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NCERT Solutions Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.2
Class 10 NCERT Solutions Maths Chapter 11 Exercise 11.2 will help students extend their question-solving expertise on the topic of combinations of plane figures. The exercise involves various practical and real-life questions related to the various geometrical figures combined with circles, sectors, and their segments. These questions are designed to enhance the analytical and problem-solving skills of students. This ultimately helps them in mathematical studies in higher classes as well as in architecture.
1.0Download NCERT Solutions Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.2 : Free PDF
2.0Introduction to Areas of Combinations of Plane Figures:
Before starting the exercise, we need to understand the fundamental concept of computing the areas of combined plane figures. In earlier classes, we were taught how to compute the areas of basic geometric shapes such as circles, squares, and rectangles. However, in real-world situations, we often encounter shapes composed of two or more basic figures. We need to know how to decompose these compound shapes into their separate parts so that we can determine the total area.
The steps to find the area of a compound shape simply involve identifying the separate figures that comprise the shape, finding their separate areas, and then adding or subtracting these areas depending on the particular combination of the shapes.
3.0Exercise 11.2 Overview: Key Concepts
In Exercise 11.2, you will encounter some important real-life scenarios where you need to find the area and perimeter of various composite figures. Let’s explore some important formulas and key concepts used in solving these questions:
Area of Shaded Region:
In exercise 11.2, most of the questions require finding the area of the shaded region, which can be done by the following approach:
- Identify the Components: The first step in solving these questions is to recognize the basic two-dimensional geometric figures that form the composite shape. For example, in an oval running track, the shape consists of two semicircles and a rectangle in the middle.
- Calculate the Area of Each Component: As we know from the previous classes, all the two-dimensional basic figures have their own formulas for calculating various physical properties like area or perimeter. Use these formulas to separately calculate the required area.
- Combine or subtract: After calculating the areas of each figure, add or subtract them as required by the problem. In other situations where one figure is cut out from another, you could be required to subtract the area of the first figure from that of the other.
Inscribed and Circumscribed Shapes:
The exercise involves various questions with inscribed and circumscribed shapes.In this exercise, inscribed shapes refer to figures placed inside a circle, whereas circumscribed shapes are those that enclose a circle. These questions generally require using the relationship between the sides and the radius of the shapes.
Concentric Circles:
The exercise also includes various questions about the concept of concentric circles. Concentric circles are circles with the same centre but varying radii. When working with concentric circles, the area of the ring-shaped area between two concentric circles is found by subtracting the area of the smaller circle from the area of the larger circle.
The concepts mentioned earlier are used throughout the exercise. To improve your knowledge and performance in this exercise, start practising the problems in the NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.2.
4.0NCERT Class 10 Maths Chapter 11 Areas Related To Circles Exercise 11.2 : Detailed Solutions
1. Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm, and O is the center of the circle.
Solution:
In the figure, ∠RPQ=90∘
(Angle subtended by a diameter on the circumference)
Therefore, △RPQ is right angled at P, RP=7 cm and PQ=24 cm
Then by Pythagoras Theorem, we have
QR2=RP2+PQ2=(7)2+(24)2=625
QR=25 cm
∴ The radius of the circle =225 cm
Now, the area of the shaded region (see figure)
= Area of semicircle - area of triangle
=21πr2−21×RP×PQ
={21×722×(225)2−21×7×24}
=(286875−84)cm2=284523
=161.535
2. Find the area of the shaded region in fig., if radii of the two concentric circles with centre 0 are 7 cm and 14 cm respectively and ∠AOC=40∘.
