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NCERT Exemplar Class 10 Maths Solutions Chapter 11 Areas Related to Circles

NCERT Exemplar Class 10 Maths Solutions Chapter 11 Areas Related to Circles

Edited By Safeer PP | Updated on Sep 06, 2022 11:49 AM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 11 is an extension to the learnings of previous classes and chapters, including circles and sectors. In this chapter, we will emphasize the area of Sectors and circles. The topics of NCERT Class 10 Maths for ‘Areas related to Circles’ can be studied and practiced by these NCERT exemplar Class 10 Maths chapter 11 solutions prepared by our accomplished mathematics experts.

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  1. NCERT Exemplar Solutions Class 10 Maths Chapter 11 Important Topics:
  2. NCERT Class 10 Exemplar Solutions for Other Subjects:
  3. NCERT Class 10 Maths Exemplar Solutions for Other Chapters:
  4. Features of NCERT Exemplar Class 10 Maths Solutions Chapter 11:

These NCERT exemplar Class 10 Maths chapter 11 solutions are highly exhaustive and accurate to hone the skills based on areas related to circles-based problems. These NCERT exemplar Class 10 Maths solutions chapter 11 are prepared on the guidelines of the CBSE Syllabus Class 10 Maths.

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
Radius of first circle= R1
Area of first circle =πR12
Radius of second circle =R2
Area of second circle =πR22
Radius of third circle = R
Area of third circle=πR2
According to question
πR12 + πR22=πR2
π(R12 + R22)=πR2
R12 + R22=R2
Hence option B is correct.

Question: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
Radius of first circle= R1
circumference of first circle =2πR1
Radius of second circle =R2
circumference of second circle =2πR2
Radius of third circle = R
circumference of third circle=2πR
According to question
2πR1 + 2πR2=2πR
2π(R1 + R2)=2πR
R1 + R2=R
Hence option A is correct.

Question: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 square.

Answer:

(B) Area of the circle > Area of the square
circumference of a circle=2\pi rLet the radius of the circle = r
perimeter of a square =4 \times sidelet the side of a square = a
According to question
circumference of a circle = perimeter of a square
\\ 2 \pi r=4a\\ \pi r=2a\\\\ 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}And \frac{14}{11}> 1Hence 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) 2r^{2} square unit (D) \sqrt{2}r^{2} square unit

Answer:

(A) r^{2} square unit1114
Base of triangle = diameter of 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 ratio of their areas is 14: 11

Question:6

It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
(A) 10 m (B) 15 m (C) 20 m (D) 24 m

Answer: (A) 10 m
Solution
Diameter of 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 radius of new park = R1
Area =\pi R_{1}^{2}
According to question
64\pi+36\pi=\pi R_{1}^{2}\\ 100\pi=\pi R_{1}^{2}\\ 100= R_{1}^{2}\\ \pi R_{1}=\pm 10R = – 10 is not possible because Radius must be positive.
Hence 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
1117
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
1118
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}=10Let Radius of 3rd circle = R cm
According to question
\\2 \pi (18)+2 \pi (10)=2 \pi R\\ 2 \pi (18+10)=2 \pi R \\ R=28 cm

Question:10

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is
(A) 31 cm (B) 25 cm (C) 62 cm (D) 50 cm

Answer:

(D) 50 cm
area of circle = \pi r^{2}
Radius of first circle (r1) = 24
Area =\pi r_{1}^{2}=\pi(24)^{2}Radius of 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 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

Question:1

Is the area of the circle inscribed in a square of side a cm, \pi a2cm2? Give reasons for your answer.

Answer:

False
1121
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 given statement is not true because area of circle inscribed in a square of side a cm is \frac{\pi a^{2}}{4} cm^{2}

Question:2

Will it be true to say that the perimeter of a square circumscribing a circle of radius a cm is 8a cm? Give reasons for your answer.

Answer:

[true]
1122
Perimeter of a square = 4 \times side
Radius of 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
Perimeter of square is 8a
Hence given statement is true.

Question:3

In Figure, a square is inscribed in a circle of diameter ‘d’ and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reasons for your answer
11231

Answer:

11232
Diameter of circle = d
Side of biggest square = d
Area of the biggest square is = side × side
=d \times d=d2
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 area of outer square is not 4 times the area of inner square.

Question:4

Is it true to say that area of a segment of a circle is less than the area of its corresponding sector? Why?

Answer:

[False]
1124
From the above figure it is clear that the given statement is true Only in case of minor segment. But area of major segment is always greatest
Hence given statement is False.

