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
r1= 19 cm
Circumference of the 1cst circle = 2π×19 = 38π cm
r2 = 9 cm
Circumference of the 2nd circle = 2π×9 = 18π cm
Now, let the radius of new circle circle = R
According to question
circumference of new circle =Circumference of the 1st circle + Circumference of the 2nd circle
2πR= 38π+18π = 56π cm
2πR =56π
R= 56π/2π
R = 28 cm.
r1 = 8 cm
Area of 1st circle = π(8)2 = 64π
Radius of 2nd circle r2 = 6 cm
Area of 2nd circle = π(6)2 = 36π
So,
The sum of 1st and 2nd circle will be = 64π+36π = 100π
let the radius of new circle = R
Area of the new circle = πR2
The area of the new circle = Area of 1st circle + Area of 2nd circle
Or, πR2 = 100πcm2
R2 = 100cm2
So, R = 10cm
Diameter of circle =21cm
The radius of gold region r1 = 21/2 cm
Area of gold region = π r12 = π(10.5)2 = 346.5 cm2
So, the radius of red region 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 blue region r3 = 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 cm2
= 3118.5 – 1386 cm2
= 1732.5 cm2
The radius of black region r4 = 31.5 cm+10.5 cm = 42 cm
∴ Area of black region = Area of fourth circle – Area of third circle
= π(42)2 – 1386 cm2
= 5544 – 3118.5 cm2
= 2425.5 cm2
The Radius of white region r5 = 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 cm2
= 8662.5 – 5544 cm2
= 3118.5 cm2
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 = (66×105) 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.
(A) 2 units (B)π units
(C) 4 units (D) 7 units
Since the perimeter of the circle = area of the circle,
2πr = πr2
Or, r = 2
So, option (A) is correct