Solution:
Radius of the outer circle = 14 cm and θ = 40°
Therefore, Area of the sector AOC = (40°/360°) × (22/7) × 14 × 14 cm²
= (1/9) × 22 × 2 × 14 cm²
= 616/9 cm²
Radius of the inner circle = 7 cm and θ = 40°
Therefore, Area of the sector BOD = (40°/360°) × (22/7) × 7 × 7 cm²
= (1/9) × 22 × 7 cm²
= 154/9 cm²
Now, area of the shaded region = Area of sector AOC - Area of sector BOD
= (616/9) - (154/9) cm²
= (1/9) (616 - 154) cm²
= (1/9) × 462 cm²
= 154/3 cm²
= 51.34 cm²
3. Find the area of the shaded region in the figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Solution:
The area of the square ABCD = (14)² cm² = 196 cm² (since the side of the square is 14 cm)
The sum of the areas of the semicircles APD and BPC
= 2 × {area of semicircle APD}
(since the areas of the two semicircles are equal)
= 2 × (1/2 πr²) = π × (AD/2)² = π × (14/2)²
(since AD is the diameter of the semicircle APD)
= (22/7) × 49 cm² = 154 cm²
The area of the shaded region
= The area of the square ABCD - The sum of the areas of the semicircles APD and BPC.
= 196 cm² - 154 cm² = 42 cm²
4. Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as center.
Solution:
Area of the circle with radius 6 cm is:
πr² = (22/7) × 6 × 6 cm² = 792/7 cm²
Area of an equilateral triangle, having side a = 12 cm, is given by:
(√3 / 4) a² = (√3 / 4) × 12 × 12 cm² = 36√3 cm²
Since each angle of an equilateral triangle = 60°
Therefore, ∠AOB = 60°
Therefore, Area of sector COD
= (θ/360°) × πr² = (60°/360°) × (22/7) × 6 × 6 cm²
= (22 × 6) / 7 cm² = 132/7 cm²
Now, area of the shaded region,
= [Area of the circle] + [Area of the equilateral triangle] - [Area of the sector COD]
= (792/7 + 36√3 - 132/7) cm²
= (660/7 + 36√3) cm²
= 156.64 cm²
5. From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining portion of the square.
Solution:
Side of the square = 4 cm
Therefore, Area of the square ABCD = 4 × 4 cm² = 16 cm²
Each corner has a quadrant of a circle of radius 1 cm.
Therefore, Area of all the 4 quadrants = 4 × (1/4)πr² = πr²
= (22/7) × 1 × 1 cm² = 22/7 cm²
Diameter of the middle circle = 2 cm
Therefore, Radius of the middle circle = 1 cm
Therefore, Area of the middle circle = πr² = (22/7) × 1 × 1 cm² = 22/7 cm²
Now, area of the shaded region = [Area of the square ABCD] - [(Area of the 4 quadrants of a circle) + (Area of the middle circle)]
= [16 cm²] - [(22/7 + 22/7) cm²]
= 16 cm² - 2 × (22/7) cm²
= 16 cm² - 44/7 cm²
= (112 - 44)/7 cm²
= 68/7 cm²
= 9.714 cm²
6. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in fig. Find the area of the design.
Solution:
0 is the centre of the circular table cover and the radius is 32 cm. Triangle ABC is equilateral. Join OA, OB, OC.
Now, ∠AOB = ∠BOC = ∠COA = 120°.
In triangle OBC, we have OB = OC.
Draw OM perpendicular to BC.
=> ∠BOM = ∠COM = 60° (because triangle OMB is congruent to triangle OMC by RHS congruence).
Now, BM/OB = sin 60°.
=> BM/32 = √3/2.
=> BM = 16√3 cm.
Then, BC = 2 × BM = 32√3 cm.
Thus, the side of the equilateral triangle ABC = 32√3 cm.
The area of the shaded region (designed) = The area of the circle - area of triangle ABC.
= {π × (32)² - (√3/4) × (32√3)²} cm²
= {(22/7) × 32 × 32 - (√3/4) × 32 × 32 × 3} cm²
= (22528/7 - 768√3) cm²
= 1888.07 cm².
7. In the figure, ABCD is a square of 14 cm. With centres A, B, C, and D, four circles are drawn such that each circle touches two of the remaining three circles. Find the area of the shaded region.
Solution:
Side of the square ABCD = 14 cm
Therefore, Area of the square ABCD = 14 × 14 cm²
= 196 cm²
Since the circles touch each other,
Radius of the circle = 14 / 2 = 7 cm
Now, the area of a sector of radius 7 cm and sector angle θ as 90° is:
(90° / 360°) × (22 / 7) × 7 × 7 cm² = (11 × 7) / 2 cm²
Area of 4 sectors = 4 × [(11 × 7) / 2] = 2 × 11 × 7 cm² = 154 cm²
Area of the shaded region = [Area of the square ABCD] - [Area of the 4 sectors]
= 196 cm² - 154 cm²
= 42 cm²
8. Fig. depicts a racing track whose left and right ends are semicircular.
The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
(i) The distance around the track along its inner edge
(ii) The area of the track.
