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From bicycle wheels to stadium designs, circles are fundamental to our lives. But have you ever stopped to think about how we measure the space they take up? Well, that is where Areas Related to Circles is important! The chapter for this topic will take our understanding of finding the area and perimeter of circular shapes, including sectors and segments. Whether we are designing a clock face, a round garden, or a curved road, it is used in a practical way!
To do well in NCERT Exemplar Class 10 Maths Chapter 11: Areas Related to Circles, you should focus on practicing questions within the different regions of a circle and their applications in the real world. Practicing questions from NCERT exemplars, past board exam papers, and sample question papers will assist in reinforcing your understanding of circles. Getting used to composite shapes or application-based problems will also help with your problem-solving skills and your marks. And also remember to practice the problems of your choice based on CBSE Class 10 Maths syllabus; if you practice on a regular basis, you will find that this chapter is among the easier and a higher-scoring sections.
Class 10 Maths Chapter 11 exemplar solutions Exercise: 11.1 Page number: 120-121 Total questions: 10 |
Question:1
If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then
(A) R1 + R2 = R
(B) R12 + R22 = R2
(C) R1 + R2 < R
(D) R12 + R22 < R2
Answer:
[B] R12 + R22 = R2
The radius of first circle = R1Question:2
If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then
(A) R1 +R2=R
(B) R1 +R2>R
(C) R1 +R2<R
(D) Nothing definite can be said about the relation among R1, R2 and R.
Answer:
[A] R1 +R2=R
R1 + R2=RQuestion:3
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 > 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 the square.
Answer:
[A] Area of the circle = Area of the square
Area of the circle > Area of the square
circumference of a circle=$2\pi r$
Let the radius of the circle = r
perimeter of a square =$4 \times side$
let the side of a square = a
According to question
circumference of a circle = perimeter of a square
$\begin{aligned} & 2 \pi r=4 a \\ & \pi r=2 a \\ & a=\frac{\pi r}{2} \\ & \therefore \frac{\text { Area of circle }}{\text { Area of square }}=\frac{\pi r^2}{\left(\frac{\pi r}{2}\right)^2}=\frac{\pi r^2}{\pi r^2} \times \frac{4}{\pi}=\frac{4 \times 7}{22}=\frac{14}{11}\end{aligned}$
And $\frac{14}{11}>1$
Hence, Area of the circle > Area of the square.
Question:4
Area of the largest triangle that can be inscribed in a semi-circle of radius r units is
(A) $r^{2}$ square unit
(B) $\frac{1}{2}r^{2}$ square unit
(C) 2$r^{2}$ square unit
(D) $\sqrt{2}r^{2}$ square unit
Answer:
[A] $r^{2}$ square unit
$r^{2}$ square unit
The base of the triangle = diameter of the triangle
= 2 x r
=2r {r is radius}
Height of triangle = r
$\\ Area=\frac{1}{2}\times base \times height\\ =\frac{1}{2} \times 2r \times r\\\\ =r^{2} \text{ Square unit}$
Question:5
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is
(A) 22: 7
(B) 14: 11
(C) 7: 22
(D) 11: 14
Answer
[B] 14:11
Solution
According to question
$2 \pi r=4a$ (Because perimeter of circle = 2πr Perimeter of square =4 $\times$ side)
$\pi r=2a$ (here side of square =a)
$a=\frac{\pi r}{2}\\\\ \therefore \frac{Area \; of \; circle}{Area \; of\; square}=\frac{\pi r^{2}}{\left (\frac{\pi r}{2} \right )^{2}}=\frac{\pi r^{2}}{\pi r^{2}}\times \frac{4}{\pi}=\frac{4 \times 7}{22}=\frac{14}{11}$ (Using area of square = a2)
Hence, the ratio of their areas is 14: 11.
Question:6
(B) 15 m
(C) 20 m
(D) 24 m
Answer:
[A] 10 m
Solution
Diameter of the first circle (D) = 16 m
Radius(R) =$\frac{16}{2}=8m$
Area =$\pi r^{2}=\pi \times 8 \times 8=64\pi$
Diameter of second circle (d) = 12 m
Radius(r) =$\frac{12}{2}=6m$
Area =$\pi r^{2}=\pi \times 6 \times 6=36\pi$
Let the radius of the new park = R1
Area =$\pi R_{1}^{2}$
According to the question
$\begin{aligned} & \quad 64 \pi+36 \pi=\pi R_1^2 \\ & 100 \pi=\pi R_1^2 \\ & 100=R_1^2 \\ & \pi R_1= \pm 10\end{aligned}$
R = – 10 is not possible because the Radius must be positive.
