Step 1: Find the circumference of the wheel.
Diameter = 70 cm, Radius (r) = 35 cm.
Circumference = 2πr = 2 × (22/7) × 35 = 220 cm.
This is the distance covered in one revolution.

Step 2: Convert the total distance to cm.
Total distance = 1.65 km = 1.65 × 1000 m = 1650 m.
= 1650 × 100 cm = 165000 cm.

Step 3: Calculate the number of revolutions.
Number of revolutions = Total Distance / Circumference
= 165000 / 220 = 750.

Answer: The wheel will make 750 revolutions.

Given: r = 14 cm, θ = 120°.

Step 1: Calculate the area of the sector.
Area_sector = (θ/360°) × πr² = (120/360) × (22/7) × 14²
= (1/3) × (22/7) × 196 = 616/3 ≈ 205.33 cm².

Step 2: Calculate the area of the triangle.
Area_triangle = (1/2)r² sinθ = (1/2) × 14² × sin(120°)
= (1/2) × 196 × sin(60°) = 98 × (√3/2) = 49√3
= 49 × 1.73 = 84.77 cm².

Step 3: Calculate the area of the segment.
Area_segment = Area_sector – Area_triangle
= 205.33 – 84.77 = 120.56 cm².

Answer: The area of the segment is approximately 120.56 cm².

Let the inner radius (park) be r = 42 m.
Let the outer radius (park + path) be R = 42 + 7 = 49 m.

The area of the path is the area of the outer circle minus the area of the inner circle (area of a ring).

Area of path = πR² - πr² = π(R² - r²)
= (22/7) × (49² – 42²) = (22/7) × (49-42)(49+42)
= (22/7) × 7 × 91 = 22 × 91 = 2002 m².

Answer: The area of the path is 2002 m².

Step 1: Find the radius of the quadrant.
The diagonal of the square OB will be the radius of the quadrant.
In right ΔOAB, OB² = OA² + AB² = 20² + 20² = 400 + 400 = 800.
Radius r = OB = √800 = 20√2 cm.

Step 2: Find the area of the quadrant.
Area_quadrant = (1/4)πr² = (1/4) × 3.14 × (20√2)²
= (1/4) × 3.14 × 800 = 3.14 × 200 = 628 cm².

Step 3: Find the area of the square.
Area_square = side² = 20² = 400 cm².

Step 4: Find the area of the shaded region.
Shaded Area = Area_quadrant – Area_square = 628 – 400 = 228 cm².

Answer: The area of the shaded region is 228 cm².

Step 1: Find the side of the equilateral triangle.
For an equilateral triangle inscribed in a circle of radius R, the side (a) is given by a = R√3.
So, side a = 6√3 cm.

Step 2: Calculate the area of the triangle.
Area_triangle = (√3/4)a² = (√3/4) × (6√3)² = (√3/4) × 108 = 27√3 cm².

Step 3: Calculate the area of the circle.
Area_circle = πR² = π(6)² = 36π cm².

Step 4: Find the required area.
Required Area = Area_circle – Area_triangle = (36π – 27√3) cm².

Answer: The area is (36π – 27√3) cm².

Step 1: Calculate the area of the square.
Area_square = 14 × 14 = 196 cm².

Step 2: Find the area of the unshaded regions.
The unshaded regions I and III can be found by: Area of Square – Area of two semicircles with diameter 14 cm.
Radius of semicircle = 7 cm. Area of two semicircles = Area of one circle = πr² = (22/7) × 7² = 154 cm².
Area (I + III) = 196 – 154 = 42 cm².

Similarly, Area (II + IV) = 196 – 154 = 42 cm².

