Chapter 3:Pair of Linear Equations in Two Variables

Chapter 5: Arithmetic Progression

Chapter 8: Introduction to Trigonometry

Chapter 9:Some Applications of Trigonometry

Chapter 12:Area Related to Circles

Area Related to Circles

r_{1}= 19 cm

Circumference of the 1c^{st} circle = 2π×19 = 38π cm

r_{2} = 9 cm

Circumference of the 2^{nd} circle = 2π×9 = 18π cm

Now, let the radius of new circle circle = R

According to question

circumference of new circle =Circumference of the 1^{st} circle + Circumference of the
2^{nd} circle

2πR= 38π+18π = 56π cm

2πR =56π

R= 56π/2π

R = 28 cm.

r_{1} = 8 cm

Area of 1^{st} circle = π(8)^{2} = 64π

Radius of 2^{nd} circle r_{2} = 6 cm

Area of 2^{nd} circle = π(6)^{2} = 36π

So,

The sum of 1^{st} and 2^{nd} circle will be = 64π+36π = 100π

let the radius of new circle = R

Area of the new circle = πR^{2}

The area of the new circle = Area of 1^{st} circle + Area of 2^{nd} circle

Or, πR^{2} = 100πcm^{2}

R^{2} = 100cm^{2}

So, R = 10cm

.

Diameter of circle =21cm

The radius of gold region r_{1} = 21/2 cm

Area of gold region = π r_{1}^{2 }= π(10.5)^{2 }= 346.5 cm^{2}

So, the radius of red region r_{2} = 10.5cm+10.5cm = 21 cm

Thus,

∴ Area of red region = Area of 2^{nd} circle − Area of gold region =
(πr_{2}^{2}−346.5) cm^{2}

= (π(21)^{2} − 346.5) cm^{2}

= 1386 − 346.5

= 1039.5 cm^{2}

Similarly,

The radius of blue region r_{3} = 21 cm+10.5 cm = 31.5 cm

∴ Area of blue region = Area of third circle – Area of second circle

= π(31.5)^{2} – 1386 cm^{2}

= 3118.5 – 1386 cm^{2 }

= 1732.5 cm^{2}

The radius of black region r_{4} = 31.5 cm+10.5 cm = 42 cm

∴ Area of black region = Area of fourth circle – Area of third circle

= π(42)^{2} – 1386 cm^{2 }

= 5544 – 3118.5 cm^{2 }

= 2425.5 cm^{2}

The Radius of white region r_{5} = 42 cm+10.5 cm = 52.5 cm

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

= π(52.5)^{2} – 5544 cm^{2 }

= 8662.5 – 5544 cm^{2 }

= 3118.5 cm^{2}

diameter of wheel d_{ 1} = 80

The radius of car’s wheel = 80/2 = 40 cm

the circumference of wheels = 2πr = 80 π cm

Since speed of car =66kmh^{-1}

66 X 1000 /60 = 1100 m min^{-1}

Distance covered by the car in 1hr = (66×10^{5}) cm

In 10 minutes, the distance covered will be = (66×10^{5}×10)/60 = 1100000 cm/s

∴ Distance covered by car = 11×10^{5} cm

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

=( 11×10^{5})/80 π = 4375.

(A) 2 units (B)π units

(C) 4 units (D) 7 units

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

2πr = πr^{2}

Or, r = 2

So, option (A) is correct