# Area of circle -Class 10 Math

CONTENT LIST

## CIRCLE

### Introduction

A circle is a set of points that are drawn equidistant from a point called “center”.

A circle has a closed two dimensional figure that is measured in terms of radius.

Circle separates the plane into two regions known as the exterior and the interior region.

### Area of circle

The area of a circle is the region bounded by the circle in a two dimensional plane.

Area of the circle: πr^2 (πd^2/4)

### Area of the sector of circle

(θ / 360°) × πr^2, if 'θ' is in degrees

### Perimeter of circle

The circle’s perimeter is simply its boundry length or circumference.

The perimeter of the circle = 2πr (πd)

The standard value of π is considered as 3.1415.

### Tangents of circle

A line that touches the circle at exactly one single point is known as the tangent to a circle.

For every point on a circle has a unique tangent passing through it.

In the below figure, PQ is a tangent that touches the circle at point A.

### Secant of circle

A line that cuts a circle at two distict point on peripheri of circle is called a secant to circle.

A secant cuts the circles at two points that forming a chord of the circle.

### Area of Segment of circle

The Area of a Segment is equal to the the area of a sector minus the triangular piece.

Area of Segment = ( θ × π/360 − sin(θ)/2 ) × r^2 (when θ is in degrees)

### Important theorems related to circle

Theorem 1: “Two equal chords of a circle subtend equal angles at the centre of the circle".

Theorem 2: “The perpendicular to a chord bisects the chord if drawn from the centre of the circle.”

Theorem 3: “Equal chords of a circle are equidistant (equal distance) from the centre of the circle.”

Theorem 4: “Measure of angles subtended to any point on the circumference of the circle from the same arc is equal to half of the angle subtended at the center by the same arc.”

Theorem 5: “The opposite angles in a cyclic quadrilateral are supplementary.”

## Area of Circle -EXERCISE

Q1: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.

Let the radius of required circle is r cm.
Therefore πr^2 =πr1^2 +πr2^2
=>πr^2 =π6^2 +π8^2
=>r^2=100
=>r=10

Q2: 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 traveling at a speed of 66 km per hour?

We know that
=> circumference = 2πr = 2π(40) = 80π
So, distance traveled in 1 revolution = 80π

Converting the speed into cm/min
=> (66 × 100000)/60 cm/min
=> 110000 cm/min

Distance traveled by car in the duration of 10 minutes
=> Speed x Time = 110000 x 10 = 1100000 cm
So, total number of complete revolution = 1100000/80π = 1100000 x 7/80 x 22 = 4375 cm

Q3: If a circular grass lawn of 35 m in radius has a path 7 m wide running around it on the outside, then find the area of the path .

Answer: Given, radius of circular grass lawn (r1) = 35 m, width of path = 7 m
So, the radius of circular grass lawn including path (r2) = 35 + 7 = 42 m

According to the question,
=> area of circular grass lawn = π(r1)² = 22/7 x 35 x 35 = 3850 m²
=> area of circular grass lawn including path = π(r2)² = 22/7 x 42 x 42 = 5544 m²

=> area of path = (area of circular grass lawn including path) - (area of circular grass lawn) = 5544 - 3850 = 1694 m²

So, the area of path is 1694 m².

Q4: A regular hexagon is inscribed in a circle of radius r. Find its perimeter of regular hexagon in terms of r. Given, radius of circle = r
One side of the regular hexagon substend an angle of 60° at center A.

Consider a △ACD, which is formed by droping a perpendicular from center to side of hexagon.
Let, CD = x (CD = Half of CC`)
So, CC` = 2x (Side of hexagon)

In right angle triangle,
=> sin 30 = x/r
=> 1/2 = x/r
=> x = r/2

Therefore, CC` = 2(r/2) = r
So, the perimeter of hexagon = 6r

Q5: Find the area of the circle that can be inscribed in a square of side 6 cm. Given, side of square = 6 cm

According to the question
Diameter of the circle = Side of square = 6 cm
So, area of circle = πd²/4 = π(6)²/4 = 36π/4 = 9π = 22x9/7 = 198/7 = 28.28 cm²

Q6: If the area of a semi-circular field is 15400 m², then find the perimeter of the field.

