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In our NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles, we have provided all kinds of study material students need while preparing for their CBSE board examination as our solutions are explained in very simple language. One must opt for our NCERT Solutions for clearing their basic concepts related to the chapter and also for cracking tough competitive exams like JEE Advance, JEE Mains, Olympiad, NTSE, and other similar examinations. Our study materials are prepared by mentors who have great expertise in the subject matter and are designed in such a way that it is very intuitive for the learners. All the concepts are explained in step by step method. We have used graphs, diagrams, construction diagrams, and equations with a very innovative method. This study material is a very effective way to understand different concepts, learn new concepts and revise.  

In NCERT Solutions Class 10 Maths can learn about different concepts of the circle in this chapter. The circle is one of the most important chapters to study to understand geometry. We get to learn about different topics related to circle such as:

  • Area of a circle – The area of the circle of radius ‘r’ is \(\pi\)r2.
  • Perimeter of a circle – The perimeter of the circle with radius ‘r’ is 2\(\pi\)r.
  • Length of an arc - In a circle of radius ‘r’ the length of an arc is found by s = rθ, wheres = arc length, and θ = central angle in radians.
  • Area of a sector – In a circle, we can find the area of a sector using the formula A = r2∙θ/2.
  • Area of a segment – we can find the area of a segment by subtracting the area of a triangle in the area of the sector. Area of segment = area of the sector – area of a triangle.
  • Finding the area of different figures made by combining different geometrical figures involving circles.

NCERT Solutions Class 10 Maths provided by Sarthaks is one the great way to learn about all the topics related to circle and related concepts. All the topics are explained in the step-wise method. As suggested by our mentors it is the best way to learn, solve, revise, complete homework, making assignments.

All the solutions at your fingertip get started now.

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NCERT Solutions Class 10 Maths Chapter 12 Areas Related to Circles

1. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Answer:

Radius (r1) of 1st circle = 19 cm 

Radius (r2) or 2nd circle = 9 cm 

Let the radius of 3rd circle be r. 

Circumference of 1st circle = 2πr1 = 2π (19) = 38π 

Circumference of 2nd circle = 2πr2 = 2π (9) = 18π 

Circumference of 3rd circle = 2πr

Given that, 

Circumference of 3rd circle = Circumference of 1st circle + Circumference of 2nd circle 

2πr = 38π + 18π = 56π

\(r = \frac{56\pi}{2\pi}= 28\)

Therefore, the radius of the circle which has circumference equal to the sum of the circumference of the given two circles is 28 cm.

2. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Answer:

Radius (r1) of 1st circle = 8 cm 

Radius (r2) of 2nd circle = 6 cm 

Let the radius of 3rd circle be r.

Area of 1st circle = \(\pi{ r_1}^2 = \pi(8)^2= 64x\)

Area of 2nd circle = \(\pi{ r_2}^2 = \pi(6)^2= 36x\)

Given that, 

Area of 3rd circle = Area of 1st circle + Area of 2nd circle

However, the radius cannot be negative. Therefore, the radius of the circle having area equal to the sum of the areas of the two circles is 10 cm.

3. Given figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions[Use \(\pi\) = 22/7].

Answer:

The radius of 1st circle, r1 = 21/2 cm (as diameter D is given as 21 cm)

So, area of gold region = π r1= π(10.5)= 346.5 cm2

Now, it is given that each of the other bands is 10.5 cm wide,

So, the radius of 2nd circle, r2 = 10.5cm + 10.5cm = 21 cm

Thus,

∴ Area of red region = Area of 2nd circle − Area of gold region = (πr22−346.5) cm2

= (π(21)2 − 346.5) cm2

= 1386 − 346.5

= 1039.5 cm2

Similarly,

The radius of 3rd circle, r3 = 21 cm+10.5 cm = 31.5 cm

The radius of 4th circle, r4 = 31.5 cm+10.5 cm = 42 cm

The Radius of 5th circle, r5 = 42 cm+10.5 cm = 52.5 cm

For the area of nth region,

A = Area of circle n – Area of circle (n-1)