Hence, the Radius is 10m
Question:7
The area of the circle that can be inscribed in a square of side 6 cm is
(A) 36 $\pi$ cm2
(B) 18 $\pi$ cm2
(C) 12 $\pi$ cm2
(D) 9 $\pi$ cm2
Answer:
(D) 9 $\pi$ cm2
Solution
Diameter of circle (d) = 6 cm
Radius (r)=$\frac{d}{2}=\frac{6}{2}=3cm$
Area =$\pi r^{2}$ (area of circle = πr2)
$=\pi \times 3 \times 3\\ =9 \pi cm^{2}$
Question:8
The area of the square that can be inscribed in a circle of radius 8 cm is
(A) 256 cm2 (B) 128 cm2 (C) 64$\sqrt{2}$ cm2 (D) 64 cm2
Answer:
(B) 128 cm2
Area of square =a2
Diagonal of square = Diameter of circle
Diagonal of square =8 $\times$ 2 =16cm
Let the side of the square = a cm
Using Pythagoras' theorem in ABC
(16)2=a2+a2
2a2=256
a2=128
Area of square ABCD = a2
=128 cm2
Question:9
The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36cm and 20 cm is
(A) 56 cm (B) 42 cm (C) 28 cm (D) 16 cm
Answer:
(C) 28 cm
Circumference of circle = $2 \pi r$
Diameter of first circle (d1) = 36
Radius (r1) =$\frac{d_{1}}{2}=\frac{36}{2}=18$
Diameter of second circle (d2) = 20 cm
Radius (r2)=$\frac{d_2}{2}=\frac{20}{2}=10$
Let the Radius of 3rd circle = R cm
According to the question
$\\2 \pi (18)+2 \pi (10)=2 \pi R\\ 2 \pi (18+10)=2 \pi R \\ R=28 cm$
Question:10
Answer:
(D) 50 cm
area of circle = $\pi r^{2}$
Radius of first circle (r1) = 24
Area $=\pi r_1^2=\pi(24)^2$
The radius of the second circle (r2) = 7 cm
Area $\pi r_{2}^{2}=\pi(7)^{2}$
Radius of third circle = R
Area of third circle = $\pi R^{2}$
According to the question
$\\ \pi (24)^{2}+\pi (7)^{2}=\pi R^{2}\\ \pi(576+49)=\pi R^{2}\\ 625=R^{2}\\ R=\pm 25\\ R=25$(Because radius is always positive)
Radius of circle = 25 cm
Diameter = 2$\times$R = 2$\times$25 = 50cm
Class 10 Maths Chapter 11 exemplar solutions Exercise: 11.2 Page number: 122-123 Total questions: 14 |
Question:1
Answer:
[False]
Use area of circle =$\pi r^{2}$
Side of square = a
Diameter of a circle = a {$\because$ circle inscribed in square}
Radius =$\frac{a}{2}$
Area =$\pi r^{2}$
= $\frac{\pi a^{2}}{4} cm^{2}$ (Because Radius = $\frac{a}{2}$ )
Hence, the given statement is not true because the area of a circle inscribed in a square of side a cm is $\frac{\pi a^{2}}{4} cm^{2}$
Question:2
Answer:
[true]
Perimeter of a square = 4 $\times$ side
The radius of the circle = a
Diameter of circle =2 $\times$ radius= 2a
Side of square =diameter of circle= 2a
Perimeter of square =4 $\times$ side=4 $\times$ 2a=8a
The perimeter of square is 8a
Hence, given statement is true.
Question:3
Answer:
Diameter of circle = d
Side of biggest square = d
The area of the biggest square is = side × side
=d $\times$ d=d2
The diagonal of smallest square = d
Let side = a
d2=a2+a2 {using Pythagoras theorem}
d2=2a2
$\frac{d}{\sqrt{2}}=a$
Area of smallest square $\frac{d}{\sqrt{2}} \times \frac{d}{\sqrt{2}}=\frac{d^{2}}{2}$
Here we found that the area of the outer square is not 4 times the area of the inner square.
Question:4
Answer:
[False]
From the above figure it is clear that the given statement is true only in the case of the minor segment. But the area of the major segment is always the greatest
Hence given statement is False.
Question:5
Answer:
[False]
Circumference of circle =2$\pi$r
Diameter = d
Radius =$\frac{d}{2}$
Circumference =2$\pi$r
= $2 \pi \frac{d}{2}=\pi d$
Here we found that the distance travelled by a circular wheel of diameter d cm in one revolution is πd, which is not equal to 2πd.
Hence, the given statement is False.
Question:6
Answer:
[True]
$\because$ Circumference of circle =$2 \pi r$
The radius of the circular wheel = r m
Circumference of wheel = $2 \pi r$
Distance covered in One revolution= circumference of wheel =$2 \pi r$
In covering a distance of s number of revolution required =$\frac{s}{2 \pi r}$
Hence, the given statement is True.
Question:7
Answer:
[False]
Area of circle =$\pi r^{2}$
Circumference of circle =$2\pi r$
Case 1:
Let r = 1
Area of circle =$\pi r^{2}$=$\pi(1)^{2}=\pi$
Circumference of circle = $2\pi r$= $2\pi (1)=2 \pi$
Case 2:
Let r = 3
Area of circle = πr2 = π(3)2 = 9π
Circumference of circle = 2πr = 2π(3) = 6π
Conclusion:- In case (1), we found that the area is less than the circumference, but in case (2,) we found that the area is greater than the circumference.
So, from the conclusion, we observe that it depends on the value of the radius of the circle.
Hence, the given statement is false.
Question:8
Answer:
[True]
$\because$ Formula of length of arc= $\frac{2 \pi r\theta }{360}$
Let the Radius of the first circle = r
Length of arc =$\frac{2 \pi r\theta_{1} }{360}$ ….. (1) {$\theta _{1}$ is the angle of first circle}
The radius of the second circle = 2r
Length of arc= $\frac{2 \pi (2r)\theta_{2} }{360}$
=$\frac{4\pi r\theta_{2} }{360}$ …..(2) {$\theta _{2}$ is the angle of second circle}
According to the question
$\\\frac{2 \pi r\theta_{1} }{360}=\frac{4\pi r\theta_{2} }{360}\\\\ \theta_{1} =2\theta_{2}$
No, this statement is True
Question:9
Answer:
[False]
Let the radius of the first circle be r1 and that of the other be r2
The length of the arcs of both circles is the same.
Let the arc length = a.
length of arc (a)=$2\pi r \times \frac{\theta }{360^{\circ}}$
Area of sector of first circle = $a \times \frac{r_{1} }{2}$ (because area of sector = $\pi r^{2}\times \frac{\theta }{360^{\circ}}=\left [ 2\pi r \times \frac{\theta }{360^{\circ}} \right ] \times \frac{r}{2}$ )
Area of sector of second circle = $a \times \frac{r_{2} }{2}$
Here we found that the area of the sector depends on the radius of the circles.
When the circle is the same, then the radius is also the same, then the given statement is true.