Question:5

Is it true that the distance travelled by a circular wheel of diameter d cm in one revolution is 2\pid cm? Why?

Answer:

False
Circumference of circle =2\pir
Diameter = d
Radius =\frac{d}{2}
Circumference =2\pir
= 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

In covering a distance s metres, a circular wheel of radius r m makes \frac{s}{2 \pi r} revolution. Is this statement true? Why?

Answer:

[True]
\because Circumference of circle =2 \pi r
Radius of 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

The numerical value of the area of a circle is greater than the numerical value of its circumference. Is this statement true? Why?

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 conclusion we observe that it depend on the value of radius of the circle.
Hence the given statement false.

Question:8

If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle. Is this statement false? Why?

Answer:

True
\because Formula of length of arc= \frac{2 \pi r\theta }{360}
Let Radius of first circle = r
Length of arc =\frac{2 \pi r\theta_{1} }{360} ….. (1) {\theta _{1} is the angle of first circle}
Radius of 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 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

The areas of two sectors of two different circles with equal corresponding arc lengths are equal. Is this statement true? Why?

Answer:

False
Let the radius of first circle is r1 and of other is r2
The length of arcs of both circles is 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 sector is depending on radius of circles.
When the circle is same then radius is also same then the given statement is true.
But in case of different circles then the radius is also different
Hence the given statement is False.

Question:10

The areas of two sectors of two different circles are equal. Is it necessary that their corresponding arc lengths are equal? Why?

Answer:

True
Let the radius of first circle is r1 and of other is r2
Let the arc length of both circles are same.
Let the arc length is 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 the circles have equal radii and equal corresponding arc length.
Hence it is necessary that their corresponding arcs lengths are equal.

Question:11

Is the area of the largest circle that can be drawn inside a rectangle of length acm and breadth b cm (a >b) is \pib2 cm2?Why?

Answer:

False
11211
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 πb2cm2.
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 circumference of 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 area of a square inscribed in a circle of diameter p cm is p2cm2? Why?

Answer:

False
11214
In the figure we see that the diameter of circle is equal to diagonal of square
Hence, diagonal of square = p cm
Let side of 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

Question:1

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.

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.
11321

Answer:

[18.24 cm2]
Area of square =(side)2 ,
Area of circle =\pi r^{2}
11322
Diagonal of square = Diameter of circle = 8 cm
Using Pythagoras theorem in \triangleABC
(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:

ANswer: 308 cm2
Solution
Area of sector =\frac{\pi r^{2} \theta }{360^{\circ}}
1133
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

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 66 km/h?

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 wheel in one revolution = 2.2m
The number of resolution per minute to keep a speed of 66 km per hour =\frac{1100}{2.2} =500 revolution

Question:5

A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m. Find the area of the field in which the cow can graze.

Answer:

154 m2
Solution
\because Area of circle =\pi r^{2}
According to question
1135
In the figure we see that the area graze by cow is in the form of 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.
11361

Answer:

458.5 cm2
Solution
If we observe the figure
11362
We found that there is a rectangle and two semi circles in it.

Length and breadth of 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

In Figure, AB is a diameter of the circle, AC = 6 cm and BC = 8 cm. Find the area of the shaded region (Use p = 3.14).
1137

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.
11381

Answer: [38.28 m2]
11382
Here length and breadth of 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.
11391

Answer:

235.44 m2
11392
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
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)
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

Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.

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

Find the area of the shaded region in Figure, where arcs drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA, respectively of a square ABCD (Use \pi = 3.14).
11311

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

In Figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm. to intesect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of the shaded region(Use \pi = 3.14).
11312

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

In Figure, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region.
11313

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}
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

A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.

Answer:

15246 m2
Area of circle =\pi r^{2}
11314
Given that AB = 105m, BC = 21m
Where AB is radius of park and BC is 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

In Figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.
11315

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

A piece of wire 20 cm long is bent into the form of an arc of a circle subtending an angle of 60° at its centre. Find the radius of the circle.

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

Question:1

The area of a circular playground is 22176 m2. Find the cost of fencing this ground at the rate of Rs 50 per metre.

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 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

The diameters of front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make in covering a distance in which the front wheel makes 1400 revolutions.

Answer:

560 revolutions.
Solution:
Circumference of circle =2 \pi r
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
Rear wheel will make 560 revolutions.

Question:3

Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by the three animals.

Answer:

\left ( 24 \sqrt{21}-72 \right )m^{2}
Solution
1143
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}
\thereforeSum 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

Find the area of the segment of a circle of radius 12 cm whose corresponding sector has a central angle of 60° (Use \pi = 3.14).