Solution:
(i)The distance around the track along the inner edge (as seen from figure)
= Circumference of APB + BC + Circumference of CQD + AD
= {π × 30 + 106 + π × 30 + 106} m
= {60π + 212} m
= {60 × 22/7 + 212} = 2804/7 m
(ii) Area of region I = (1/2)π × (40)² - (1/2)π × (30)²
{∵ outer radius = 30 m + 10 m = 40 m}
= (1/2)π × 700 m² = (1/2) × (22/7) × 700 m²
= 1100 m²
Similarly, area of the region II = 1100 m²
Area of the region III (rectangle)
= 106 × 10 = 1060 m²
Similarly, the area of the region IV = 1060 m²
Then, the total area of the track
= 2 × 1100 m² + 2 × 1060 m²
= (2200 + 2120) m² = 4320 m²
9. In fig., AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
Solution:
Let O be the center of the circle, and OA = 7 cm.
Therefore, AB = 2(OA) = 2 × 7 = 14 cm.
Also, OC = OA = 7 cm.
Since AB and CD are perpendicular to each other, OC ⊥ AB.
Thus, the area of triangle ABC is:
Area of ΔABC = (1/2) × AB × OC = (1/2) × 14 cm × 7 cm = 49 cm².
The radius of the small circle is:
Radius of the small circle = (1/2)(OD) = (1/2) × 7 = 7/2 cm.
Therefore, the area of the small circle is:
Area of the small circle = πr² = (22/7) × (7/2) × (7/2) cm² = (11 × 7)/2 = 77/2 cm².
The radius of the big circle is:
Radius of the big circle = 14/2 cm = 7 cm.
The area of the semicircle OACB is:
Area of semicircle OACB = (1/2)πr² = (1/2) × (22/7) × 7 × 7 cm² = 11 × 7 cm² = 77 cm².
Now, the area of the shaded region is:
Area of the shaded region = [Area of the small circle] + [Area of the big semicircle OACB] - [Area of ΔABC]
Area of the shaded region = (77/2) cm² + 77 cm² - 49 cm²
Area of the shaded region = (77 + 154 - 98)/2 cm²
Area of the shaded region = (231 - 98)/2 cm² = 133/2 cm² = 66.5 cm².
10. The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)
Solution:
Area of the triangle ABC (equilateral) = 17320.5 cm²
Let the side of the equilateral triangle ABC be x cm.
Then, (√3 / 4) × x² = 17320.5
=> (1.73205 / 4) × x² = 17320.5 (since √3 = 1.73205)
=> x² = 40000
=> x = 200 cm
Then, the radius of each circle = 100 cm.
Area of the sector APR = (60 / 360) × π × (100)² cm² = (π / 6) × 10000 cm²
Similarly, area of the sector BPQ = area of the sector CQR = (π / 6) × 10000 cm²
Total area of regions I, II and III (i.e., non-shaded region of triangle ABC)
= 3 × (π / 6) × 10000 cm²
= (1 / 2) × 3.14 × 10000 cm²
= 15700 cm²
Then, the required area of the shaded region of triangle ABC
= Area of triangle - Area of unshaded region
= 17320.5 cm² - 15700 cm²
= 1620.5 cm²
11. Designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.
Solution:
Since the circles touch each other, the side of the square ABCD is equal to 3 times the diameter of a circle.
Therefore, the side of the square ABCD = 3 × (2 × radius of a circle) = 3 × (2 × 7 cm) = 42 cm.
The area of the square ABCD = 42 cm × 42 cm = 1764 cm².
Now, the area of one circle = πr².
This is equal to (22/7) × 7 cm × 7 cm = 154 cm².
Since there are 9 circles, the total area of the 9 circles = 154 cm² × 9 = 1386 cm².