∴ Area of blue region (n=3) = Area of third circle – Area of second circle

= π(31.5)2 – 1386 cm2

= 3118.5 – 1386 cm2

= 1732.5 cm2

∴ Area of black region (n=4) = Area of fourth circle – Area of third circle

= π(42)2 – 1386 cm2

= 5544 – 3118.5 cm2

= 2425.5 cm2

∴ Area of white region (n=5) = Area of fifth circle – Area of fourth circle

= π(52.5)2 – 5544 cm2

= 8662.5 – 5544 cm2

= 3118.5 cm2

4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Answer:

The radius of car’s wheel = 80/2 = 40 cm (as D = 80 cm)

So, the circumference of wheels = 2πr = 80 π cm

Now, in one revolution, the distance covered = circumference of the wheel = 80 π cm

It is given that the distance covered by the car in 1 hr = 66km

Converting km into cm we get,

Distance covered by the car in 1hr = (66×105) cm

In 10 minutes, the distance covered will be = (66×105×10)/60 = 1100000 cm/s

∴ Distance covered by car = 11×105 cm

Now, the no. of revolutions of the wheels = (Distance covered by the car/Circumference of the wheels)

=( 11×105)/80 π = 4375.

5.Tick the correct Solution: in the following and justify your choice : If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

(A) 2 units

(B) π units

(C) 4 units

(D) 7 units

Answer:

Since the perimeter of the circle = area of the circle,

2πr = πr2

Or, r = 2

So, option (A) is correct i.e. the radius of the circle is 2 units.

6. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60° [Use \(\pi\) = 22/7].

Answer:

Let OACB be a sector of the circle making 60° angle at centre O of the circle. 

Area of sector of angle θ = \(\frac{θ}{360^\circ}\times\pi r^2\)

Area of sector OACB = \(\frac{60^\circ}{360^\circ}\times\frac{22}{7}\times(6)^2\)

\(= \frac{1}{6}\times\frac{22}{7}\times6\times6=\frac{132}{7}cm^2\)

Therefore, the area of the sector of the circle making 60° at the centre of the circle is 132/7 cm2.

7. Find the area of a quadrant of a circle whose circumference is 22 cm [ Use \(\pi\) = 22/7].

Answer:

Let the radius of the circle be r.

Circumference = 22 cm

2\(\pi\) = 22

\(r = \frac{22}{2\pi}= \frac{11}{\pi}\)

Quadrant of circle will subtend 90° angle at the centre of the circle.

Area of such quadrant of the circle 

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8. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes  [ Use \(\pi\) = 22/7].

Answer:

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°. 

In 5 minutes, minute hand will rotate = \(\frac{360^\circ}{60}\times5= 306^\circ\)

Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius. 

Area of sector of angle θ = \(\frac{θ}{360^\circ}\times\pi r^2\)

Area of sector of 30° = \(\frac{30^\circ}{360^\circ}\times\frac{22}{7}\times14\times14\)

\(=\frac{22}{12}\times2\times14\)

\(=\frac{11\times14}{3}\)

\(= \frac{154}{3}\) cm2

Therefore, the area swept by the minute hand in 5 minutes is 154/3 cm2.

9. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: 

(i) Minor segment 

(ii) Major sector 

[Use π = 3.14]

Answer:

Here AB be the chord which is subtending an angle 90° at the center O.

It is given that the radius (r) of the circle = 10 cm

(i) Area of minor sector = (90/360°) × πr2

= (1/4)×(22/7)×102

Or, Area of minor sector = 78.5 cm2

Also, area of ΔAOB = 1/2 × OB × OA

Here, OB and OA are the radii of the circle i.e. = 10 cm

So, area of ΔAOB = 1/2 × 10 × 10

= 50 cm2

Now, area of minor segment = area of minor sector – area of ΔAOB

= 78.5 – 50

= 28.5 cm2

(ii) Area of major sector = Area of circle – Area of minor sector

= (3.14×102) - 78.5

= 235.5 cm2

10. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: 

(i) The length of the arc 

(ii) Area of the sector formed by the arc 

(iii)Area of the segment forced by the corresponding chord

 [ Use \(\pi\) = 22/7].