But in the case of different circles then the radius is also different
Hence, the given statement is False.
Question:10
Answer:
[True]
Let the radius of the first circle is r1 and of the other is r2
Let the arc length of both circles are same.
Let the arc length be a.
length of arc (a)= $2\pi r \times \frac{\theta }{360^{\circ}}$
Area of sector of first circle = $a \times \frac{r_{1} }{2}$
(because area of sector = $\pi r^{2}\times \frac{\theta }{360^{\circ}}=\left [ 2\pi r \times \frac{\theta }{360^{\circ}} \right ] \times \frac{r}{2}=\frac{r \times a}{2}$ )
Area of sector of other circle = $a \times \frac{r_{2} }{2}$
Here we found that both areas are equal in the case of when r1 = r2
Hence, the area of two sectors of two different circles would be equal only in case of both circles have equal radii and equal corresponding arc lengths.
Hence, it is necessary that their corresponding arc lengths are equal.
Question:11
Answer:
[False]
Diameter of circle = b
Radius =$\frac{b}{2}$
Area =$\pi r^{2}=\pi \left (\frac{b}{2} \right )^{2}=\frac{1}{4}\pi b^{2}cm^{2}$
Here we found that the area of the largest circle is not equal πb2 cm2.
Hence, the given statement is False.
Question:12
Circumferences of two circles are equal. Is it necessary that their areas be equal? Why?
Answer:
[True]
Use circumference of circle =$2 \pi r$
Let two circles having radius r1 and r2
Here, it is given that their circumferences are equal
$2 \pi r_{1}=2 \pi r_{2}\\ \Rightarrow r_{1}= r_{2}$
We know that area of circle = $\pi r^{2}$
Area of circle with radius r1 = πr12
Area of circle with radius r2 = πr22 ……..(1)
Put r2 = r1 in (1) we get
πr22 = πr12
Hence the area of given circles are also equal because two circles with equal radii will also have equal areas.
Hence the given statement is True.
Question:13
Areas of two circles are equal. Is it necessary that their circumferences are equal? Why?
Answer:
[True]
Solution
Use area of circle $\pi r^{2}$
Let two circles having radius r1 and r2
Here it is given that their areas are equal
$\pi r_{1}^{2}=\pi r_{2}^{2}\\ \\ r_{1}^{2}= r_{2}^{2}\\ \\ r_{1}= r_{2}\\$
We know that the circumference of the circle =$2\pi r$
Circumference of circle with radius r1 = 2πr1
Circumference of circle with radius r2 = 2πr2 …..(1)
Put r2 = r1 in (1) we get
2πr1 = 2πr2
Hence the circumference of given circle are also equal because two circles with equal radii will also have equal circumference.
Therefore, the given statement is True.
Question:14
Is it true to say that the area of a square inscribed in a circle of diameter p cm is p2cm2? Why?
Answer:
[False]
In the figure, we see that the diameter of the circle is equal to the diagonal of a square
Hence, the diagonal of square = p cm
Let side of the square = a cm Using Pythagoras' theorem we get
$\\p^{2}=a^{2}+a^{2}\\ p^{2}=2a^{2}\\ \frac{p^{2}}{2}=a^{2}\\ a=\frac{p}{\sqrt{2}}$
Area of square = side × side
$=\frac{p}{\sqrt{2}} \times \frac{p}{\sqrt{2}}=\frac{p^{2}}{2}cm^{2}$
Here we found that area of square is not equal to p2cm2.
Hence the given statement is False
Class 10 Maths Chapter 11 exemplar solutions Exercise: 11.3 Page number: 125-128 Total questions: 16 |
Question:1
Answer:
Radius = 33 cm
Solution
$\because$ Circumference of circle =$2 \pi r$
The radii of the two circles are $r_{1}=15 cm$ and $r_{2}=18 cm$
Let the circumference of these two are $C_{1}$ and $C_{2}$ respectively.
Let the circumference of required circle is C with radius R.
So according to question $C=C_{1}+C_{2}$
$2 \pi R=2 \pi r_{1}+2 \pi r_{2}\; \; \; \; \; \; \; \; (\because C=2 \pi r)\\ 2 \pi R=2 \pi(15+18)\\ R=33 cm$
Hence the radius of the required circle is 33cm.
Question:2
In Figure, a square of diagonal 8 cm is inscribed in acircle. Find the area of the shaded region.
Answer:
[18.24 cm2]
Area of square =(side)2 ,
Area of circle =$\pi r^{2}$
Diagonal of square = Diameter of circle = 8 cm
Using Pythagoras theorem in $\triangle$ABC
$(AB)^{2}+(BC)^{2}=(8)^{2}\\ a^{2}+a^{2}=(8)^{2}\\ 2a^{2}=(8)^{2}\\ a^{2}=\frac{64}{2}\\ a=\sqrt{32}=4\sqrt{2}$
Area of square ABCD =a2
$=4\sqrt{2}\\ =32 cm^{2}$
Diameter of circle = 8 cm
Radius (r) =$\frac{8}{2}=4cm$
Area of circle = $\pi r^{2}\\$
$= 3.14 \times 4 \times 4=50.24 cm^{2}$
Area of shaded region = Area of circle – Area of square
=50.24 - 32
=18.24 cm2
Question:3
Find the area of a sector of a circle of radius 28 cm and central angle 45°.
