44\times7}{2\times22}=7\text{cm}$Hence area of the circle $=\pi\text{r}^2$
$=\frac{22}{7}\times(7)^2\text{cm}$
$=\frac{22}{7}\times7\times7\text{cm}^2=154\text{cm}^2$
View full question & answer→Question 142 Marks
Find, in terms of $\pi,$ the length of the arc that subtends an angle of 30° at the centre of a circle of radius 4cm.
Answer
Length of arc $=\frac{\theta}{360^\circ}\times2\pi\text{r}$
Radius = r = 4cm
$\theta=$ angle subtended at centre = 30°
Arc length $=\frac{30^\circ}{360^\circ}\times2\times\pi\times4$
$=\frac{2\pi}{3}\text{cm}$ View full question & answer→Question 152 Marks
The area of a sector of a circle of radius 2cm is $\pi\text{cm}^2.$ Find the angle contained by the sector.
AnswerWe know that the area A of a sector of an angle $ θ$ in the circle of radius r is given by
$\text{A}=\frac{\theta}{360^\circ}\times\pi\text{r}^2$
It is given that r = 2cm and area $\text{A}=\pi\text{cm}^2$
Now we substitute the value of r and A in above formula to find the value of $ θ,$
$\pi=\frac{\theta}{360^\circ}\times\pi\times2\times2$
$\theta=\frac{360^\circ\times\pi}{\pi\times2\times2}$
$=90^\circ$
View full question & answer→Question 162 Marks
In the following figure, from a rectangular region $A B C D$ with $A B=20 cm$, a right triangle $A E D$ with $A E=9 cm$ and $D E$ $=12 cm$, is cut off.
On the other end, taking BC as diameter, a semicircle is added on outside' the region. Find the area of the shaded region. [Use $\pi=\frac{22}{7}$ ]
AnswerIn right triangle $A E D$
$A D^2=A E^2+D E^2$
$=(9)^2+(12)^2$
$=81+144$
$=225$
$\therefore AD^2=225$
$\Rightarrow AD=15 cm$
We know that the opposite sides of a rectangle are equal
$A D=B C 15 cm$
Area of the shaded region $=$ Area of rectangle - Area of triangle $A E D+$ Area of semicircle
$=\text{AB}\times\text{BC}-\frac{1}{2}\times\text{AE}\times\text{DE}+\frac{1}{2}\pi\Big(\frac{\text{BC}}{2}\Big)^2$
$=20\times15-\frac{1}{2}\times9\times12+\frac{1}{2}\times\frac{22}{7}\times\big(\frac{15}{2}\big)^2$
$=300-54+88.3982$
$=334.3928\text{cm}^2$
Hence, the area of shaded region is $334.39 \text{cm}^2$
View full question & answer→Question 172 Marks
In a circle of radius 6cm, a chord of length 10cm makes an angle of 110° at the centre of the circle. Find:
The area of the circle.
AnswerArea of circle $=\pi\text{r}^2=\frac{22}{7}\times6\times6$
$=\frac{792}{7}\text{cm}^2$
View full question & answer→Question 182 Marks
A horse is tied to a pole with 28m long string. Find the area where the horse can graze.
$\big(\text{Take } \pi = \frac{22}{7}\big)$
AnswerWe know that the horse is tied to a pole with 28m long string. So the horse can graze the area of a circle of radius 28m. Area of circle is,
$\text{A}=\pi\text{r}^2$
$=\frac{22}{7}\times28\times28\times\text{m}^2$
$=2464\text{m}^2$
Hence the horse can graze $2464\text{m}^2$ area.
View full question & answer→Question 192 Marks
Figure shows a sector of a circle of radius r cm containing an angle $\theta$.