Step 3: Find the area of the shaded region.
Shaded Area = Area_square – [Area (I+III) + Area(II+IV)]
This is incorrect. A simpler method is: Area of 2 shaded regions = Area of 2 semicircles – Area of Square + Area of 2 unshaded regions. Let’s find the area of the two “leaf” shapes (shaded) plus the two unshaded regions. This equals the area of two full circles with diameter 14cm. Wait, that’s not right.
Let’s use a simpler logic: Area of shaded region = [Area(Semicircle on AB) + Area(Semicircle on CD)] – [Area of Square – Area(Semicircle on AD) – Area(Semicircle on BC)]
Let’s try: Area of the two leaves = Area(Semicircle APD) + Area(Semicircle BPC) – Area(Square ABCD)
Wait, the logic is simpler: Area Shaded = Area(Semicircle AD) + Area(Semicircle BC) – Area(Square without leaves). Let’s denote the shaded parts S1, S2 and unshaded U1, U2. Area(Square) = S1 + S2 + U1 + U2 = 196. Area(Semicircle on AB) + Area(Semicircle on CD) = U1 + S1 + S2 + U2 = 2 * (1/2 * pi * 7^2) = 154. Area(Semicircle on AD) + Area(Semicircle on BC) = U2 + S1 + S2 + U1 = 154. Summing these: 2(S1+S2+U1+U2) = 308 => 2 * Area(Square) = 308 => Area(Square) = 154. This is a contradiction.
Correct Logic: Area of Vertical Leaves = Area(Semi on AD) + Area(Semi on BC) – Area(Square) = 154 – 196 = -42 (this is impossible). The area of the region is the sum of areas of two segments. Let’s reconsider. Area of Shaded Region = Area(Sector ADC) – Area(Triangle ADC) * 2 Let’s try a different way. Area of the four leaf-like unshaded regions = Area(Square) – Area(Inscribed Circle) = 196 – pi*7^2 = 196 – 154 = 42. No, that’s for a circle, not semicircles. Correct logic: Area of shaded region = Area of 2 circles with diameter 14 – Area of square. No. Area of Shaded Region = Area(Square) – Area of 4 unshaded corners. Let’s find the area of the unshaded regions. Area(Unshaded) = 2 * [Area(Square) – Area(Semicircle on AB) – Area(Semicircle on CD)]… No. Area(Shaded) = 2 * [Area(Semicircle on AB) – Area(Segment of quadrant)]… Final method: Area(2 leaves) = 2 * Area(Semicircle) – Area(Square) = pi * r^2 – (2r)^2. No. Area(Region I + III) = Area(Square) – Area of two semicircles on AD and BC = 196 – 154 = 42 cm². Area(Region II + IV) = Area(Square) – Area of two semicircles on AB and CD = 196 – 154 = 42 cm². Total Unshaded Area = 42 + 42 = 84 cm². Shaded Area = Area(Square) – Total Unshaded Area = 196 – 84 = 112 cm². Wait, this double counts the central overlap.
Let’s mark the four unshaded regions as R1, R2, R3, R4. Area(Square) = Area(Semicircle AD) + Area(Semicircle BC) – Area(Shaded). 196 = 154 – Area(Shaded) -> This is wrong. Area(Shaded) = Area(Semi AD) + Area(Semi BC) – Area(Square) = 154 – 196 -> impossible. Let’s rethink: Area(Sector DAB) + Area(Sector DCB). Area of the region bounded by arc DC and chord DC. Area of the shaded region = Area of square – 2 * [Area of quadrant – Area of isosceles triangle]. Shaded Area = 2 × (Area of Semicircle) – Area of Square = πr² – (2r)² = 154 – 196 -> wrong. Let’s call the unshaded regions U1, U2, U3, U4. Area(Square) = Area(Semi AB) + Area(Semi CD) – Area(overlap) => Area of overlap is the shaded region. 196 = 154 + 154 – Shaded Area => Shaded Area = 308 – 196 = 112 cm². No. Area(Semi AD) + Area(Semi BC) = Area(Square). Shaded area is not involved here. Area (Unshaded regions 1 and 3) = Area(Square) – Area(Semicircle on AD and BC) = 196 – 154 = 42 cm^2. Shaded area = Area(Semicircle on AB and CD) – Area(Unshaded regions 2 and 4). Correct logic: Area(Shaded) = Area(Semi 1) + Area(Semi 2) – Area of Square = πr² – (2r)² — this is for a different figure. Final attempt with the simplest logic: Area of Shaded Region = Area(Semicircle on AB) + Area(Semicircle on DC) – Area(Unshaded Region). No. Area(Shaded) + Area(Unshaded) = Area(Square). Area(Unshaded) = 2 * (Area of Square – Area of 2 Quadrants) = 2 * (14^2 – 2 * 1/4 * pi * 7^2) Let’s find Area of 2 vertical segments + 2 horizontal segments. Area of Shaded Region = 2 × (Area of Semicircle) – Area of Square. This is a known result. = 2 × (1/2 π r²) – (2r)² = πr² – 4r² = r²(π-4). Here r=7. 49(22/7-4) = 49(-6/7) -> wrong. Area of shaded region = 4/7 * a^2 = 4/7 * 14^2 = 4/7 * 196 = 4 * 28 = 112 cm^2. Let’s prove it. Area(Shaded) = 2*Area(Segment of quadrant). Area(quadrant with r=14) = 1/4 * pi * 14^2 = 154. Area(triangle) = 1/2 * 14 * 14 = 98. Segment = 154-98 = 56. 2*56=112. Let’s try another method. Area of vertical semicircles = 154. Area of horizontal semicircles = 154. Sum = 308. This sum covers the square once, and the shaded region twice. So, Sum = Area(Square) + Area(Shaded) => 308 = 196 + Area(Shaded) => Area(Shaded) = 112 cm². This is the correct logic.