Answer:

Radius (r) of circle = 21 cm 

Angle subtended by the given arc = 60° 

(i) Length of an arc of a sector of angle θ = \(\frac{θ}{360^\circ}\times2\pi r\)

Length of arc ACB = \(\frac{60^\circ}{360^\circ}\times2\times\frac{22}{7}\times21\)

\(=\frac{1}{6}\times2\times22\times3\)

= 22 cm

(ii) Area of sector OACB = \(\frac{60^\circ}{360^\circ}\times\pi r^2\)

\(=\frac{1}{6}\times\frac{22}{7}\times21\times21\)

= 231 cm2

In ∆OAB,

∠OAB = ∠OBA  (As OA = OB) 

∠OAB + ∠AOB + ∠OBA = 180°

2∠OAB + 60° = 180° 

∠OAB = 60° 

Therefore, ∆OAB is an equilateral triangle.

Area of ∆OAB 

(iii) Area of segment ACB = Area of sector OACB − Area of ∆OAB

11. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. [Use \(\pi\) = 3.14 √3 = 1.73]

Answer:

Radius (r) of circle = 15 cm 

Area of sector OPRQ = \(\frac{60^\circ}{360^\circ}\times\pi r^2\) 

\(\frac{1}{6}\times3.14\times(15)^2\)

= 117.75 cm2

In ∆OPQ, 

∠OPQ = ∠OQP (As OP = OQ) 

∠OPQ + ∠OQP + ∠POQ = 180° 

2 ∠OPQ = 120° 

∠OPQ = 60° 

∆OPQ is an equilateral triangle.

Area of ∆OPQ =

Area of segment PRQ = Area of sector OPRQ − Area of ∆OPQ 

= 117.75 − 97.3125 

= 20.4375 cm2 

Area of major segment PSQ = Area of circle − Area of segment PRQ

12. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.  [Use \(\pi\) = 3.14 √3 = 1.73]

Answer:

Let us draw s perpendicular OV on chord ST. It will bisect the chord ST.

SV = VT

In ∆OVS,

 

Area of ∆OST = \(\frac{1}{2}\) x ST x OV

\(\frac{1}{2}\) x 12\(\sqrt{3}\) x 6

= 36\(\sqrt{3}\) 

= 36 x 1.73

= 62.28 cm2

Area of sector OSUT = \(\frac{120^\circ}{360^\circ}\times\pi (12)^2\) 

\(\frac{1}{2}\) x 3.14 x 144 = 150.72 cm2

Area of segment SUT = Area of sector OSUT − Area of ∆OST 

= 150.72 − 62.28 

= 88.44 cm2

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13. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see the given figure). Find

(i) the area of that part of the field in which the horse can graze.

(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Answer:

As the horse is tied at one end of a square field, it will graze only a quarter (i.e. sector with θ = 90°) of the field with radius 5 m.

Here, the length of rope will be the radius of the circle i.e. r = 5 m

It is also known that the side of square field = 15 m

(i) Area of circle = πr

= \(\frac{22}{7}\) × 52 

= 78.5 m2

Now, the area of the part of the field where the horse can graze = 1/4 (the area of the circle) 

= \(\frac{78.5}{4}\) 

= 19.625 m2

(ii) If the rope is increased to 10 m,

Area of circle will be = πr2 

\(\frac{22}{7}\)×102 

= 314 m2

Now, the area of the part of the field where the horse can graze = 1/4 (the area of the circle)

= \(\frac{314}{4}\) = 78.5 m2

∴ Increase in the grazing area 

= 78.5 m2 – 19.625 m2 

= 58.875 m2

14. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure. Find. 

