Circumference Of A Circle And Area

Given that radius of the circle = 3.5 cm

We know that circumference of the circle = 2πr

Also, the area of a circle = πr²

Given that circumference of the circle = 44 m

We know that circumference of the circle = 2πr

⇒ r = 7 m

Also, the area of a circle = πr²

A semicircle is a half of the circle.

Given that radius of the semicircular plot = 21 m

To find the perimeter of the semicircular plot, we need to find the perimeter of the semicircle and add the diameter of the circle to it.

Diameter = AB = 2×radius = 2× 21 = 42 m

We know that circumference of the circle = 2πr

⇒ Perimeter of semicircle = πr

⇒ Perimeter of semicircle

Total perimeter of the semicircular plot = 66 +42 = 108 m

Also, the area of a circle = πr²

⇒ Area of semicircle

⇒ Area of semicircular plot

⇒ Area of semicircular plot

Distance travelled by the wheel of the scooter = 88 m in 100 revolutions

Distance travelled in 1 revolution

Distance travelled in 1 revolution of a wheel = circumference of the wheel

We know that circumference of the circle = 2πr

⇒ r = 14 cm

Given that the area of a circle = 154 cm²

We know that the area of a circle = πr²

⇒ r = 7 cm

Also, the circumference of the circle = 2πr

Given area of the square = 484 m²

We know that area of the square = (Side)²

⇒ (Side)² = 484

⇒ side = 22 m

Also, given that circumference of a circle = perimeter of a square

We know that circumference of the circle = 2πr

And the perimeter of a square = 4× length of the side

⇒ Perimeter of square = 4×22 = 88 m

⇒ Circumference of circle = 88 m

⇒ r = 14 m

Also, the area of a circle = πr²

Given that cost of fencing a circular field at the rate of Rs 24 per meter = Rs 5280

Fencing takes place at the circumference of the circular field.

So, the circumference of the field

We know that circumference of the circle = 2πr

⇒ r = 35 m (radius of the field)

Also, the area of a circle = πr²

⇒ Area of the field

The field is to be ploughed at the rate of 0.50 per m².

Cost of ploughing the field = 3850×0.50 = Rs 1925

Given: Radius of the smaller circle AB = 35 m

Also, CD = 7 m

⇒ Radius of the larger circle AC = AD + CD = 35 + 7 = 42 m

Area of the path = Area of the larger circle – area of the smaller circle

∵The area of a circle = πr²

⇒ Area of the path = πR² - πr²

Using (a+b)(a-b) = a² - b²

⇒ Area of the path = 22× 77 = 1694 m²

Let the radius of the smaller circle be r and radius of the larger circle be R.

The area enclosed between two concentric circles = Area of the larger circle – area of the smaller circle

∵The area of a circle = πr²

⇒ Required area = πR² - πr²

⇒ Required area = π (R² - r²)

Using (a+b)(a-b) = a² - b²

⇒ Required area = π (R+r)(R-r)

The radius of the smaller circle AB = 3 cm

And radius of the larger circle AC = 4 cm

The area enclosed between two concentric circles = Area of the larger circle – area of the smaller circle

∵The area of a circle = πr²

⇒ Required area = πR² - πr²

⇒ Required area = π (R² - r²)

Using (a+b)(a-b) = a² - b²

⇒ Required area = π (R+r)(R-r)

Given: Radius of the circle = 7 cm and angle subtended by the arc = 60°

We know that the length of the arc

⇒ Length of the arc

Given: Radius of the circle = 10.5 cm and angle subtended by the arc = 45°

We know that the area of the minor sector

⇒ Area of the minor sector

⇒ Area of the minor sector = 43.3125 cm²

Given: Radius of the circle = 7 cm and length of an arc = 12 cm

We know that the length of the arc

We know that the area of the minor sector

⇒ Area of the minor sector

⇒ Area of the minor sector = 42cm²

Given: Radius of the circle = 21 cm and angle subtended by the arc = 60°

(i) We know that the length of the arc

⇒ Length of BDC

(ii) We know that the area of the minor sector

⇒ Area of ABDC

⇒ Area of ABDC = 231cm²

(iii) Area of the segment BDC = area of sector ABDC – area of triangle ABC

In ∆ABC,

∠A = 60°, AB = AC = 21 cm {radius of the circle}

⇒ ∠ABC = ∠ACB {angles opposite to equal sides are equal}

By the angle sum property of the triangle,

∠ABC + ∠ACB + ∠A = 180°

⇒ 2∠ABC = 180° - 60°

⇒ ∠ABC = 60°

Hence, ∆ABC is an equilateral triangle.