The area of the sector is $A cm ^2$ and perimeter of the sector is $50\ cm .$
$$Prove that $A =25 r - r ^2$
AnswerRadius of the sector of the circle $= rcm$ and angle at the centre $=\theta$ Area of sector $OAB = A cm { }^2$ and
perimeter of sector $O A B=50\ cm$
perimeter of sector $=2 \times$ radius + arc
$50=2\text{r}+2\pi\text{r}\Big(\frac{\theta}{360^\circ}\Big)$
$50-2\text{r}=2\pi\text{r}\Big(\frac{\theta}{360^\circ}\Big)$
$\Big(\frac{50-2\text{r}}{2\pi\text{r}}\Big)=\frac{\theta}{360^\circ}$
$\frac{50}{2\pi\text{r}}-\frac{2\text{r}}{2\pi\text{r}}=\frac{\theta}{360^\circ}$
$\Big(\frac{25}{\pi\text{r}}-\frac{1}{\pi}\Big)=\frac{\theta}{360^\circ}$
Putting value of $=\frac{\theta}{360^\circ}$
Area of the sector $=\pi\text{r}^2\times\frac{\theta}{360^\circ}$
$\therefore\text{A}=\pi\text{r}^2\times\Big(\frac{25}{\pi\text{r}}-\frac{1}{\pi}\Big)=\pi\text{r}^2\times\frac{1}{\pi}\Big(\frac{25}{\text{r}}-1\Big)$
$=\text{r}^2\Big(\frac{25}{\text{r}}-1\Big)=\frac{25}{r}\times\text{r}^2-\text{r}^2$
$=25\text{r}-\text{r}^2$ Hence proved. View full question & answer→Question 202 Marks
A sector of a circle of radius 8cm contains an angle of 135°. Find the area of the sector.
AnswerRadius (r) = 8cm
$\theta=$ angle subtended at centre = 135° Area of sector $=\frac{\text{x}}{360^\circ}\times\pi\text{r}^2$ $=\frac{135}{360^\circ}\times\frac{22}{7}\times8\times8$ $=\frac{528}{7}\text{cm}^2$ View full question & answer→Question 212 Marks
Find the angle subtended at the centre of a circle of radius 'a' by an arc of length $\Big(\frac{\text{a}\pi}{4}\text{cm}\Big)$
AnswerWe know that the arc length l of a sector of an angle $θ$ in a circle of radius r is
$\text{l}=\frac{\theta}{360^\circ}\times2\pi\text{r}$
It is given $\text{l}=\frac{\text{a}\pi}{4}\text{cm}$ and radius r = a cm.
Now we substitute the value of l and r in above formula to find the value of angle $θ$ subtended at the centre of circle.
$\frac{\text{a}\pi}{4}\text{cm}=\frac{\theta}{360^\circ}\times2\pi\times\text{a}$
$\theta=\frac{\text{a}\pi\times360^\circ}{2\pi\text{a}\times4}$
$\theta=45^\circ$
View full question & answer→Question 222 Marks
An arc of length $20\pi\text{ cm}$ subtends an angle of 144° at the centre of a circle. Find the radius of the circle.
AnswerLength of an arc $=20\pi\text{ cm}$
Angle subtended by the arc = 144°
Let r be the radius of the circle, then
$\therefore\ 2\pi\text{r}\big(\frac{\theta}{360^\circ}\big)=20\pi$
$\Rightarrow2\pi\text{r}\times\frac{144}{360^\circ}=20\pi$
$\Rightarrow2\pi\text{r}\times\frac{2}{5}=20\pi$
$\Rightarrow\text{r}=\frac{20\pi\times5}{2\pi\times2}=5\times5=25$
$\therefore$ Radius = 25cm
View full question & answer→Question 232 Marks
An arc of length 15cm subtends an angle of 45° at the centre of a circle. Find in terms of $\pi$, the radius of the circle.