Answer:
308 cm2
Solution
Area of sector =$\frac{\pi r^{2} \theta }{360^{\circ}}$
The radius of the circle r = 28 cm
Angle (q) = 45°
Area of sector = $\frac{\pi r^{2} \theta }{360^{\circ}}$
$\\=\frac{22 \times 28 \times 28 \times 45 }{7 \times 360}\\ =308 cm^{2}$
Question:4
Answer:
500 revolutions
Circumference of circle =$2 \pi r$
The speed of wheel = 66 km per hour =$\frac{66 \times 1000}{60}$
= 1100 m/min
Radius =$= 35m=\frac{35}{100}=0.35m$ (because 1m = 100cm)
Circumference of wheel =$2 \pi r=2 \times \frac{22}{7} \times 0.35$
=2.2 m
The distance covered by the wheel in one revolution = 2.2m
The number of resolutions per minute to keep a speed of 66 km per hour =$\frac{1100}{2.2} =500$ revolution
Question:5
Answer:
154 m2
Solution
$\because$ Area of circle =$\pi r^{2}$
According to the question
In the figure we see that the area grazed by the cow is in the form of the fourth part of a circle
Hence area grazes by cow =$\frac{\theta}{360^{\circ}}\times \pi r^{2}$ here $\theta=90^{\circ}$
$=\frac{1}{4} \times \pi r^{2}$ (Because Area of circle $\pi r^{2}$
$=\frac{1}{4} \times \frac{22}{7}\times 14 \times 14$
$=11 \times 14=154 m^{2}$
Question:6
Find the area of the flower bed (with semi-circular ends) shown in Figure.
Answer:
458.5 cm2
Solution
If we observe the figure
We found that there is a rectangle and two semi-circles in it.
The length and breadth of the rectangle is 38cm and 10cm respectively,
Area of rectangle =$l \times b =38 \times 10=380 cm^{2}$
Diameter of semi-circle = 10cm
Radius of semi circle=$\frac{10}{2}=5 cm$
Area of semi circle=$\frac{1}{2}\pi r^{2}=\frac{1}{2} \times 3.14 \times 25=39.25 cm^{2}$
Hence the total Required area = Area of rectangle + 2(Area of semi-circle)
$\\=380+2 \times 39.25\\ =380+78.5\\ =458.5 cm^{2}$
Question:7
Answer:
[54.5 cm2]
Given: AC = 6cm and BC = 8cm
In the figure $\triangle$ ABC is a right angle triangle.
Hence using Pythagoras' theorem
$\\(AB)^{2}=(AC)^{2}+(BC)^{2}\\ =(6)^{2}+(8)^{2}\\ =36+64\\ =100\\ AB=\sqrt{100}=10\\ AB=10 cm$
Diameter of circle = AB = 10 cm
Radius =$\frac{10}{2}=5 cm$
Area of circle =$\pi r^{2}$
$=3.14 \times (5)^{2}=78.5 cm^{2}$
Area of $\triangle ABC=\frac{1}{2}\times AC \times BC\\$
$=\frac{1}{2} \times 6 \times 8=24m^{2}$
Area of shaded region = Area of circle – Area of DABC
=78.5-24=54.5 cm2
Question:8
Find the area of the shaded field shown in Figure.
Answer:
[38.28 m2]
Here length and breadth of the rectangle ABCD is 8m and 4m respectively.
Are of rectangle $ABCD=l \times b=8 \times 4 = 32 m^{2}$
Radius of semi-circle = 2m
Area of semi circle=$\frac{1}{2}\pi r^{2}$
=$\frac{1}{2}\times 3.14 \times (2)^{2}=6.28 m^{2}$
Area of shaded field = Area of rectangle ABCD + Area of semi-circle
= 32+6.28
= 38.28 m2
Question:9
Find the area of the shaded region in Figure.
Answer:
235.44 m2
There are two semi-circle with diameter (d) 4 cm.
Radius(r) =$\frac{d}{2}=\frac{4}{2}=2m$
Area of semi-circle =$\frac{1}{2}\times \pi \times (r)^{2}=\frac{1}{2}\times \pi \times (2)^{2}=2 \pi$
The length and breadth of rectangle ABCD is 16m and 4m respectively
Area of ABCD=16 x 4=64 m2 ($\because$ Area of rectangle = length× breadth)
The length and breadth of rectangle UVWX is 26m and 12m respectively
Area of UVWX=26 x 12 =312 m2 ($\because$ Area of rectangle = length× breadth)
Area of shaded region = Area of UVWX – Area of ABCD – 2 × Area of semi-circle
$\\=312 -64-2(2 \pi)\;\;\;\;\;\;\;\;\;\;(here \; \pi=3.14)\\ =312-64-12.56\\ =235.44 m^{2}$
Question:10
Answer:
$\left [ \frac{308-147\sqrt{3}}{3} \right ]$
Solution
Here $\theta=60^{\circ}$
r=14 cm
Area of segment = $\frac{\pi r^{2} \theta}{360}-\frac{1}{2}r^{2}\sin \theta$
$\\=\frac{\frac{22}{7}\times 14 \times 14 \times 60}{360}-\frac{1}{2}\times 14 \times 14 \times \sin 60\\ =\frac{22 \times 28 \times 60}{360}-\frac{1}{2}\times 14 \times 14 \times \frac{\sqrt{3}}{2}\\ =\frac{308}{3}-49\sqrt{3}\\ =\frac{308-147\sqrt{3}}{3}cm^{2}$
Question:11
Answer:
[30.96 cm2]
Solution
Here ABCD is a square of side 12 cm
Area of ABCD= (side)2=(12)2=144 cm2
Area of sector =$\frac{\theta }{360^{\circ}} \times \pi r^{2}$ here $\theta=90^{\circ}$
Here PSAP, PQBP, QRCQ, RSDR all sectors are equal
Area of 4 sectors =$4 \times \frac{\theta }{360^{\circ}} \times \pi r^{2}$
$=4 \times \frac{1 }{4} \times \pi r^{2}\\ =3.14 \times 36\\ =113.04 cm^{2}$
Area of shaded region = Area of square – Area of 4 sectors
= 144-113.04
=30.96 cm2
Question:12
Answer:
39.25 cm2
Solution
Angle made by vertices A, B and C = 60° { In equilateral triangle all angles = 60°}
Diameter of circle = 10
Radius =$\frac{10}{2}=5 cm$
Area of shaded region = 3 × Area of sector