Area of a equilateral trianglewhere a is the side of it.

Area of ∆ABC

⇒ Area of ∆ABC = 190.95cm²

∴ Area of the segment BDC = area of sector ABDC – area of triangle ABC

⇒ Area of the segment BDC = 231 – 190.95

⇒ Area of the segment BDC = 40.05cm²

Let us consider clock as a circle with radius 10.5 cm.

Also, the 60 minutes of the clock constitute the 360° angle of the circle.

So, the angle formed by the minute hand in 10 minutes

We know that the area of the minor sector

⇒ Area of the minor sector

⇒ Area of the minor sector = 57.75cm²

Given: Radius of the circle = 3.5 cm and angle subtended by the arc = 90°

Area of the segment BDC = area of sector ABDC – area of triangle ABC

We know that the area of the minor sector

⇒ Area of ABDC

⇒ Area of ABDC = 9.625cm²

In ∆ABC,

∠A = 90°, AB = AC = 3.5 cm {radius of the circle}

∵ ∆ABC is a right angled triangle.

Area of ∆ABC

⇒ Area of ∆ABC

⇒ Area of ∆ABC = 6.125cm²

∴ Area of the segment BDC = area of sector ABDC – area of triangle ABC

⇒ Area of the segment BDC = 9.625 – 6.125

⇒ Area of the segment BDC = 3.5cm²

Given that circumference of the circle = 22 cm

We know that circumference of the circle = 2πr

⇒ r = 3.5cm

Quadrant of a circle is the one fourth part of a circle.

Also, the area of a circle = πr²

⇒ Area

Area of a quadrant

Let us consider clock as a circle with radius 5 cm.

Also, the hour hand of the clock makes 360° when takes 12 hours.

So let us convert 7 minutes to degrees.

There are 12×60 minutes in 12 hours.

So, the angle formed by the hour in 7 minutes

We know that the area of the minor sector

⇒ Area of the minor sector

⇒ Area of the minor sector = 0.7638cm²

Given: AB = 10 cm, BC = 7 cm, Radius of the quadrants = 3.5 cm

In the given figure, the four quadrants together form a circle of radius 3.5 cm

∴ Area of the shaded region = Area of the rectangle – Area of the circle formed by the 4 quadrants

Area of rectangle = Length × breadth = 10× 7 = 70 cm²

Area of a circle = πr²

Area of the shaded region = 70 – 38.5 = 31.5 cm²

Let ABCD be the square with side 14 cm.

⇒ BC = 14 cm

Let the circle centered at E be the incircle of the ABCD.

BC = diameter of the circle = 14 cm

⇒ Radius of the circle = 7 cm

Given that radius of the circle = 7 cm

We know that circumference of the circle = 2πr

⇒ Circumference

Given that difference between the circumference of the circle and radius of it = 74 cm

We know that circumference of the circle = 2πr

⇒ 2πr – r = 74

Area of a circle = πr²

Given: Radius of the circle = 3 cm and angle subtended by the arc = 90°

Area of the segment ARB = area of sector OARB – area of triangle OAB

We know that the area of the minor sector

⇒ Area of ABDC

⇒ Area of ABDC = 7.071428cm²

In ∆ABC,

∠A = 90°, AB = AC = 3 cm {radius of the circle}

∵ ∆ABC is a right angled triangle.

Area of ∆ABC

⇒ Area of ∆ABC

⇒ Area of ∆ABC = 4.5cm²

∴ Area of the segment BDC = area of sector ABDC – area of triangle ABC

⇒ Area of the segment BDC = 7.071428 – 4.5

⇒ Area of the segment BDC = 2.57cm²

Given the circumference of a circle is equal to the perimeter of a square. We know that circumference of the circle = 2πr and the perimeter of a square = 4× length of the side.