Answer
Length of arc = 15cm
$\theta=$ angle subtended at centre = 45°
Let radius = r cm
arc length $=\frac{\theta}{360^\circ}\times2\pi\text{r}$
$=\frac{45^\circ}{360^\circ}\times2\pi\text{r}$
$=\frac{45^\circ}{360^\circ}\times2\pi\text{r}=15$
$\text{r}=\frac{15\times360}{45\times2\pi}=\frac{60}{\pi}\text{cm}$
Radius $=\frac{60}{\pi}\text{cm}$ View full question & answer→Question 242 Marks
From a circular piece of cardboard of radius 3cm two sectors of 90° have been cut off. Find the perimeter of the remaining portion nearest hundredth centimeters $\Big(\text{Take }\pi =\frac{22}{7}\Big). $
AnswerRadius of the circular piece of cardboard (r) = 3cm
$\therefore$ Two sectors of 90° each have been cut off $\therefore$ We get a semicular cardboard piece. $\therefore$ Perimeter of arc ACB $=\frac{1}{2}(2\pi\text{r})=\pi\text{r}$ $=\frac{22}{7}\times3=\frac{66}{7}=9.428\text{cm}$ View full question & answer→Question 252 Marks
The minute hand of a clock is 10cm long. Find the area of the face of the clock described by the minute hand between 8:00 AM and 8:25 AM.
Answer
Radius of minute hand (r) = 10cm
For 1 hr = 60 min, minute hand completes one revolution = 360°
60 min = 360°
1 min = 6°
From 8:00 AM.to 8:25 AM it is 25 min angle subtended $ = 6^\circ \times 25 = 150^\circ =\theta$
Area described $=\frac{\theta}{360^\circ}\times\pi\text{r}^2$
$=\frac{150}{360}\times\frac{22}{7}\times10\times10$
$=\frac{250\times11}{3}$
$=\frac{2750}{3}\text{cm}^2$ View full question & answer→Question 262 Marks
If the diameter of a semi-circular protractor is 14cm, then find its perimeter.
Question 272 Marks
A square water tank has its side equal to $40 \text {m}$ . There are four semi-circular grassy plots all round it.
Find the cost of turfing the plot at $\text {Rs}.1.25$ per square metre (Take $\pi=3.14$ ).
AnswerSide of square tank $(a)=40 m$
Radius of each semicircular grassy plots $=\frac{40}{2}=20$
$\therefore$ Area of $4$ semicircular plots
$=4 \times \frac{1}{2} \pi r^2=2 \times 3.14 \times(20)^2 m^2=2 \times 3.14 \times 400 m^2=800 \times 3.14=2512 m^2$
Rate of turfing the plots $=₹ 1.25$ per $m ^2$ .
Total cost $=2512 \times 1.25=₹ 3140$ View full question & answer→Question 282 Marks
The circumference of two circles are in the ratio $2 : 3$. Find the ratio of their areas.
AnswerLet radius of two circles be $r_1$_ and $r_2$_ then their circumferences will be $2\pi\text{r}_1:2\pi\text{r}_2=\text{r}_1:\text{r}_2$
But circcumference ratio is given as $2 : 3$
$r_1 : r_2 = 2 : 3$
Ratio of area $=\pi\text{r}^2_2:\pi\text{r}^2_2$
$=\Big(\frac{\text{r}_1}{\text{r}_2}\Big)^2$
$=\Big(\frac{2}{3}\Big)^2$
$=\frac{4}{9}$
$=4:9$
$\therefore$ ratio of areas =$ 4 : 9$
View full question & answer→Question 292 Marks
Write the area of the sector of a circle whose radius is r and length of the arc is l.
AnswerLet arc l subtends angle 9 at the centre of the circle
Now radius of a circle = r
and length of arc = l
$\therefore \text{l}=2\pi\text{r}\times\frac{\theta}{360^\circ}$
$\Rightarrow\frac{\pi\theta }{360^\circ}=\frac{\text{l}}{2\text{r}}\ \dots(\text{i})$
Now area of the sector $=\pi\text{r}^2\frac{\theta}{360^\circ}$
$=\frac{\pi\theta}{360^\circ}\times\text{r}^2$
$=\frac{\text{l}}{2\text{r}}\times\text{r}^2=\frac{\text{lr}}{2}$ [From (i)]
$=\frac{1}{2}\text{lr}$
View full question & answer→Question 302 Marks
Find the area of the largest triangle that can be inscribed in a semi-circle of radius r units.