$\\=3 \times \frac{\pi r ^{2} \theta}{360}\\ =\frac{3 \times (5)^{2} \times 3.14 \times 60 }{360}\\ =\frac{25 \times 3.14}{2}\\ =\frac{25 \times 314}{2 \times 100}\\ =\frac{25 \times 157}{100}\\ =39.25 cm^{2}$
Question:13
Answer:
[ 308 cm2]
Solution
Area of sector with angle $\theta = \frac{\pi r^{2} \theta}{360^{\circ}}$
Here $\angle p, \angle Q, \angle R=60^{\circ}$
The radius of each circle = 14 cm
There are three sectors
Area of each sector = $\frac{\pi \times (14)^{2} \times 60}{360}\\$
$\\=\frac{\frac{22}{7} \times 196}{6}\\ =\frac{616}{6}cm^{2}$
Area of shaded region = 3 x (Area of one sector)
$\\=3 \times \frac{616}{6}\\ =308cm^{2}$
Question:14
Answer:
15246 m2
Area of circle =$\pi r^{2}$
Given that AB = 105m, BC = 21m
Where AB is the radius of the park and BC is the wide of road
AC=AB+BC
AC=105+21=126 m
Area of big circle=$\pi r^{2}$
$\\=\pi (126)^{2}\\ =49896 m^{2}$
Area of small circle =$\pi r^{2}$
$\\=\pi (105)^{2}\\ =34650 m^{2}$
Area of road =Area of big circle - Area of small circle
=49896-34650=15246 m2
Question:15
Answer:
[1386cm2]
Area of sector =$\frac{\pi r^{2}\theta}{360^{\circ}}$
Here $\theta=90^{\circ}$
Radius = 21 cm
There are four sectors in the figure
Area of sector =$\frac{\pi \times (21)^{2} \times 90}{360}$
=$\frac{\frac{22}{7} \times 441}{4}=346.5 cm^{2}$
Area of shaded region = 4 × Area of one sector
= 4 × 346.5
= 1386 cm2
Question:16
Answer:
$\frac{60}{\pi}cm$
Solution
Given $\theta=60^{\circ}$
Length of arc = 20 cm
We know that
Length of arc =$\frac{\theta}{360}\times 2 \pi r$
$\\20=\frac{\theta}{360}\times 2 \pi r\\\\ \frac{20 \times 360}{60 \times 2 \pi}=r\\ \\r=\frac{60}{\pi}cm$
Class 10 Maths Chapter 11 exemplar solutions Exercise: 11.4 Page number: 132-135 Total questions: 20 |
Question:1
Answer:
[26400]
Given:- Area of circular playground = 22176 m2
Rate of fencing = 50 Rs. per meter
Circumference of circle =$2 \pi r$
We have to find the radius (r) of the playground
Area of playground =$\pi r^{2}$
$\\22176=\frac{22}{7}\times r^{2}\\\\ r^{2}=\frac{22176 \times 7}{22}\\\\ r^{2}=\frac{155232}{22}\\\\ r^{2}=7056\\ r=\sqrt{7056}\\ r=84m$
Circumference of playground =$2 \pi r$
$\\=2 \times \frac{22}{7}\times 84\\ =528 m$
Cost of fencing the playground = 528 $\times$ 50
= 26,400 Rs.
Hence the cost of fencing the playground at the rate of 50 per meter is 26400 Rs.
Question:2
Answer:
560 revolutions.
Solution:
Circumference of circle =$2 \pi r$
The diameter of front wheel = 80 cm
Radius =$\frac{80}{2}=40cm$
Diameter of rear wheel = 2m = 200 cm
Radius = $\frac{200}{2}=100cm$
We know that distance covered in one revolution =$2 \pi r$
distance covered by front wheel in 1400 revolution =$1400 \times 2 \pi(40)$
Let rear wheel take x revolutions = x × 2π(100)
According to question
$\\1400 \times 2\pi(40)= x \times 2\pi(100)\\\\ x=\frac{1400 \times 40}{100}=560$
The rear wheel will make 560 revolutions.
Question:3
Answer:
$\left ( 24 \sqrt{21}-72 \right )m^{2}$
Solution
Let their angles of triangle are $\angle A,\angle B$ and $\angle C$ rope’s length (radius) = 7 cm
Area of sector with angle A = $\frac{\pi r^{2} \times \angle A }{360}=\frac{\pi \times (7)^{2} \times \angle A}{360}$
Area of sector with angle B= $\frac{\pi r^{2} \times \angle B }{360}=\frac{\pi \times (7)^{2} \times \angle B}{360}$
Area of sector with angle C= $\frac{\pi r^{2} \times \angle C }{360}=\frac{\pi \times (7)^{2} \times \angle C}{360}$
$\therefore$Sum of the areas are
$=\frac{\left ( \angle A +\angle B+\angle C \right )\times \pi \times (7)^{2}}{360}$
$=\frac{180}{360}\times\frac{22}{7}\times 49$ {$\because$ Sum of angles of a triangle = $180^{\circ}$ }
=77 m2
Sides of triangular field are 15m, 16m and 17m
Let a =15m, b =16m, c = 17m
$\\S=\frac{(a+b+c)}{2}\\ =\frac{(15+16+17)}{2}\\ =\frac{48}{2}\\ =24m$
Area of triangular field
$\\=\sqrt{s(s-a)(s-b)(s-c)}\\ =\sqrt{24(24-15)(24-16)(24-17)}\\ =\sqrt{24 \times 9 \times 8 \times 7}\\ =\sqrt{8 \times 3 \times 9 \times 8 \times 7}\\ =8\sqrt{3 \times 3 \times 3 \times 7}\\ =24 \sqrt{21}m^{2}$
So area of the field which cannot be grazed by the animals = Area of triangular field – Area of sectors of field
$=\left (24 \sqrt{21}-77 \right )m^{2}$
Question:4
Answer:
$(75.36-36\sqrt{3})cm^{3}$
Solution
Area of sector – Area of triangle
Radius = 12 cm
Angle = 60°
Area of sector OAB =$\frac{\pi r^{2} \theta}{360}$
$\\=\frac{3.14 \times 12 \times 12}{360}\times 60\\\\ =75.36 cm^{2}$
DAOB is isosceles triangles
Let $\angle OAB=OBA=X$
$\\OA=OB=12 cm\\ \angle AOB=60^{0}$
$\angle OAB +\angle OBA+\angle AOB=180^{0}$ {$\because$ Sum of all interior angles of a triangle is 180°}
x + x +60=180
2x =120
x=60
Here all the three angles are 60° $\therefore$given triangle is an equilateral triangle.