Therefore, the area of the remaining portion of the handkerchief = (1764 - 1386) cm² = 378 cm².
12. In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the:
(i) quadrant OACB
(ii) shaded region.
Solution:
(i) Area of the quadrant OACB (radius = 7/2 cm)
= (1/4) × π × r² = (1/4) × (22/7) × (7/2)² cm²
= (1/4) × (22/7) × (49/4) cm² = (11 × 7) / 8 cm²
= 77/8 cm²
(ii) In right angled triangle OBD,
OB = 7/2 cm, OD = 2 cm
The area of triangle OBD = (1/2) × OB × OD
= (1/2) × (7/2) × 2 cm² = 7/2 cm²
Then area of the shaded region = The area of quadrant OACB - The area of triangle OBD
= 77/8 cm² - 7/2 cm² = (77 - 28) / 8 cm² = 49/8 cm²
= 6.125 cm²
13. In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
Solution:
OABC is a square such that its side OA = 20 cm.
Therefore, OA = 20 cm.
Therefore, OB² = OA² + AB²
Therefore, OB² = 20² + 20² = 400 + 400 = 800
OB = √800 = 20√2 cm.
Now, area of the quadrant OPBQ = (1/4)πr²
= (1/4) × (314/100) × 800 cm² = 314 × 2 = 628 cm²
Area of the square OABC = 20 × 20 cm² = 400 cm²
Therefore, Area of the shaded region = Area of the quadrant OPBQ - Area of the square OABC
= 628 cm² - 400 cm² = 228 cm²
14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O. If ∠AOB = 30°, find the area of the shaded region.
Solution:
Radius of the bigger circle R = 21 cm and sector angle θ = 30°.
Therefore, Area of the sector OAB
= (30°/360°) × (22/7) × 21 × 21 cm²
= (11 × 21) / 2 cm²
= 231 / 2 cm²
Again, radius of the smaller circle, r = 7 cm.
Also, the sector angle is 30°.
Therefore, Area of the sector OCD
= (30°/360°) × (22/7) × 7 × 7 cm²
= 77 / 6 cm²
Therefore, Area of the shaded region = Area of the sector OAB - Area of the sector OCD
= (231/2 - 77/6) cm²
= (693 - 77) / 6 cm²
= 616 / 6 cm²
= 308 / 3 cm²
= 102.67 cm²
15. In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Solution:
BC = √(14² + 14²) = 14√2 cm
Area of region II = Area of sector ABC - Area of triangle ABC
= { (1/4) x π x (14)² - (1/2) x 14 x 14 } cm²
= { (1/4) x (22/7) x 196 - 98 } cm²
= 56 cm²
The area of the shaded region III = The area of the semicircle drawn on BC as diameter - The area of region II
= { (1/2) x π x (14√2 / 2)² - 56 } cm²
= { (1/2) x (22/7) x 98 - 56 } cm²
= (154 - 56) cm²
= 98 cm²
16. Calculate the area of the designated region in the figure, common between the two quadrants of circles of radius 8 cm each.
Solution:
Side of the square = 8 cm
Therefore, Area of the square (ABCD) = 8 × 8 cm² = 64 cm²
Now, radius of the quadrant ADQB = 8 cm
Therefore, Area of the quadrant ADQB = (90°/360°) × (22/7) × 8 × 8 cm²
= (1/4) × (22/7) × 64 cm²
= (22 × 16)/7 cm²
Similarly, area of the quadrant BPDC = (22 × 16)/7 cm²
Sum of the two quadrants = 2 × ((22 × 16)/7) cm² = 704/7 cm²
Now, area of design = (Sum of the area of two quadrants) - (Area of the square ABCD)
= 704/7 cm² - 64 cm²
= (704 - 448)/7 cm²
= 256/7 cm²
= 36.57 cm²
5.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.2.
- Helps in solving problems involving composite shapes with circles, sectors, and segments.
- Enhances the ability to apply circle, sector, and segment area formulas in combination.
- Strengthens logical thinking by breaking down complex figures into simpler parts.
- Aids in competitive exams like NTSE and Olympiads where geometry plays a crucial role.
- Builds a strong base for higher-level geometry and trigonometry.