AnswerRadius of the semi circle r units
When a largest triangle inscribed in a semicircle, then base = r + r = 2r units
Thus, Area of triangle is given by
$\frac{1}{2}\times\text{base}\times\text{height}$
$=\frac{1}{2}\times(2\text{r})\times\text{r}$
$=\text{r}^2$ Square units
View full question & answer→Question 312 Marks
A park is in the form of a rectangle $120\ m \times 100\ m$. At the centre of the park there is a circular lawn, The area of park excluding lawn is $8700\ m^2.$
Find the radius of the circular lawn. $\big(\text{Use }\pi=\frac{22}{7}\big)$
AnswerLet the radius of circular lawn be r. Then,
Area of circular lawn $=\pi\text{r}^2$
It is given that
Area of park excluding lawn = Area of rectangle-Area of circular lawn
$8700 = 120 \times100 $- $\pi\text{r}^2$
$\pi\text{r}^2$ $= 12000 - 8700$
$\frac{22}{7}\text{r}^2=3300$
$\text{r}^2=\frac{3300\times7}{22}$
$\text{r}^2=1050$
$\text{r}=\sqrt{1050}$
$\text{r}=32.40\text{m}$
Hence, radius of circular lawn is $32.40\ m$
View full question & answer→Question 322 Marks
A sector of$ 56^\circ$ cut out from a circle contains area $4.4\ cm^2$. Find the radius of the circle.
AnswerWe know that the area A of a sector of circle at an angle $\theta$ of radius r is given by.
$\text{A}=\frac{\theta}{360^\circ}\pi\text{r}^2$
It is given that, Area of a sector A $= 4.4\ cm^2$ and angle $\theta=56^\circ$
We can find the value of r by substituting these values in above formula,
$\text{A}=\frac{56^\circ}{360^\circ}\times\frac{22}{7}\text{r}^2$
$4.4=\frac{56^\circ}{360^\circ}\times\frac{22}{7}\text{r}^2$
$\text{r}^2=\frac{360^\circ}{56^\circ}\times\frac{7}{22}\times4.4$
$\text{r}^2=9$
$\text{r}=\sqrt{9}$
$\text{r}=3\text{cm}$
View full question & answer→Question 332 Marks
Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?
AnswerA circle has two segments, a major segment and a minor segment.
The area of the minor segment is less than the area of the corresponding sector. But same is not true for the major segment.
View full question & answer→Question 342 Marks
An arc subtends an angle of 90° at the centre of the circle of the radius 14cm. Write the area of minor sector thus formed in terms of $\pi.$
AnswerAB is an arc of the circle with centre O and radius 14cm and subtends an angle of 90° at the centre O.
$\therefore$ Area of the sectore AOB $=\pi\text{r}^2\times\frac{\theta}{360^\circ}$ $=\pi\times14\times14\times\frac{90^\circ}{360^\circ}\text{cm}^2=49\pi\ \text{cm}^2$ View full question & answer→Question 352 Marks
A sector of a circle of radius 4cm contains an angle of 30°. Find the area of the sector.
AnswerRadius of the sector of a circle (r) = 4cm
Angle at the centre $ (\theta) = 30^\circ$
$\therefore$ Area of the sector $=\pi\text{r}^2\times\frac{\theta}{360^\circ}$
$=\pi(4)^2\times\frac{30^\circ}{360^\circ}\text{cm}^2$
$=16\pi\times\frac{1}{12}\text{cm}^2$
$=\frac{4}{3}\pi\text{cm}^2$
View full question & answer→Question 362 Marks
In the following figure, OABC is a square of side $7\ cm$. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. $\Big(\text{Use }\pi=\frac{22}{7}\Big).$
AnswerArea of shaded rigion = Area of square OABC - Area of quadrant OAPC
$=\text{(Side)}^2-\frac{1}{4}\pi\text{r}^2$
$=(7)^2-\frac{1}{4}\times\frac{22}{7}\times7\times7$
$=49-38.5$
$=10.5\text{cm}^2$
Hence, the area of the shaded region is $10.5\ cm^2$
View full question & answer→Question 372 Marks
The area enclosed between the concentric circles is $770\ cm^2$^. If the radius of the outer circle is $21\ cm$, find the radius of the inner circle.