Answer: The area of the shaded region is 112 cm².

Step 1: Calculate the area of the minor sector.
r = 10 cm, θ = 90°.
Area_{minor_sector} = (90/360) × 3.14 × 10² = (1/4) × 314 = 78.5 cm².

Step 2: Calculate the area of the triangle.
Area_triangle = (1/2) × base × height = (1/2) × 10 × 10 = 50 cm².

Step 3: Calculate the area of the minor segment.
Area_{minor_segment} = Area_{minor_sector} – Area_triangle = 78.5 – 50 = 28.5 cm².

Step 4: Calculate the area of the major segment.
Area_circle = πr² = 3.14 × 10² = 314 cm².
Area_{major_segment} = Area_circle – Area_{minor_segment} = 314 – 28.5 = 285.5 cm².

Answer: The area of the major segment is 285.5 cm².

Step 1: Find the length of the straight sections.
Inner radius r = 35 m. Circumference of 2 inner semicircles = 2πr = 2 × (22/7) × 35 = 220 m.
Total inner length = 2 × (Straight section) + 220 = 400 m.
Length of one straight section = (400 – 220)/2 = 90 m.

Step 2: Calculate the area of the two straight rectangular parts of the track.
Area_rectangles = 2 × (length × width) = 2 × (90 × 7) = 1260 m².

Step 3: Calculate the area of the two semicircular parts of the track.
Outer radius R = 35 + 7 = 42 m.
Area of 2 semicircular ends = Area of outer ring = π(R² – r²)
= (22/7) × (42² – 35²) = (22/7) × (42-35)(42+35) = (22/7) × 7 × 77 = 1694 m².

Step 4: Find the total area of the track.
Total Area = Area_rectangles + Area_{semicircular_ends} = 1260 + 1694 = 2954 m².

Answer: The area of the track is 2954 m².

Perimeter of a sector = Length of arc + 2 × radius.
16.4 = Length of arc + 2(5.2) => 16.4 = Length of arc + 10.4.
Length of arc (l) = 16.4 – 10.4 = 6 cm.

There is a direct formula for the area of a sector given arc length and radius:
Area = (1/2) × l × r.

Area = (1/2) × 6 × 5.2 = 3 × 5.2 = 15.6 cm².

Answer: The area of the sector is 15.6 cm².

The largest triangle that can be inscribed in a semi-circle will have the diameter of the semi-circle as its base.

Base of the triangle = Diameter = 2r.

The height of the triangle will be maximum when the third vertex is at the highest point of the semi-circle, which is at a distance of the radius from the diameter.

Height of the triangle = radius = r.

Area = (1/2) × base × height = (1/2) × (2r) × (r) = r².

Answer: The area of the largest triangle is r².