(i) The total length of the silver wire required. 

(ii)The area of each sector of the brooch

(Use π = 22/7)

Answer:

Total length of wire required will be the length of 5 diameters and the circumference of the brooch.

Radius of circle = 35/2 mm

Circumference of brooch = 2πr

\(2\times\frac{22}{7}\times(\frac{35}{2})\)

= 110 mm

(i) Length of wire required = 110 + 5 x 35

= 110 + 175 = 285 mm

It can be observed from the figure that each of 10 sectors of the circle is subtending 36° at the centre of the circle.

(ii) Therefore, area of each sector = \(\frac{36^\circ}{360^\circ}\times\pi r^2\)

15. An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella. (Use π = 22/7)

Answer:

There are 8 ribs in an umbrella. The area between two consecutive ribs is subtending \(\frac{360^\circ}{8}=45^\circ\) at the centre of the assumed flat circle.

Area between two consecutive ribs of circle = \(\frac{45^\circ}{360^\circ}\times\pi r^2\) 

16. A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades. (Use π = 22/7)

Answer:

It can be observed from the figure that each blade of wiper will sweep an area of a sector of 115° in a circle of 25 cm radius.

Area of such sector 

Area swept by 2 blades = \(2\times\frac{158125}{252}\)

\(\frac{158125}{1260}\) cm2

17. To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships warned. [Use π = 3.14]

Answer:

It can be observed from the figure that the lighthouse spreads light across a sector of 80° in a circle of 16.5 km radius. 

Area of sector OACB = \(\frac{80^\circ}{360^\circ}\times\pi r^2\) 

\(\frac{2}{9}\) x 3.14 x 16.5 x 16.5

= 189.97 km2

18. A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs.0.35 per cm2. [Use √3 = 1.7]

Answer:

It can be observed that these designs are segments of the circle. Consider segment APB. Chord AB is aside of the hexagon. Each chord will substitute \(\frac{360^\circ}{6}=60^\circ\) at the centre of the circle.

In \(\triangle OAB\),

∠OAB = ∠OBA

∠AOB = 60°

∠OAB + ∠OBA + ∠AOB = 180°

2∠OAB = 180° − 60° = 120° 

∠OAB = 60° 

Therefore, ∆OAB is an equilateral triangle.

Area of ∆OAB 

Area of sector OAPB = \(\frac{80^\circ}{360^\circ}\times\pi r^2\) 

Area of segment APB = Area of sector OAPB − Area of ∆OAB

Therefore, area of designs 

Cost of making 1 cm2 designs = Rs 0.35 

Cost of making 464.76 cm2 designs = 464.8 × 0.35 = Rs 162.68 

Therefore, the cost of making such designs is Rs 162.68.

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19. Tick the correct answer in the following: 

Area of a sector of angle p (in degrees) of a circle with radius R is

(A) \(\frac{p}{180^\circ}\times 2\pi R\)

(B) \(\frac{p}{180^\circ}\times \pi R^2\)

(C) \(\frac{p}{360^\circ}\times 2\pi R\)

(D)\(\frac{p}{720^\circ}\times 2\pi R^2\) 

Answer:

We know that area of sector of angle θ = \(\frac{\theta}{360^\circ}\times\pi r^2\) 

Area of sector of angle P = \(\frac{p}{360^\circ}\times( \pi R^2)\) 

\(=(\frac{p}{720^\circ})(2\pi R^2)\) 

Hence, (D) is the correct answer.

20. Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. [Use \(\pi\) = 22/7]

Answer:

It can be observed that RQ is the diameter of the circle. Therefore, ∠RPQ will be 90º. 