AnswerRadius of outer circle = $21\ cm$
Radius of inner circle $= R_2$
Area between concentric circles = area of outer circle - area of inner circle
$\Rightarrow770=\frac{22}{7}(21^2-\text{R}^2_2)$
$\Rightarrow21^2-\text{R}^2_2=35\times7=245$
$\Rightarrow441-245=\text{R}^2_2$
$\Rightarrow\text{R}_2=\sqrt{196}=14\text{cm}$
Radius of inner circle $= 14\ cm.$ View full question & answer→Question 382 Marks
Find the area of a sector of circle of radius 21cm and central angle 120°.
AnswerArea of the sector $=\frac{\theta}{360^\circ}\times\pi\text{r}^2$
$=\frac{120}{360^\circ}\times\frac{22}{7}\times(21)^2\text{cm}^2$
$=22\times21\text{cm}^2=462\text{cm}^2$
View full question & answer→Question 392 Marks
In a circle of radius 6cm, a chord of length 10cm makes an angle of 110° at the centre of the circle. Find:
The length of the arc AB,
AnswerLength of arc $=\frac{\theta}{360^\circ}\times2\pi\text{r}$
$=\frac{110}{360}\times2\times\frac{22}{7}\times6$
$=\frac{242}{21}\text{cm}$
View full question & answer→Question 402 Marks
In a circle of radius 6cm, a chord of length 10cm makes an angle of 110° at the centre of the circle. Find:
The circumference of the circle.
Question 412 Marks
In the following figure, PQRS is a square of side 4cm. Find the area of the shaded square.
AnswerArea of the shaded region is equal to the area of the square minus area of four sectors with the same radius minus area of the circle.
We know that area of the four sectors with radius is equal to area of one circle.
$\therefore$ Area of the shaded region $=4^2-\pi\times1^2-\pi\times1^2$
$\therefore$ Area of the shaded region $=4^2-2\times\pi\times1$
$\therefore$ Area of the shaded region $=16-2\pi$
Therefore, area of the shaded region is $(16-2\pi)\text{cm}^2$
View full question & answer→Question 422 Marks
The area of a circular playground is $22176 m^2$. Find the cost of fencing this ground at the rate of $₹ 50$ per metre.
AnswerGiven, area of a circular playground $=22176 m^2$
$\therefore\pi\text{r}^2=22176 \ [\therefore\text{area of circle=}\pi\text{r}^2]$
$\Rightarrow\frac{22}{7}\text{r}^2=22176\Rightarrow\text{r}^2=1008\times7$
$\Rightarrow\text{r}^2=7056\Rightarrow\text{r}=84\text{m}$
$\therefore$ Circumference of a circle
$=2\pi\text{r}=2\times\frac{22}{7}\times84$
$=44\times12=528\text{m}$
$\therefore$ Cost of fencing this ground $=528\times50$
$=₹ 26400$
View full question & answer→Question 432 Marks
What is the angle subtended at the centre of a circle of radius 6 cm by an arc of length $3\pi\text{ cm}?$
AnswerLet the arc subtends angle $\theta$ at the centre of a circle
Radius of circle (r) = 6cm
Length of arc = $3\pi\text{ cm}$
$\therefore 3\pi=2\pi\text{r}\times\frac{\theta}{360^\circ}$
$\Rightarrow3\pi=2\pi\times6\times\frac{\theta}{360^\circ}$
$\Rightarrow\frac{\theta}{360^\circ}=\frac{3\pi}{12\pi}=\frac{1}{4}$
$\Rightarrow\theta=\frac{360}{4}=90$
$\therefore \text{Angle}=90^\circ$
View full question & answer→Question 442 Marks
What is the ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal?