Answer:

(75.36-36\sqrt{3})cm^{3}
Solution
1144
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° \thereforegiven 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

A circular pond is 17.5 m is of diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs 25 per m2

Answer:

[3064.28 RS ]
1145
Given Diameter of 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

In Figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and distance between AB and DC = 14 cm. If arcs of equal radii 7 cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.
1146

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

Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

Answer:

1.966 cm2
Radius = 3.5
Diameter = 3.5 + 3.5 = 7 cm
1147
Here ABC is equilateral triangle because AB = BC = CA = 7 cm
\thereforeArea 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 equilateral triangle each angle = 60°
All the sectors are 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

Four circular cardboard pieces of radii 7 cm are placed on a paper in such a way that each piece touches other two pieces. Find the area of the portion enclosed Between these pieces.

Answer:

[42 cm2]
Area of shaded portion = Area of square – Area of 4 sectors
1149
Radius = 7 cm
Side of square = 14 cm
AB=BC=CD=DA=14 cm
\therefore ABCD is a square
\thereforeArea 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

On a square cardboard sheet of area 784 cm2, four congruent circular plates of Maximum size are placed such that each circular plate touches the other two Plates and each side of the square sheet is tangent to two circular plates. Find The area of the square sheet not covered by the circular plates.

Answer:

[168 cm2]
11410
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

Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Figure. Find the area of floor that remains uncovered with tiles. (Use \pi = 3.14)
11411

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

All the vertices of a rhombus lie on a circle. Find the area of the rhombus, if area of the circle is 1256 cm2. (Use \pi = 3.14).

Answer: [800 cm2]
Solution
11412
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 required area of rhombus is =800cm2

Question:13

An archery target has three regions formed by three concentric circles as shown in Figure. If the diameters of the concentric circles are in the ratio 1 : 2 : 3, then find the ratio of the areas of three regions
114131

Answer:

[1 : 3 : 5]
Solution
114132
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

The length of the minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6 : 05 am and 6 : 40 a m.

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
In 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

Area of a sector of central angle 200° of a circle is 770 cm2. Find the length of the corresponding arc of this sector.

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

The central angles of two sectors of circles of radii 7 cm and 21 cm are respectively 120° and 40°. Find the areas of the two sectors as well as the lengths of the corresponding arcs. What do you observe?

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.
114171

Answer:

[Area of shaded area=154.88cm2 ]
114172
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

Find the number of revolutions made by a circular wheel of area 1.54 m2 in Rolling a distance of 176 m.

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

Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

Answer [32.16cm2]
Solution
11419
By using Pythagoras in \triangleABC
\\(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

Find the difference of the areas of a sector of angle 120° and its corresponding major sector of a circle of radius 21 cm.

Answer: [462 cm2]
Solution
11420
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

NCERT Exemplar Solutions Class 10 Maths Chapter 11 Important Topics:

The important topics covered in NCERT exemplar class 10 maths solutions chapter 11 are:

  • Area of Sector
  • Area of Circles
  • NCERT exemplar Class 10 Maths solutions chapter 11 discusses the method to find out perimeter and area for any given shape, which can be seen as a combination of circles, sectors, rectangles, triangles.

NCERT Class 10 Exemplar Solutions for Other Subjects:

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NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Maths Solutions 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, A textbook of Mathematics by Monica Kapoor, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths.

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 specially caters to the students learning in a low internet connectivity environment or those who plan to be offline while studying NCERT exemplar Class 10 Maths chapter 11.

Check Chapter-Wise Solutions of Book Questions

Must, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. For a given perimeter, which two-dimensional shape has the maximum area?

The circle will have the maximum area among all the two-dimensional shapes for given circumference.

2. What will be the largest square area drawn in a circle of radius R?

For the largest square area the diameter must be equal to diagonal of the square.

Hence, the area will be 2R2

3. Is the chapter Area related to circles important for Board examinations?

The chapter Area related to circles is vital for Board examinations as it holds around 2-3% weightage of the whole paper.

4. What type of questions are expected from Area related to Circles?

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.

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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

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If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

Read Complete Answer
sai raja 01 September,2024
show the previous year exams ?

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

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SWETCHCHA MISKA 10 July,2024
as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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.

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Manasvini Gupta 23 July,2024
i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

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!

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Ashvadha 03 July,2024
i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

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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sadafreen356 03 July,2024
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Popular CBSE Class 10th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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October 24, 2024 - 09:54 PM IST
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