Answer: Given, area of semi-circular field (πr²/2) = 15400 m²
=> πr²/2 = 15400 => r² = 15400(2)(7)/(22) = 700 x 14 = 9800 => r = √9800 = √4900 x √2 = 70√2
=> perimeter of semi-circular feild = πr + 2r = r(π + 2) = 70√2(3.14 + 2) = 70√2 x 5.14 = 359.80√2 = 360√2 m
So, the perimeter of semi-circular feild is 360√2 m

Q7: The area of a circular ring formed by two concentric circles whose radii are 5.7 cm and 4.3 cm respectively is ..........

According to the question
=> area of inner circle = π(r1)² = π(4.3)² = π(18.49) = 3.14 x 18.49 = 58.05 cm²
=> area of outer circle = π(r2)² = π(5.7)² = π(32.49) = 3.14 x 32.49 = 102.01 cm²

=> area of circular ring = (area of outer circle) - (area of inner circle) = 102.01 - 58.05 = 43.96 cm²

Q8: The sum of the areas of two circle, which touch each other externally, is 153 π . If the sum of their radii is 15, then find the ratio of the larger to the smaller radius.

Answer: Let, radius of the bigger circle is R and that of smaller circle is r. R+r=15 ⇒ R=15-r

According to the question
=> Sum of are of two circles = πR² + πr² = 153 π
=> π(R² + r²) = 153π

Replacing R with r in the above equation, we get
=> r² - 15r + 36 = 0
=> r² - 12r - 3r + 36 = 0
=> r(r - 12) -3(r - 12) = 0
=> (r - 3)(r - 12) = 0

So, the value of r is 3 and 12.
By taking r = 3, we get, R = 15 - 3 = 12

So, the ratio of the larger to the smaller radius = 12 / 3 = 4:1

Q9: In a circle of radius 14 cm, an arc subtends an angle of 45° at the centre, then find the area of the sector.

According to the question
=> Area of sector = θ/360 x πr²
=> Area of sector = 45/360 x 22/7 x 14 x 14
=> Area of sector = 1/8 x 22/7 x 14 x 14
=> Area of sector = 11 x 7 = 77 cm²

So, the area of the sector is 77 cm².

Q10: 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 find relation between R,R1 and R2.

Area of circle with radius R2 = π(R2)²
Area of circle with radius R = π(R)²

According to the question
=> (Area of circle with radius R1) + (Area of circle with radius R2) = Area of circle with radius R
=> π(R1)² + π(R2)² = π(R)²

=> (R1)² + (R2)² = (R)²

So, the relation between R, R1 and R2 is (R1)² + (R2)² = (R)².

Q11: Find the area of a circular path of uniform width d surrounding a circular region of radius r.

Answer: Given, Radius of a circular region of a circular path (inner circle) = r , Uniform width of a circular path = h
Radius of outer circle, R = radius of inner circle + width = r + h ----(i)

According to the question
=> Area of a circular path (A) = (Area of outer circle) - (Area of inner circle)
=> A = πR² - πr²
=> A = π(R² - r²)
=> A = π[(r + h)² - r²]
=> A = π[r² + h² + 2rh - r²]
=> A = π[r² - r² + h² + 2rh]
=> A = π[h² + 2rh]
=> A = πh(2r + h)

So, the area of a circular path is πh(2r + h)

Q12: Find the area of the square that can be inscribed in a circle of radius 8 cm.

Answer: Let ABCD be the square inscribed by the circle.
=> OA = OB = OC = OD

ABC is a right angled triangle, as OA=8,OB=8
=> AB = 8 + 8 = 16

According to Pythagoras theorem
=> Square of hypotenuse = Sum of squares of other two sides
=> AC² = AB² + BC²
=> AC² = AB² + AB² = 2AB² [AB = BC]
=> 16² = 2AB²
=> AB = 8√2

Therefore side of the square is 8√2.
So, area of square = (8√2)² = 64 x 2 = 128 cm²

Q13: Find the radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm.

Answer: Given, the diameter of two circles are 36 cm and 20 cm.

=> Circumference of circle with diameter of 36 cm = 36π
=> Circumference of circle with diameter of 20 cm = 20π

=> Sum of circumference of circles = 36π + 20π = 56π

According to the question
=> Circumference of circles with radius r = Sum of circumference of two circles = 56π
=> 2πr = 56π
=> 2r = 56
=> r = 28 cm

Q14: Find 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.

Answer: Given, the radii of two circle are 24 cm and 7 cm.

=> Area of circle with radius of 24 cm = (24)²π = 576π
=> Area of circle with radius of 7 cm = (7)²π = 49π
=> Sum of area of circles = 576π + 49π = 625π

According to the question
=> Area of circles with radius r = Sum of area of two circles = 625π
=> πr² = 625π
=> r² = 625
=> r = √625 = 25 cm

So, the diameter of circle with radius r is 50 cm.