AnswerDiameter of a circle and side of an equilateral triangle are same
Let the diameter of the circle = a
Then radius (r) $=\frac{\text{a}}{2}$
$\therefore$ Area of the circle $=\pi\text{r}^2=\pi\times\frac{\text{a}^2}{4}=\frac{\text{a}^2\pi}{4}$
Side of an equilateral triangle = a
$\therefore$ Area $=\frac{\sqrt{3}}{4}\text{a}^2$
$\therefore$ Ratio between their areas $=\frac{\text{a}^2}{4} \pi:\frac{\sqrt{3}\text{a}^2}{4}=\pi :\sqrt{3}$
View full question & answer→Question 452 Marks
Find the area of a circle whose circumference is 44cm.
AnswerCircumference of a circle = 44cmLet r be the radius,
then $2\pi\text{r}=\text{circumference}$
$\Rightarrow2\times\frac{22}{7}\text{r}=44\Rightarrow\text{r}=\frac{44\times7}{2\times22}=7$
$\therefore$ Area of the circle $=\pi\text{r}^2=\frac{22}{7}(7)^2\text{cm}^2$
$=\frac{22}{7}\times7\times7=154\text{cm}^2$
View full question & answer→Question 462 Marks
The area of a sector of a circle of radius 5cm is $5\pi \text{ cm}^2.$ Find the angle contained by the sector.
AnswerArea of the sector of a circle $=5\pi\text{ cm}^2$
Radius of the circle (r) = 5cm
Let 9 be the angle at the centre, then
$\pi\text{r}^2\times\frac{\theta}{360^\circ}=5\pi$
$\pi(5)^2\times\frac{\theta}{360^\circ}=5\pi\Rightarrow25\pi\times\frac{\theta}{360^\circ}=5\pi$
$\Rightarrow\frac{\theta}{360^\circ}=\frac{5\pi}{25\pi}=\frac{1}{5}\Rightarrow\theta=\frac{1}{5}\times360^\circ$
$\Rightarrow\theta=72$
$\therefore$ Angle at the centre = 72°
View full question & answer→Question 472 Marks
What is the length (in terms of $\pi$ ) of the arc that subtends an angle of $36^\circ$ at the centre of a circle of radius 5cm?
AnswerWe have
r = 5cm
$\theta=36^\circ$
We have to find the length of the arc.
Length of the arc $= \frac{\theta}{360}\times2\pi\text{r}$
Substituting the values we get,
Length of the arc $= \frac{36}{360}\times2\pi\times5\ \dots(1)$
Now we will simplify the equation (1) as below,
Length of the arc $=\frac{1}{10}\times2\pi\times5$
Length of the arc $=\frac{1}{2}\times2\pi$
Length of the arc $=\pi$
Therefore, length of the arc is $\pi\text{ cm}$
View full question & answer→Question 482 Marks
If the adjoining figure is a sector of a circle of radius 10.5cm, what is the perimeter of the sector? $\Big(\text{Take }\pi=\frac{22}{7 }\pi=\frac{22}{7}\Big)$
AnswerRadius of the circle = 10.5cm
Angle at the centre of the circle = 60°
Length of the arc AB $=2\pi\text{r}\times\frac{\theta}{360^\circ}$
$=2\times\frac{22}{7}\times10.5\times\frac{60}{360}$
$=\frac{44}{7}\times\frac{105}{10}\times\frac{1}{6}=11\text{cm}$
Perimeter of the sectore = 2r + l
= 2 × 10.5 + 11
= 21 + 11 = 32cm
View full question & answer→Question 492 Marks
In the following figure, ABCD is a rectangle, having $AB = 20\ cm$ and $BC = 14\ cm$. Two sectors of $180^\circ $ have been cut off. Calculate: The area of the shaded region.