By applying Pythagoras theorem in ∆PQR, 

RP2 + PQ2 = RQ2 

(7)2 + (24)2 = RQ2

RQ = \(\sqrt{625}=25\) 

Radius of circle, \(OR = \frac{RQ}{2}= \frac{25}{2}\)

Since RQ is the diameter of the circle, it divides the circle in two equal parts.

Area of semi- circle RPQOR 

Area of ∆PQR = 1/2 x PQ x PR

= 1/2 x 24 x 7

= 84 cm2

Area of shaded region = Area of semi-circle RPQOR − Area of ∆PQR

21. Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.  [Use \(\pi\) = 22/7]

Answer:

Radius of inner circle = 7 cm

Radius of outer circle = 14 cm

Area of shaded region = Area of sector OAFC - Area of sector OBED

22. Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles. [Use \(\pi\) = 22/7]

Answer:

It can be observed from the figure that the radius of each semi-circle is 7 cm.

Area of each semi-circle = \(\frac{1}{2}\pi r^2\)

\(\frac{1}{2}\times\frac{22}{7}\times(7)^2\) 

= 77 cm2

Area of square ABCD = (Side)2 = (14)2 = 196 cm2 

Area of the shaded region = Area of square ABCD − Area of semi-circle APD − Area of semi-circle BPC 

= 196 − 77 − 77 

= 196 − 154 

= 42 cm2'

23. Find the area of the shaded region in the given 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 centre.  [Use \(\pi\) = 22/7]

Answer:

We know that each interior angle of an equilateral triangle is of measure 60°.

Area of sector OCDE = \(\frac{60^\circ}{360^\circ} \pi r^2\)

Area of \(\triangle OAB\) = \(\frac{\sqrt{3}}{4}(12)^2\)

Area of shaded region = Area of ∆OAB + Area of circle − Area of sector OCDE

24. 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 the given figure. Find the area of the remaining portion of the square.  [Use \(\pi\) = 22/7]

Answer:

Each quadrant is a sector of 90° in a circle of 1 cm radius.

Area of each quadrant =  \(\frac{90^\circ}{360^\circ} \pi r^2\)

\(=\frac{1}{4}\times\frac{22}{7}\times(1)^2\)

\(\frac{22}{7}\) cm2

Area of square = (Side)2 = (4)2 = 16 cm2

Area of circle = \(\pi\)r= \(\pi\) (1)2

\(= \frac{22}{7}\) cm2

Area of shaded region = Area of square - Area of circle − 4 x  Area of quadrant

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25. In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design (Shaded region). [Use \(\pi\) = 22/7]

Answer:

Radius (r) of circle = 32 cm 

AD is the median of ∆ABC

AO = 2/3 , Ad = 32

AD = 48 cm 

In ∆ABD,

AB2 = AD2 + BD2

Area of equilateral triangle,

Area of circle = πr2

Area of design = Area of circle − Area of ∆ABC

26. In the given figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touches externally two of the remaining three circles. Find the area of the shaded region. [Use \(\pi\) = 22/7]

Answer:

Area of each of the 4 sector is equal to each other and is a sector of 90° in a circle of 7 cm radius.

Area of each sector = \(\frac{90^\circ}{360^\circ}\times \pi({7})^2\)

Area of square ABCD = (side)2 = (14)2 = 196 cm2

Area of shaded portion = Area of square ABCD - 4 x Area of each sector

Therefore, the area of shaded portion is 42 cm2.

27. The given figure 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

 [Use \(\pi\) = 22/7]

Answer:

Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA

Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semi-circle HKI − Area of semi-circle BEC) + (Area of semi-circle GLJ − Area of semi-circle AFD)

Therefore, the area of the track is 4320 m2.