Q15: The radii of two circles are 19 cm and 9 cm respectively. Find the radius of a circle which has circumference equal to sum of their circumferences.

Answer: Given, the radii of two circles are 19 cm and 9 cm.

=> Circumference of circle with diameter of 19 cm = 38π
=> Circumference of circle with diameter of 9 cm = 18π

=> Sum of circumference of circles = 19π + 9π = 56π

According to the question
=> Circumference of circles with radius r = Sum of circumference of two circles = 56π
=> 2πr = 56π
=> 2r = 56
=> r = 28 cm

Q16: Two coins of diameter 2 cm and 4 cm respectively are kept one over the other. Find the area of the shaded ring shaped region in square cm.

Answer: Given, the diameter of two circles are 2 cm and 4 cm.

=> Area of circle of diameter of 4 cm = 4π
=> Area of circle of diameter of 2 cm = π

=> Area of shaded ring shaped region = (Area of circle of diameter of 4 cm) - (Area of circle of diameter of 2 cm)
=> Area of shaded ring shaped region = 4π - π = 3π = 3 x 3.16 cm² = 9.42 cm²

Q17: The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.

Answer: Given, length of minute hand of a clock = 12 cm

We known that
=> Angle substended by 1 min of a clock = 360/60 = 6°
=> Angle substended by 35 min of a clock = 35 x 6° = 210°

According to the question
=> Angle of sector = (θ/360) x πr²
=> Angle of sector = (210/360) x π x 144
=> Angle of sector = (7/12) x π x 144
=> Angle of sector = 7 x π x 12 = 84π

So, the area of the face of the clock described by the minute hand in 35 minutes is 84π.

Q18: The length of the minute hand of clock is 14 cm. Find the area swept by the minute hand in 15 minutes.

Answer: Given, length of minute hand of a clock = 14 cm

We known that
=> Angle substended by 1 min of a clock = 360/60 = 6°
=> Angle substended by 35 min of a clock = 15 x 6° = 90°

According to the question
=> Angle of sector = (θ/360) x πr²
=> Angle of sector = (90/360) x π x 196
=> Angle of sector = (1/4) x π x 196
=> Angle of sector = π x 49 = 49π

So, the area of the face of the clock described by the minute hand in 35 minutes is 49π.

Q19: The radius of a circle is 17.5 cm. Find the area of the sector of the circle enclosed by two radii and an arc 44 cm in length.

Answer: Given, Radius of a circle (r) = 17.5 cm, length of arc = 44 cm

According to the question
=> Area of Sector = 1/2 × Arc Length × Radius
=> Area of Sector = 1/2 × 44 × 17.5 = 385

So, the area of the sector is 385 cm².

Q20: The diameter of two circle with centre A and B are 16 cm and 30 cm respectively. If area of another circle with centre C is equal to the sum of areas of these two circles, then find the circumference of the circle with centre C.

Answer: Given, Diameter of two circle with centre A and B are 16 cm and 30 cm respectively.

=> area of circle with centre A = π(16)²/4 = 64π
=> area of circle with centre B = π(30)²/4 = 225π

According to the question
=> area of circle with centre C = (area of circle with centre A) + (area of circle with centre B) = 64π + 225π = 289π
=> πr² = 289π
=> r² = 289
=> r = √289 = 17 cm

So, the circumference of the circle with centre C = 2πr = 2 x π x 17 = 34π cm

Q21: A thin wire is in the shape of a circle of radius 77 cm. It is bent into a square. Find the side of the square (Take π =22/7 ).

=> Circumference of the circle = 2πr = 2 x 22/7 x 77 = 2 x 22 x 11 = 484cm

According to the question
=> Perimetre of square = Circumference of the circle = 484 cm
=> 4(side) = 484 cm
=> side = 484/4 = 121 cm

So, the length of each side of square is 121 cm

Q22: What is the perimeter of a sector of a circle whose central angle is 90° and radius is 7 cm?

According to the question
=> Perimetre of the sector = (θ/360) x 2πr + 2r
=> Perimetre of the sector = (90/360) x 2π(7) + 2(7)
=> Perimetre of the sector = (1/4) x 14π + 14
=> Perimetre of the sector = (1/4) x (22 x 14)/7 + 14
=> Perimetre of the sector = 11 + 14 = 25 cm

So, the perimetre of the sector is 25 cm.