AnswerABCD is a rectangle whose Length $AB = 20\ cm$
and width $BC = 14\ cm$
$\therefore$ Area of the rectangle $= l \times b = 20 \times 14 = 280\ cm^2$
Radius of each semicurcle (r) $=\frac{\text{BC}}{2}=\frac{14}{2}$
$=7$
$\therefore$ Area of two semicircles $=2\times\frac{1}{2}\pi\text{r}^2=\pi\text{r}^2$
$=\frac{22}{7}\times7\times7=154\text{cm}^2$
Area of shaded region = area of rectangle - area of $2$ semicircles
$=280-154=126\text{cm}^2$
View full question & answer→Question 502 Marks
The wheel of a motor cycle is of radius $35\ cm$. How many revolutions per minute must the wheel make so as to keep a speed of 66km/hr?
AnswerGiven, radius of wheel, $r = 35\ cm$
Circumference of the wheel $=2\pi\text{r}$
$=2\times\big(\frac{22}{7}\big)\times35=220\text{cm}$
But speed of the wheel $= 66\ kmh^{-1}$
$=\big(\frac{66\text{x}1000}{60}\big)\text{m}/\text{mm}$
$= 1100 \times 100cm min^{-1}$
$= 110000cm min^{-1}$
$\therefore$ Number of revolutions in $1$ min
$=\big(\frac{110000}{220}\big)=500\text{ revolition}$
Hence, required number of revolutions per minute is $500$.
View full question & answer→Question 512 Marks
What is the area of a sector of a circle of radius $5\ cm$ formed by an arc of length $3.5\ cm$?
AnswerWe have
$r = 5\ cm$
Lenght of the arc $(l) = 3.5\ cm$
Now we will find the area of the sector.
Area of the sectore $=\frac{1}{2}\times\text{lr}$
Substituting the values we get,
Area of the sectore $=\frac{1}{2}\times3.5\times5\ \dots(1)$
Now we will simplify the equation (1) as below,
Area of the sectore $=\frac{1}{2}\times17.5$
Area of the sectore $= 8.75$
Therefore, area of the sector is $8.75\ cm^2$
View full question & answer→Question 522 Marks
Find the circumference and area of a circle of radius 4.2cm.
AnswerRadius(r) = 4.2cm
Circumference $=2\pi\text{r}$
$= 2 \times \frac{22}{7} \times 4.2$
$= \Big(\frac{44}{10}\times 6 \Big) = \frac{264}{10}$
$= 26.4\text{cm}$
$\text{Area} = \pi\text{r}^{2} = \frac{22}{7}\times 4.2\times 4.2$
$=\frac{22\times6\times42}{10\times10}=\frac{5544}{100}=55.44\text{cm}^2$
View full question & answer→Question 532 Marks
The area of a sector is one-twelfth that of the complete circle. Find the angle of the sector.
AnswerWe Know, $\text{A}=\frac{\theta}{360}\times$ Area of the circle .....(i)
Let the area of the circle be Ar.
Thus area of the sector $=\frac{1}{12}\text{Ar}\ ....(\text{ii})$
From (i) and (ii) we have
$\frac{1}{12}\text{Ar}=\frac{\theta}{360}\times\text{Ar}$
$\Rightarrow\frac{360}{12}=\theta$
$\Rightarrow\theta=30^\circ$
View full question & answer→Question 542 Marks
In a circle of radius 10 cm, an arc subtends an angle of 108° at the centre. what is the area of the sector in terms of $\pi$ ?
AnswerRadius of the circle = 10cm
Angle at the centre = $108^\circ$
$\therefore$ Area of the sector $=\pi\text{r}^2\times\frac{\theta}{360^\circ}$
$=\pi (10)^2\times\frac{108}{360}\text{cm}^2$
$=100\pi \times \frac{3}{10}\text{cm}^2=30\pi\text{cm}^2$
View full question & answer→Question 552 Marks
A calf is teid with a rope of length $6\ m$ at the corner of a square grassy lawn of side $20\ m$. If the length of the rope is increased by $5.5\ m$, find the increase in area of the grassy lawn in which the calf can graze.