Area of $\triangle AOB=\frac{\sqrt{3}}{4}(side)^{2}$ {$\because$ Area of equilateral triangle$=\frac{\sqrt{3}}{4}(side)^{2}$ }
$=\frac{\sqrt{3}}{4}(12 \times 12)$
$=36\sqrt{3}cm^{2}$
Area of segment = Area of sector OBCA – Area of DAOB
= $\left (75.36-36\sqrt{3} \right )cm^{2}$
Question:5
Answer:
[3064.28 RS ]
Given the Diameter of the circular pond = 17.5 m
Radius (r)=$\frac{17.5}{2}m$
Width of Path = 2m
R=r+width of path
$\\=\frac{17.5}{2}m+2m\\\\=\frac{17.5+4}{2}=10.75m$
Area of path =$\pi(R^{2}-r^{2})$
$\\=\frac{22}{7}\times(10.75^{2}-8.75^{2})\\\\ =\frac{22}{7}\times(115.5625-76.5625)\\\\ =\frac{22}{7}\times(39)\\\\ =122.57m^{2}$
Area of path = 122.57m2
Rate of construction = Rs 25 per m2
Cost of construction $=25\times 122.57=3064.28Rs$
Question:6
Answer:
[196 m2]
Given AB = 18 cm
DC = 32 cm
Radius of circle = 7 cm
Area of trapezium =$\frac{1}{2}$ × sum of the parallel sides × distance between parallel sides.
= $\frac{1}{2}$ ×(AB+CD) × 14
=$\frac{1}{2}$ ×(18+32) × 14
=(50) × 7
=350 cm2
$\theta=\angle A+\angle B+\angle C+\angle D$ = 360° (sum of interior angles of quadrilateral)
Radius = 7 cm
Area of all sectors =$\frac{\theta}{360^{\circ}}\times \pi r^{2}$ here $\theta=360^{\circ}$
=$\pi r^{2}$
=$\frac{22}{7}\times 7 \times 7$
=154 cm2
Area of shaded region = Area of trapezium – Area of all sectors
=350-154=196 cm2
Question:7
Answer:
1.966 cm2
Radius = 3.5
Diameter = 3.5 + 3.5 = 7 cm
Here ABC is an equilateral triangle because AB = BC = CA = 7 cm
$\therefore$Area of $\triangle ABC=\frac{\sqrt{3}}{4}a^{2}$
$\\=\frac{\sqrt{3}}{4} \times 7 \times 7\\ =\frac{49\sqrt{3}}{4}\\ =21.217 cm^{2}$
In an equilateral triangle each angle = 60°
All the sectors are the same
$\therefore$ Area of all there sectors $=3 \times \frac{\theta}{360^{\circ}} \times \pi r^{2}$
$\\=3 \times \frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times (3.5)^{2}\\ =19.251 cm^{2}$
Area of enclosed region = Area of DABC – Area of their sectors
=21.217-19.251
= 1.966cm2
Question:8
Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.
Answer:
[8.75 cm2]
Length of the Arc = 3.5 cm
Radius (r)= 5 cm
Area of sector=$\frac{1}{2}$ $\times$ radius $\times$ length of arc
$\\=\frac{1}{2} \times r \times \frac{\theta}{360^{\circ}}\times 2\pi r\\\\ =\frac{1}{2} \times 5\times3.5=8.75cm^{2}$
Question:9
Answer:
[42 cm2]
Area of shaded portion = Area of square – Area of 4 sectors
Radius = 7 cm
Side of square = 14 cm
AB=BC=CD=DA=14 cm
$\therefore$ ABCD is a square
$\therefore$Area of square= 14 $\times$ 14 (Area of square = (side)2 )
=196 cm2
We know that each angle of square = $90^{\circ}$
Area of 4 sectors =$4 \times \frac{\pi r^{2} \theta}{360}\\$
$\\=\frac{4 \times 7 \times 7 \times 90}{360} \times \frac{22}{7}\\\\ =154 cm^{2}$
Area of shaded portion = Area of square – Area of 4 sectors
=196-154=42cm2
Question:10
Answer:
[168 cm2]
Given area of sheet =784 cm2
Let the side of the sheet = a
a2=7842 (Area of square = (side) 2)
a=$\sqrt{784}=28cm$
Diameter of each circular plate =$\frac{a}{2}=\frac{28}{2}=14cm$
Radius =$\frac{d}{2}=\frac{14}{2}=7cm$
Area of 4 circular plates =$4 \times \pi r^{2}$
$\\=4 \times \frac{22}{7} \times 7 \times 7\\ =616cm^{2}$
Area of sheet not covered with circular plates = Area of sheet – Area of 4 circular plates.