⇒ 2πr = 4×side

… (i)

We know that area of the square = (Side)²

Also, the area of a circle = πr²

⇒ Ratio of their areas

⇒ Ratio of their areas

⇒ Ratio of their areas

⇒ Ratio of their areas

Given: Radius of the smaller circle AB = 3.5 m

Also, CD = 1.4 m

⇒ Radius of the larger circle AC = AD + CD = 3.5 + 1.4 = 4.9 m

Area of the path = Area of the larger circle – area of the smaller circle

∵The area of a circle = πr²

⇒ Area of the path = πR² - πr²

⇒ Area of the path

⇒ Area of the path

Using (a+b)(a-b) = a² - b²

⇒ Area of the path

⇒ Area of the path

⇒ Area of the path = 36.96 m²

Given: Radius of the circle = 8 cm

Diameter of the circle = 16 cm

For the largest square inscribed in it i.e. BCDE,

Diagonal of the square = diameter of the circle

⇒ BD = 16 cm

Area of a square

⇒ Area of BCDE

Given: ABMC is a quadrant of radius 14 cm.

In ∆ABC,

∠A = 90°, AB = AC = 14 cm

By Pythagoras theorem,

BC² = AB² + AC²

⇒ BC² = 2× 14×14

⇒ BC = 14√2 cm

Now,
Area of ∆ABC

⇒ Area of ∆ABC

⇒ Area of ∆ABC = 98cm²

We know that the area of the minor sector

⇒ Area of ABMC

⇒ Area of ABMC = 154cm²

So, area of ABMC = area of ∆ABC + area of BCM

⇒ Area of BCM = area of ABMC - area of ∆ABC

⇒ Area of BCM = 154 – 98 = 56cm²

Also, we have a semicircle made at BC in the figure.

Diameter of the semicircle = BC = 14√2 cm

⇒ Radius of the semicircle = 7√2 cm

Area of semicircle

⇒ Area of semicircular plot

⇒ Area of semicircular plot = 154 cm²

Area of the shaded region = Area of the semicircle - Area of BCM

⇒ Area of the shaded region = 154 – 56 = 98 cm²

Given:

In ∆ABC,

AC = 6 cm, BC = 8 cm

Also, ∠C = 90° because angle subtended by the diameter is 90° and AB is the diameter.

By Pythagoras theorem,

BA² = CB² + AC²

⇒ BC² = 36 + 64

⇒ BC = 10 cm

⇒ Area of ∆ABC = 24 cm²

Radius of the circle = 5 cm

Area of a circle = πr²

Area of the shaded region = Area of circle –Area of the triangle

⇒ Area of the shaded region = 78.57 -24 = 54.57 cm²

Given side of the square ABCD = 10 cm

Area of the square = side × side

⇒ Area of ABCD = 10 × 10 = 100 cm²

Also given that semicircles are drawn with each side of the square as diameter

Diameter = 10 cm

Radius = 5 cm

Let us mark the four unshaded areas as I, II, III and IV.

Area of I + Area III

= Area of square - Area of two semi-circles

= 21.5 cm²

Similarly, Area of II + Area of IV = 21.5 cm²

So, area of shaded region = ar of ABCD - ar (I + II + III + IV)

= (100 – 2 × 21.5) cm²

= 100 - 43

= 57 cm²

In the given figure the radius of the semicircle = 7 cm

The diameter of the circle inscribed in the semicircle = 7 cm

⇒ Radius of the circle = 3.5 cm

Area of a circle = πr²

Given: The sum of circumferences of two circles of radii R₁ and R₂ is equal to the circumference of a circle of radius R.

We know that circumference of the circle = 2πr

⇒ 2πR₁+2πR₂ = 2πR

⇒ 2π (R₁+R₂) = R

⇒ R₁+R₂ = R

Let ABCD be the square with side 14 cm.

⇒ BC = 14 cm

Let the circle centered at E be the incircle of the ABCD.

BC = diameter of the circle = 14 cm

⇒ Radius of the circle = 7 cm

Given that radius of the circle = 7 cm

We know that circumference of the circle = 2πr