Q23: If the perimeter and the area of the circle are numerically equal, then find the radius of the circle.

According to the question
=> (perimeter of the circle) = (area of the circle)
=> 2πr = πr²
=> 2 = r

So, the radius of the circle is 2 units.

Q24: In the given figure, AB is the diameter where AP = 12 cm and PB = 16 cm. Taking the value of π as 3, find the perimeter and area of the shaded region.

Answer: Given AP = 12 cm, BP = 16 cm, π = 3

For finding the area of the shaded region.
In △APB, angle APB is 90° because PQ is a diametre substanding angle 180° at centre.

According to Pythagoras theorem
=> (AB)² = (AP)² + (BP)²
=> (AB)² = (12)² + (16)²
=> (AB)² = 144 + 256 = 400
=> AB = √400 = 20 cm

According to the question
=> Area of triangle APB = 1/2 x AP x BP
=> Area of triangle APB = 1/2 x 12 x 16
=> Area of triangle APB = 12 x 8 = 96 cm²

=> Area of semi-circle = πr²/2
=> Area of semi-circle = π(20/2)²/2 = π(10)²/2
=> Area of semi-circle = 50 x 3 = 150 cm²

=> Area of shaded region = (Area of semi-circle) - (Area of triangle APB)
=> Area of shaded region = 150 - 96 = 54 cm²

For finding the perimatre of the shaded region.
=> Perimetre of shaded region = length of arc APB + AP + BP
=> Perimetre of shaded region = π(10) + 12 + 16
=> Perimetre of shaded region = 10 x 3 + 28
=> Perimetre of shaded region = 30 + 28 = 58 cm

Q25: If the radius of a circle is doubled, what about its area?

Area (A1) = πr²

Then, area (A2) = π(2r)² = 4πr² = 4(A1)

Therefore, area became 4 times when the radius is doubled.

Q26: If the radius of the circle is 6 cm and the length of an arc 12 cm. Find the area of the sector.

Answer: Given, radius of the circle = 6 cm, length of arc = 12 cm

We know that
=> area of sector = (length of arc) x (radius/2)
=> area of sector = 12 x 6/2 = 12 x 3 = 36 cm²

So, the area of sector is 36 cm².

Q27: The perimeter of a sector of a circle of radius 5.2 cm is 16.4 cm. Find the area of the sector.

Answer: Let AOB be the sector with O as center.
Given, Radius (r) = 5.2 cm, Perimetre of sector = 16.4

According to the question
=> OA + AB + OB = 16.4
=> 5.2 + 5.2 + AB = 16.4
=> AB = 6 cm

=> area of sector = (length of arc) x (radius/2)
=> area of sector = 6 x 5.2/2 = 3 x 5.2 = 15.6 cm²

So, the area of sector is 15.6 cm².

Q28: Find the perimeter of the shaded region if ABCD is a square of side 21 cm and APB and CPD are semicircle. Use π = 22/7.

Answer: Given, lenght of side of square = 21 cm, APB and CPD are semicircle

According to the question
=> Perimeter of the shaded region = AD + BC + lengths of the arc of semi-circles APB
=> Perimeter of the shaded region = 21 + 21 + 2(22/7)(21/2) = 42 + 66 = 108 cm

So, the perimeter of the shaded region is 108 cm.

Q29: In Figure, PQ and AB are two arcs of concentric circles of radii 7 cm and 3.5 cm respectively, with centre O. If ∠POQ = 30°, then find the area of shaded region. Answer: Given, Radii of two circles = 7 cm and 3.5 cm, ∠POQ = 30°

=> area of arc with radius 7 cm = 30/360 x 22/7 x 7 x 7
=> area of arc with radius 3.5 cm = 30/360 x 22/7 x 3.5 x 3.5

According to the question
=> area of shaded region = (area of arc with radius 7 cm) - (area of arc with radius 3.5 cm)
=> area of shaded region = [30/360 x 22/7 x 7 x 7] - [30/360 x 22/7 x 3.5 x 3.5]
=> area of shaded region = 30/360 x 22/7 x [(7 x 7) - (3.5 x 3.5)]
=> area of shaded region = 30/360 x 22/7 x [49 - 12.25]
=> area of shaded region = 30/360 x 22/7 x 36.75
=> area of shaded region = 1.83 x 5.35
=> area of shaded region = 9.79 cm²

So, the area of shaded region is 9.79 cm².

