Areas Related to Circles

∴ p² = side² + side²

⇒ p² = 2side²

or side² = $\frac{p^2}{2}$ cm² = area of the square

∴ Its area = d

Diagonal of inner square = d

∴ Side = $\frac{d}{\sqrt{2}}$

⇒ Area = $\frac{d^2}{2}$

Area of outer square = 2 × Area of inner square.

⇒ 2r = r²

⇒ r(r – 2) = 0 or r = 2

i.e. d = 4 units

= (2r + πr) cm

⇒ 2r + πr = 36

$\Rightarrow r\left(2+\frac{22}{7}\right)=36$

⇒ r = 7cm

Diameter 2r = 2 × 7 = 14 cm.

= $\frac{22}{7}$ × 7 + 2 × 7 = 22 + 14 = 36cm

Then, diagonal of the square = 2r

Area of largest ∆ABC = $\frac{1}{2}$ × AB × CD

$\frac{1}{2}$ × 2r × r = r² sq. units

rectangle = b cm

∴ Radius = $\frac{b}{2}$ cm

or r (2π – 1) = 37

πr² = p(8)2 + p(6)²

⇒ πr² = 64p + 36p

⇒ pr² = 100p

∴ r² = 100pp = 100

⇒ r = 10cm

Hence, radius of required circle is 10 cm.

Perimeter = 66 cm

= πr + 2r = 66 [∴ Perimeter of a semicircle = πr + 2r]

Diameter = 2r cm and Circumference = 2πr cm

According to question,

Circumference = Diameter + 16.8

⇒ 2πr = 2r + 16.8

⇒ 2 × $\frac{22}{7}$ × r = 2r + 16.8

⇒ 44r = 14r + 16.8 × 7

⇒ 44r – 14r = 117.6 or 30r = 117.6

⇒ r = $\frac{117.6}{30}$ = 3.92

Hence, radius = 3.92 cm.

2πR = 396 and 2πr = 352

Hence, width of the track = (R – r) m = (63 – 56) m = 7 m

Let the inner and outer radii of the circular track berm and R m respectively. Then,

Inner circumference = 2πr = 220 m

Since the track is 7 m wide everywhere. Therefore,

R = Outer radius = r + 7 = (35 + 7)m = 42 m

∴ Outer circumference = 2πR = 2 × $\frac{22}{7}$ × 42m = 264m

Rate of fencing = ₹ 2 per metre

∴ Total cost of fencing = (Circumference × Rate) = ₹(264 × 2) = ₹ 528

radius of the wheel = 40 cm.

Now, distance travelled in one complete revolution of wheel = 2π × 40 = 80π

Since, speed of the car is 66 km/h

So, distance travelled in 10 minutes $=\frac{66\times 100000\times 10}{60}$

= 11 × 100000 cm = 1100000 cm.  

So, Number of complete revolutions in 10 minutes

Area of the shaded region

= ar (square) – 4 (ar of quadrant)

= 2(3.5) = 7 cm

∴ ∆ABC is an equilateral ∆

Area of the shaded region

= ar(square) – 4(circle)

= (side)² – 4 (πr²)

= (28)² – 4 × $\frac{22}{7}$ × 7 × 7 = 784 – 616 = 168 cm²

Area of shaded region

= 3(ar of minor segment) = 3[ar(minor sector) – ar(∆ABC)]

v.

^{​​​​​​​}​​​​​​​^{​​​​​​​}​​​​​​​Solution:

Area of circular path formed by two concentric circles 

Let r and R be the radii of each smaller circle and larger circle, respectively.

We have, 

Solution:

Area of smaller circle πr²

Solution:

Area of larger circle πR²

​​​​​​​​​​​​​​Solution:

Area of the black color region = Area of larger circle – Area of 4 smaller circles.

= 804.57 – 4 × 50.28 = 603.45cm²

^{​​​​​​​​​​​​​​​​​​​​​}Solution:

Area of quadrant of a smaller circle

^{​​​​​​​}Solution:

Area between two concentric circles