Chapter 11: Areas Related to Circles

According to the question,

Area of the circular playground = 22176 m² (Given)

Let r be the radius of the circle.

∴ πr² = 22176

⇒ (22/7)r² = 22176

⇒ r² = 22176 × (22/7)

⇒ r² = 7056

⇒ r = 84

∴ The radius of the circular playground = 84 m

Now, the circumference of the circle = 2πr

= 2×(22/7)×84

= 528 m

Cost of fencing 1 meter of ground = Rs 50

∴ Cost of fencing the total ground = Rs 528 × 50 = Rs 26,400

According to the question,

Diameter of front wheels = d₁ = 80 cm

Diameter of rear wheels = d₂ = 2 m = 200 cm

Let r₁ be the radius of the front wheels = 80/2 = 40 cm

Let r₂ be the radius of the rear wheels = 200/2 = 100 cm

Now, the circumference of the front wheels = 2πr

= 2 ×(22/7) × 40

= 1760/7 cm

Circumference of the rear wheels = 2πr = 2 × (22/7) × 100 = 4400/7 cm

No. of revolutions made by the front wheel = 1400

∴ Distance covered by the front wheel = 1400 × (1760/7) = 352000 cm

Number of revolutions made by the rear wheel in covering a distance in which the front wheel makes 1400 revolutions,

= 560 revolutions.

According to the question,

The sides of the triangle are 15 m, 16 m, and 17 m.

Now, perimeter of the triangle = (15+16+17) m = 48 m

∴ Semi-perimeter of the triangle = s = 48/2 = 24 m

By Heron’s formula,

Area of the triangle = √(s(s – a)(s – b)(s – c)), where a, b and c are the sides of the triangle

= √(24(24 – 15)(24 – 16)(24 – 17))

= 109.982 m²

Let B, C and H be the corners of the triangle on which buffalo, cow and horse are tied, respectively, with ropes of 7 m each.

So, the area grazed by each animal will be in the form of a sector.

∴ Radius of each sector = r = 7 m

Let x, y, z be the angles at corners B, C, H, respectively.

∴ Area of sector with central angle x,

= ½ (x/180) × πr² = (x/360) × π(7)²

Area of sector with central angle y,

= ½ (y/180) × πr² = (y/360) × π(7)²

Area of sector with central angle z,

= ½ (z/180) × πr² = (z/360) × π(7)²

Area of field not grazed by the animals = Area of triangle – (area of the three sectors)

= 109.892 – 77

= 32.982 cm²

According to the question,

Radius of the circle = r = 12 cm

∴ OA = OB = 12 cm

∠ AOB = 60° (given)

Since triangle OAB is an isosceles triangle, ∴ ∠ OAB = ∠ OBA = θ (say)

Also, the sum of interior angles of a triangle is 180°,

∴ θ + θ + 60° = 180°

⇒2θ = 120° ⇒ θ = 60°

Thus, the triangle AOB is an equilateral triangle.

∴ AB = OA = OB = 12 cm

Area of the triangle AOB = (√3 /4)× a², where a is the side of the triangle.

= (√3 /4) × (12)²

= (√3 /4) × 144

= 36√3 cm²

= 62.354 cm²

Now, Central angle of the sector AOBCA = ∅ = 60° =  (60π / 180) = (π/3) radians

Thus, area of the sector AOBCA = ½ r² ∅

= ½ × 12² × π/3

= 12² × (22 / (7×6))

= 75.36 cm²

Now, Area of the segment ABCA = Area of the sector AOBCA – Area of the triangle AOB

= (75.36 – 62.354) cm² = 13.006 cm²

Diameter of the circular pond = 17.5 m

Let r be the radius of the park = (17.5/2) m = 8.75 m

According to the question,

The circular pond is surrounded by a path of width 2 m.

So, Radius of the outer circle = R = (8.75+2) m = 10.75 m

Area of the road = Area of the outer circular path – Area of the circular pond

= πr² – πR²

= 3.14 × (10.75)² – 3.14 × (8.75)²

= 3.14 × ((10.75)² – (8.75)²)

= 3.14 × ((10.75 + 8.75) × (10.75 – 8.75))

= 3.14 × 19.5 × 2 = 122.46 m²

Hence, the area of the path is 122.46 m².