Q30: Find the area of shaded region shown 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. Answer: Given, radius of circle (r) = 6 cm, side of equilateral triangle OAB (a) = 12 cm, angle (θ) = 60°

=> Area of minor arc = θ/360 x πr² = 60/360 x π x 6 x 6 = 1 x π x 6 = 6π
=> Area of circle = πr² = π x 6 x 6 = 36π
=> Area of major arc = (Area of circle) - (Area of minor arc) = 36π - 6π = 30π = 30 x 22/7 = 660/7 = 94.28 cm²

=> Area of equilateral triangle = (√3/4)a² = (√3/4) x 12 x 12 = √3 x 12 x 3 = 36√3 = 62.35 cm²
=> Area of shaded triangle = (Area of equilateral triangle) - (Area of minor arc) = 62.35 - 18.84 = 43.51 cm²

According to the question
=> Area of shaded region = (Area of major arc) + (Area of shaded triangle)
=> Area of shaded region = 94.28 + 43.51
=> Area of shaded region = 137.79 cm²

Q31: In the fig., PSR, RTQ and PAQ are three semi-circles of diameters 10 cm, 3 cm and 7 cm region respectively. Find the perimeter of shaded region. Use π = 22/7.

Answer: Given, diametre of PSR = 10 cm, diametre of RTQ = 3 cm, diametre of PAQ = 7 cm

=> Radius of semicircle PSR = 1/2 x 10 = 5 cm
=> Radius of semicircle RTQ = 1/2 x 3 = 1.5 cm
=> Radius of semicircle PAQ = 1/2 x 7 = 3.5 cm

According to the question
=> Perimeter of shaded region = (Circumference of semicircle PSR) + (Circumference of semicircle RTQ) + (Circumference of semicircle PAQ)
=> Perimeter of shaded region = (π × 5) + (π × 1.5) + (π × 3.5) = 10π = 10 x 22/7 = 31.4

So, the perimeter of shaded region is 31.4 cm.

Q32: Find the area of the adjoining diagram.

Answer: Given, Radius of two semi-circles = 7 m, length of rectangle = 16 m, breadth of rectangle = diametre of two semi-circle = 14 cm

=> area of rectangle = (length of rectangle) x (breadth of rectangle) = 16 x 14 = 224 cm²
=> area of two semi-circle = 2(πr²/2) = πr² = 22/7 x 7 x 7 = 22 x 7 = 154 cm²

According to the question
=> Area of adjoining diagram = (area of rectangle) + (area of two semi-circle)
=> Area of adjoining diagram = 224 + 154 = 378 cm²

So, the area of adjoining diagram is 378 cm².

Q33: Find the area of the shaded region. Use π = 22/7.

Answer: Given, sides of square = 14 cm, radius of circle inside the square = 7/2 = 3.5 cm, radius of 2 arc of circle = 3.5 cm, radius of semi-circle in the bottom = 7 cm

=> Area of sqaure = (side) x (side) = 14 x 14 = 196 cm²
=> Area of circle inside the square = πr² = 22/7 x 3.5 x 3.5 = 22 x 3.5 x 0.5 = 38.5 cm²
=> Area of 2 arc of circle = 2(θ/360 x πr²) = 2(90/360 x π x 3.5 x 3.5) = 1/2 x 22/7 x 3.5 x 3.5 = 11 x 3.5 x 0.5 = 19.25 cm²
=> Area of of semi-circle in the bottom = πr²/2 = 22/7 x 7 x 7 x 1/2 = 11 x 7 = 77 cm²

According to the question
=> Area of the shaded region = (Area of sqaure) - (Area of circle inside the square) + (Area of 2 arc of circle) + (Area of of semi-circle in the bottom)
=> Area of the shaded region = 196 - 38.5 + 19.25 + 77 = 253.75 cm²

So, the area of the shaded region is 253.75 cm².

Q34: A road which is 7 m wide surrounds a circular park whose circumference is 88 m. Find the area of the road.

Answer: Given, width of road = 7 m, circumference of circular park (2πr) = 88 m

For radius of the circular park
=> 2πr = 88
=> r = 88/2π = 44 x 7/22 = 2 x 7 = 14 m

=> Radius of circular park inculding road (R) = 14 + 7 = 21 m

According to the question
=> Area of circular park = πr² = 22/7 x 14 x 14 = 616 m²
=> Area of circular park inculding road = πR² = 22/7 x 21 x 21 = 1386 m²

=> Area of road = (Area of circular park inculding road) - (Area of circular park)
=> Area of road = 1386 - 616 = 770 m²

So, the area covered by the road along the park is 770 m².