=784-616=168cm2
Question:11
Answer:
[4.3 m2]
Diameter of tile =50cm=0.5m (1m = 100cm)
Radius =$\frac{50}{2}=25=0.25m$
Number of tiles lengthwise =$\frac{5}{0.5}=10$ tiles
Number of tiles widthwise =$\frac{4}{0.5}=8$ tiles
Total tiles =$10 \times 8=80$
Area of floor not covered by tiles = Area of rectangular floor – Area of 80 tiles
$\\=5 \times 4-80\pi r^{2}\\ $
$=20-80 \times \pi \times 0.25 \times 0.25\\ $
$=20-\frac{8 \times 314 \times 25 \times 25}{100 \times100\times100}\\$
$ =20-\frac{157}{10}\\$
$ =20-15.7\\ =4.3m^{2}$
Question:12
Answer: [800 cm2]
Solution
Given that area of circle =1256cm2
$\\\pi r^{2}=1256\\\\ r^{2}=\frac{1256}{314}\times 100\\\\ r^{2}=400 \; \; \; \; \; \; \; \; \left ( \pi=3.14 \right )\\\\ r=\sqrt{400}\\\\ r=20cm$
Diameter of circle = 40 cm
As we know that the diameter of circle is equal
Diagonals of rhombus = Diameters of circle = 40 cm
Each diagonals of rhombus = 40 cm
Area of rhombus =$\frac{1}{2}$ $\times$ product of digonals
= $\frac{1}{2}$ $\times$ 40 $\times$ 40
= 800cm2
Hence the required area of rhombus is =800cm2
Question:13
Answer:
[1 : 3 : 5]
Solution
d1:d2:d3 = 1: 2 : 3 [multiplying by s]
= s : 2s : 3s
Radius of inner circle (r1)=$\frac{s}{2}$
Radius of middle circle (r2)=$\frac{2s}{2}=s$
Radius of outer circle (r3)=$\frac{3s}{2}$
Area of region enclosed between second and first circle
$\\=\pi r_{2}^{2}-\pi r_{1}^{2}\\ =\pi s^{2}-\frac{\pi s^{2}}{4}\\ =\frac{3 \pi s^{2}}{4}$
Area of region enclosed between third and second circle
$\\=\pi r_{3}^{2}-=\pi r_{2}^{2}\\ =\frac{\pi 9s^{2}}{4}-\pi s^{2}\\ =\frac{ 5\pi s^{2}}{4}$
Area of first circle $=\pi r_{1}^{2}=\frac{\pi s^{2}}{4}$
Ratio of area of three regions
$\\=\frac{\pi s^{2}}{4}:\frac{3\pi s^{2}}{4}:\frac{5\pi s^{2}}{4}\\ =\pi s^{2}:3 \pi s^{2}:5\pi s^{2}\\ =1:3:5$
Question:14
Answer:
$\left [ 45\frac{5}{6}cm^{2} \right ]$
Solution
We know that minute hand revolving in 60 min =$360^{\circ}$
In 1 minute it is revolving =$\frac{360}{60}=6^{\circ}$
Time difference =(6:40am -6:05am) =35 min
At 6:05 am and 6.40 am, there is 35 minutes
In 35 minutes angle between min hand and hour hand =$\left (6\times 35 \right )^{\circ} =210^{\circ}$
Length of minute hand (r)=5cm
Area of sector =$\frac{\pi r^{2}\theta}{360}$
$\\=\frac{22}{7}\times \frac{5 \times 5 \times 210^{\circ}}{360}\;\;\;\left \{\because \theta = 210^{\circ} \right \}\\ =\frac{11\times 5\times 5}{6}\\ =\frac{275}{6}\\ =45\frac{5}{6}cm^{2}$
Hence required area is $45\frac{5}{6}cm^{2}$
Question:15
Answer:
$\left [ 73\frac{1}{3}cm \right ]$
Area of sector=$\frac{\pi r^{2}\theta}{360}$
Angle = 200°
Area of sector = 770 cm2
$\\\frac{\pi r^{2}\theta}{360}=770\\\\ \pi r^{2} \times 200^{\circ}=770 \times 360 ^{\circ}\\\\ r^{2}=\frac{770 \times 360 ^{\circ} \times 7}{200^{\circ \times 22}} \;\;\; \left [ here \pi=\frac{22}{7} \right ]\\\\ r^{2}=49 \times 9\\ r=\sqrt{49 \times 9}=7 \times 3=21cm$
Length of the corresponding arc =$\frac{\theta \times 2\pi r}{360}$
$\\=\frac{200^{\circ} \times 2 \times \pi \times 21}{360^{\circ}}\\\\ =\frac{10 \times 7}{3}\times\frac{22}{7}\\\\ =\frac{220}{3}\\\\ =73\frac{1}{3}cm$
Question:16
Answer:
$\because$ Area of sector=$\frac{\pi r^{2} \theta}{360}$
Radius of first sector(r1) = 7 cm
Angle ($\theta_{1}$ ) = 120°
Area of first sector(A1) =$\frac{\pi r_{1}^{2} \theta}{360}$
$\\=\frac{22 \times 7 \times 7 \times 120^{\circ} }{360^{\circ}}\\ =\frac{154}{3}cm^{2}$
Radius of second sector(r2) = 21 cm
Angle ($\theta_{2}$ ) = 40°
Area of sector of second circle (A2)=$\frac{\pi r_{2}^{2} \theta}{360}$
$\\=\frac{22}{7}\times \frac{21\times 21}{360}\times40\\\\ =154 cm^{2}$
Corresponding arc length of first circle =$\frac{2\pi r_{1} \theta}{360}$
=$\frac{\pi r_{1} \theta}{180}$
$\\=\frac{22}{7}\times \frac{7\times 120}{180}\\=\frac{44}{3}cm$
Corresponding arc length of second circle =$\frac{2\pi r_{2}\theta}{360}$
=$\frac{\pi r_{2}\theta}{180}$
$\\=\frac{22}{7}\times \frac{21\times 40}{180}\\=\frac{44}{3}cm$
We observe that the length of arc of both circle are equal.