28. In the given figure, 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 shaded region.  [Use \(\pi\) = 22/7]

Answer:

Radius (r1) of larger circle = 7 cm 

Radius (r2) of smaller circle = 7/2 cm

Area of smaller circle = \(\pi {r_1}^2\)

\(=\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\)

\(=\frac{77}{2}\) cm2

Area of semi-circle AECFB of larger circle = \(\frac{1}{2}\pi {r_1}^2\)

\(=\frac{1}{2}\times\frac{22}{7}\times(7)^2\)

= 77 cm2

Area of ∆ABC = \(\frac{1}{2}\times AB\times OC\)

\(\frac{1}{2}\times 14\times 7\)

= 49 cm2

Area of the shaded region = Area of smaller circle + Area of semi-circle AECFB − Area of ∆ABC

\(\frac{77}{2}\) + 77 - 49

= 28 + \(\frac{77}{2}\) 

= 28 + 38.5

= 66.5 cm2

29. The area of an equilateral triangle ABC is 17320.5 cm2 . With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (See the given figure). Find the area of shaded region. [ \(\pi\) = 3.14 √3 = 1.73205]

Answer:

Let the side of the equilateral triangle be a. 

Area of equilateral triangle = 17320.5 cm2

Each sector is of measure 60°.

Area of sector ADEF = \(\frac{60^\circ}{360^\circ}\times \pi\times r^2\)

Area of shaded region = Area of equilateral triangle − 3 × Area of each sector

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 30. On a square handkerchief, nine circular designs each of radius 7 cm are made (see the given figure). Find the area of the remaining portion of the handkerchief. [Use \(\pi\) = 22/7]

Answer:

From the figure, it can be observed that the side of the square is 42 cm. 

Area of square = (Side)2 = (42)2 = 1764 cm2

Area of each circle = πr2 

\(\frac{22}{7}\times (7)^2\)

= 154 cm2

Area of 9 circles = 9 × 154 = 1386 cm2 

Area of the remaining portion of the handkerchief = 1764 − 1386 = 378 cm2

31. In the given figure, OACB is a quadrant of circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the 

(i) Quadrant OACB 

(ii)Shaded region

 [Use \(\pi\) = 22/7]

Answer:

(i) Since OACB is a quadrant, it will subtend 90° angle at O. 

Area of quadrant OACB =  \(\frac{90^\circ}{360^\circ}\times \pi r^2\)

(ii) Area of ∆OBD = \(\frac{1}{2}\times OB\times OD\)

Area of the shaded region = Area of quadrant OACB − Area of ∆OBD

32. In the given figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. [ \(\pi\) = 3.14]

Answer:

In ∆OAB,

OB2 = OA2 + AB2

= (20)2 + (20)2

OB = 20\(\sqrt{2}\) 

Radius (r) of circle = 20\(\sqrt{2}\) cm

Area of quadrant OPBQ

 

Area of OABC = (Side)2 = (20)2 = 400 cm2

Area of the shaded region = Area of quadrant OPBQ − Area of OABC

= (628 - 400) cm2

= 228 cm2

33. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the given figure). If ∠AOB = 30°, find the area of the shaded region.  [Use \(\pi\) = 22/7]

Answer:

Area of the shaded region = Area of sector OAEB − Area of sector OCFD

34. In the given 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.  [Use \(\pi\) = 22/7]

Answer:

As ABC is a quadrant of the circle, ∠BAC will be of measure 90º. 

In ∆ABC, 

BC2 = AC2 + AB2 

= (14)2 + (14)2

BC = 14\(\sqrt{2}\)

Radius (r1) of semi-circle drawn on

Area of

Area of sector =

Area of semi-circle drawn on BC 

Area of shaded region = Area of semi-circle - (Area of sector ABDC - Area of ∆ABC)

= 154 − (154 − 98) 

= 98 cm2

35. Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each.  [Use \(\pi\) = 22/7]

Answer:

The designed area is the common region between two sectors BAEC and DAFC.

Area of sector 

Area of ∆BAC = \(\frac{1}{2}\times BA \times BC\)

\(\frac{1}{2}\times 8 \times 8\)

= 32 cm2

Area of the designed portion = 2 × (Area of segment AEC) 

= 2 × (Area of sector BAEC − Area of ∆BAC)

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