Answer
he area grazed by the calf is in the form of a quadrant of a circle with radius 6m.Area grazed by the calf with rope 6m $=\frac{1}{4}\pi(6)^2$
$= 28.28m^2$
When the rope length is increased then total rope length $= 6 + 5.5\ m = 11.5\ m$
Area covered by the calf for grazing with rope 11.5m $=\frac{1}4{}\pi(11.5)^2$
$= 103.91\ m^2$
Hence, increase in area of grassy lawn that is grazed $= 103.91 - 28.28 = 75.63\ m^2$ View full question & answer→Question 562 Marks
In the following figure, the square ABCD is divided into five equal parts, all having same area. The central part is circular and the lines AE, GC, BF and HD lie along the diagonals AC and BD of the square. If AB = 22cm, find:
The circumference of the central part.
AnswerWe have a square ABCD.
We have,
AB = 22cm
We have to find the perimeter of the triangle. We have a relation as,
Area of circular region $=\frac{1}{5}$ (Area of ABCD)
So,
$\pi\text{r}^2=\frac{1}{2}(22)^2$
$\text{r}=\frac{22}{\sqrt{5\pi}}$
$=5.56$
So perimeter of the circular region,
$=2\pi\text{r}$
$=(2)\frac{22}{7}\Big(\frac{22}{\sqrt{5\pi}}\Big)$
$=34.88\text{cm}$ View full question & answer→Question 572 Marks
Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii $15\ cm$ and $18\ cm$.
AnswerLet the radius of a circle be $r$.
Circumference of a circle $=2\pi\text{r}$
Let the radii of two circles are $r_1$ and $r_2$ whose
values are $15\ cm$ and $18\ cm$ respectively.
i.e.,$r_1= 15\ cm$ and $r_2= 18\ cm$
Now, by given condition,
Circumference of circle = Circumference of first circle + Circumference of second circle
$\Rightarrow2\pi\text{r}=2\pi\text{r}_1+2\pi\text{r}_2$
$\Rightarrow\text{r}=\text{r}_1+\text{r}_2$
$\Rightarrow\text{r}=15+18$
$\therefore\text{r}=33\text{cm}$
Hence, required radius of a circle is $33\ cm$.
View full question & answer→Question 582 Marks
Four cows are tethered at four corners of a square plot of side 50m, so that they just cannot reach one another. What area will be left ungrazed?
AnswerSide of square (a) = 50m
$\therefore$ Area of the square field $=\text{a}^2=(50)^2\text{m}^2=2500\text{m}^2$
Radius of each quadrant (r) $=\frac{50}{2}=25\text{m}$
$\therefore$ Area of 4 quadrants $=4\times\frac{1}{4}\pi\text{r}^2$
$=\pi\text{r}^2=\frac{22}{7}\times(25)^2\text{m}^2$
$=\frac{22}{7}\times625=1964.29\text{m}^2$
Area of ungrazed place $=2500-1964.29$
$=535.71\text{m}^2$ View full question & answer→Question 592 Marks
If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.
AnswerLet ABCD be the square inscribed in a circle of radius r.
Here, OA = OB = r.
$\therefore\text{OA}^2+\text{OB}^2=\text{AB}^2$
$\Rightarrow\text{r}^2+\text{r}^2=\text{AB}^2$
$\Rightarrow2\text{r}^2=\text{AB}^2$
Now, area of square ABCD $=\text{AB}^2=2\text{r}^2$
Area of circle $=\pi\text{r}^2$
Now we will find the ratio of area of the circle and the square.
$\frac{\text{Area of circle}}{\text{Area of square}}=\frac{\pi\text{r}^2}{2\text{r}^2}=\frac{\pi}{2}$
Hence, the ratio area of the circle and square is $\pi:2.$ View full question & answer→