Now, the cost of constructing the path per m² = Rs. 25

∴ cost of constructing 122.46m² of the path = Rs. 25 × 122.46 = Rs. 3061.50

AB = 18 cm, DC = 32 cm

Distance between AB and DC = Height = 14 cm

Now, Area of the trapezium = (1/2) × (Sum of parallel sides) × Height

= (½) × (18+32) × 14 = 350cm²

As AB ∥ DC, ∴ ∠ A +∠ D = 180°

And ∠ B +∠ C = 180°

Also, radius of each arc = 7 cm

Therefore,

Area of the sector with central angle A = (½) × (∠A/180) × π × r²

Area of the sector with central angle D = (½) × (∠D/180) × π × r²

Area of the sector with central angle B = (½) × (∠B/180) × π × r²

Area of the sector with central angle C = (½) × (∠C/180) × π × r²

The total area of the sectors,

= 77 + 77

= 154

∴ Area of shaded region = Area of trapezium – (Total area of sectors)

= 350 – 154 = 196 cm²

Hence, the required area of the shaded region is 196 cm².

The three circles are drawn such that each of them touches the other two.

By joining the centres of the three circles,

We get,

AB = BC = CA = 2(radius) = 7 cm

Therefore, triangle ABC is an equilateral triangle with each side 7 cm.

∴ Area of the triangle = (√3 /4) × a², where a is the side of the triangle.

= (√3 /4) × 7²

= (49/4) √3 cm²

= 21.2176 cm²

Now, Central angle of each sector = ∅ = 60° (60π/180)

= π/3 radians

Thus, area of each sector = (1/2) r²θ

= (1/2) × (3.5)² × (π/3)

= 12.25 × (22/ (7×6))

= 6.4167 cm²

Total area of three sectors = 3 × 6.4167 = 19.25 cm²

∴ The area enclosed between three circles = Area of triangle ABC – Area of the three sectors

= 21.2176 – 19.25

= 1.9676 cm²

Hence, the required area enclosed between these circles is 1.967 cm² (approx.).

Radius of the circle = r = 5 cm

Arc length of the sector = l = 3.5 cm

Let the central angle (in radians) be θ.

As, Arc length = Radius × Central angle (in radians)

∴ Central angle (θ) = Arc length / Radius = l / r = 3.5/5 = 0.7 radians

Now, Area of the sector = (½) × r²θ = (½) × 25 × 0.7 = 8.75 cm²

Hence, the required area of the sector of a circle is 8.75 cm².

According to the question,

The four circles are placed such that each piece touches the other two pieces.

By joining the centres of the circles by a line segment, we get a square ABDC with sides,

AB = BD = DC = CA = 2(Radius) = 2(7) cm = 14 cm

Now, Area of the square = (Side)² = (14)² = 196 cm²

ABDC is a square,

Therefore, each angle has a measure of 90°.

i.e., ∠ A = ∠ B = ∠ D = ∠ C = 90° = π/2 radians = θ (say)

Given that,

Radius of each sector = 7 cm

Area of the sector with central angle A = (½)r²θ

= ½ r² θ

= ½ × 49 × π/2

= ½ × 49 × (22/(2×7))

= (77/2) cm²

Since the central angles and the radius of each sector are same, area of each sector is 77/2 cm²

∴ Area of the shaded portion = Area of square – Area of the four sectors

= 196 – (4 × (77/2))

= 196 – 154

= 42 cm²

Therefore, the required area of the portion enclosed between these pieces is 42 cm².

According to the question,

Area of the square = 784 cm²

∴ Side of the square = √Area = √784 = 28 cm

The four circular plates are congruent,

Then diameter of each circular plate = 28/2 = 14 cm

∴ The radius of each circular plate = 7 cm

We know that,

Area of the sheet not covered by plates = Area of the square – Area of the four circular plates

As all four circular plates are congruent,

We have the area of all four plates equal.

∴ Area of one circular plate = πr² = (22/7) × 7² = 154 cm²

Then,

Area of four plates = 4×154 = 616 cm²

Area of the sheet not covered by plates = 784 – 616 = 168 cm²