Question:17
Find the area of the shaded region given in Figure.
Answer:
[Area of shaded area=154.88cm2 ]
Area of square PQRS =(side)2=(14)2
=196 cm2
Area of ABCD (let side a) =(side)2=(a)2
Area of 4 semi-circle $\left ( r=\frac{a}{2} \right )=4 \times \frac{1}{2}\pi\left ( \frac{a}{2} \right )^{2}$
Area of semi-circle=$\frac{1}{2}\times \pi \times r^{2}$
$\\=\frac{2\pi a^{2}}{4}\\ =\frac{\pi a^{2}}{2}$
Total inner area = Area of ABCD + Area of 4 semi-circles
$\\=a^{2}+\frac{\pi a^{2}}{2}\\$
$\\EF=8cm\\ EF=\frac{a}{2}+a+\frac{a}{2}\\ 8=\frac{a+2a+a}{2}\\ 4a=16\\ a=4cm$
Area of inner region =$4^{2}+\frac{\pi 4^{2}}{2}\\$
$\\=16+\frac{ 16 \pi}{2}\\ =16+8 \pi$
Area of shaded area = Area of PQRS – inner region area
$\\=196-16-8 \pi\\ =180-8 \times 3.14\\ =180-25.12\\ =154.88 cm^{2}$
Question:18
Answer:
[40] revolutions
Circumference of circle =$2\pi r$
Area of wheel = 1.54m2
Distance = 176 m
$\\\pi r^{2}=1.54\\\\ r^{2}=\frac{154}{100}\times \frac{7}{22}\\\\ r=\sqrt{\frac{539}{1100}}$
r = 0.7m
Circumference =$2\pi r$
$\\=2 \times \frac{22}{7}\times0.7\\ =2 \times \frac{22}{7}\times \frac{7}{10}\\ =\frac{44}{10}\\ =4.4m$
Number of revolution $=\frac{\text{distance}}{\text{circumference}}\\$
$=\frac{176}{44} \times 10=40$ revolutions
Question:19
Answer [32.16cm2]
Solution
By using Pythagoras in $\triangle$ABC
$\\(AB)^{2}+(BC)^{2}=(AC)^{2}\\ r^{2}+r^{2}=25\\ 2r^{2}=25\\ r=\frac{5}{\sqrt{2}}cm$
Area of circle =$\pi r^{2}=\frac{22}{7} \times \frac{5}{\sqrt{2}}\times\frac{5}{\sqrt{2}}=39.28cm^{2}$
Area of sector =$\frac{\pi r^{2} \theta}{360}$
$\\=3.14 \times \frac{25}{2}\times\frac{90}{360}\\ =9.81cm^{2}$
Area of $\triangle ABC=\frac{1}{2}\times base \times height$
$\\=\frac{1}{2}\times \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}}\\ =\frac{25}{4} =6.25cm^{2}$
Area of minor segment = Area of sector – Area of DABC
=9.81-6.25
=3.56cm2
Area of major segment = Area of circle – Area of minor segment
=39.28-3.56=35.72cm2
Required difference = Area of major segment – Area of minor segment
=35.72-3.56=32.16cm2
Question:20
Answer: [462 cm2]
Solution
Radius = 21 cm
Angle = 120°
Area of circle = $\pi r^{2}$
$\\=\frac{22}{7}\times 21\times 21\\ =1386 cm^{2}$
Area of minor sector with angle 120° OABO =$\frac{\pi r^{2} \theta}{360}$ $\left [ \theta =120^{\circ} \right ]$
$\\=\frac{22}{7}\times \frac{21\times 21}{360}\times120\\ =462cm^{2}$
Area of major sector AOBA= Area of circle – area of minor sector
= 1386-462=924cm2
Required area =924-462=462cm2
The students can download the pdf version by using on NCERT Exemplar Class 10 Maths solutions chapter 11 pdf download feature of online tools. This feature especially caters to students learning in a low internet connectivity environment or those who plan to be offline while studying the NCERT Exemplar Class 10 Maths chapter 11.
These Class 10 Maths NCERT exemplar Chapter 11 solutions emphasise the area of sectors and circles.
In this chapter, NCERT exemplar problems are very tricky and will help develop an excellent logical brain. Sometimes the composite figure will be complicated to resolve, but the perfect analysis will resolve it to determine the perimeter and area.
Class 10 students can use these detailed solutions on area-related to Circles-based practice problems as reference content.
These Class 10 Maths NCERT exemplar solutions Chapter 11 Area related to Circles are sufficient to solve the problems of NCERT Class 10 Maths.
Here are the subject-wise links for the NCERT solutions of class 10:
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
Given below are the subject-wise NCERT Notes of class 10 :
Here are some useful links for NCERT books and NCERT syllabus for class 10:
Given below are the subject-wise exemplar solutions of class 10 NCERT:
The circle will have the maximum area among all the two-dimensional shapes for given circumference.
For the largest square area the diameter must be equal to diagonal of the square.
Hence, the area will be 2R2
The chapter Area related to circles is vital for Board examinations as it holds around 2-3% weightage of the whole paper.
Generally, the paper consists of either a Long Short Answer question or multiple short answer questions from this chapter. NCERT exemplar Class 10 Maths solutions chapter 11 can help the students develop high-order thinking skills and ace the Area related to Circles related problems.
Hello
Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.
1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.
2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.
3. Counseling and Seat Allocation:
After the KCET exam, you will need to participate in online counseling.
You need to select your preferred colleges and courses.
Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.
4. Required Documents :
Domicile Certificate (proof that you are a resident of Karnataka).
Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
https://medicine.careers360.com/neet-college-predictor
Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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As per latest 2024 syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
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