Question

Prakash went to a park in his colony and sat on a bench. The park is circular in shape. There is a pond in the park which touches the centre of the park. A rope ST is tied on the circumference of the park to separate the shallow part from the deep part of the pond. The pond is deep near the circular edge The radius of the circular park is 420 m and the pond covers the th part of the park. The rest of the circular park is covered with green grass 41. पार्क का केंद्रीय कोण है The central angle of the pond is: (a) 360 (b) 300 (c) 90 c) 60 4.पार्क का कुल क्षेत्रफल है The total area of the pond is: (a) 0.0924km (b)0.0654km (c) 0.1386 km (d) 0.0462 km (d) दीर्घ विज्यपंड 43.घास तथा उथला तालाब पार्क के हिस्से को उकते हैं (a) लघु विज्यखंड (b) दीर्घ वृत्तखंड (c) लघु वृत्तखड The grass and shallow pond together covers part of the park: (a) Minor sector (b) Major segment (c) Minor segment (d) Major sector 44. पार्क के हरे भाग की परिधि है। The circumference of the green part of the park is : (a)3040m (b) 2640m (c) 840m (d) 2200m 45. तालाब की कुल लम्बाई है: The total length of the pond is: (a) 1.32 km (b) 1.26 km (e) 1.28 km (d) 1.3 km केस-स्टडी IL/Cuse-Study II​

Answer 1

Answer:

1-c)60

2-a)0.0924

3-b)major seg

4-d)2200

5-c)1.28

Step-by-step explanation:

i think im right

Answer 2

1. The central angle of the pond is equal to 60° .
2. The total area of the pond is equal to 0.0924 km² .
3. The grass and shallow pond together covers Major segment of the park .
4. The circumference of the green part of the park is equal to 3040 m .
5. The total length of the pond is equal to 1.28 km .

Given :-

• Radius of circular park = 420 m
• The pond covers the 1/6th part of the park .
• The rest of the circular park is covered with green grass .

To Find :-

1) The central angle of the pond is ?

2) The total area of the pond is ?

3) The grass and shallow pond together covers part of the park ?

4) The circumference of the green part of the park is ?

5) The total length of the pond is ?

Formula used :-

• Area of circle = πr²
• Area of sector = (θ/360°)•πr² { where θ is angle at centre)
• Circumference of the circle = 2πr
• Circumference of the sector = (θ/360°)•2πr + 2r
• 1 km = 1000 m
• 1 km² = 1000000 m²
• Complete angle = 360°

Solution :-

1)

Since the pond covers the 1/6th part of the park,

→ Complete angle of a circle = 360°

So,

→ central angle of the pond = (1/6) of 360° = (1/6) × 360° = 60° (d) (Ans.)

2)

we have,

→ Central angle of the pond = θ = 60°

→ Radius of circle = r = 420 m

So,

→ Total area of pond = (θ/360°) × π × r × r

→ Total area of pond = (60°/360°) × (22/7) × 420 × 420

→ Total area of pond = (1/6) × 22 × 60 × 420

→ Total area of pond = 22 × 10 × 420

→ Total area of pond = 92400 m²

now,

→ 1000000 m² = 1 km²

→ 1 m² = (1/1000000) km²

→ 92400 m² = (1/1000000) × 92400 = 0.0924 km² (a) (Ans.)

3) The grass covers Major sector of the park and the pond covers Minor sector of the park . Now, ST separate the shallow part from the deep part of the pond .

So,

→ The grass and shallow pond together area = Major sector area + Area of ∆SOT = Major segment (b) (Ans.)

4)

→ Angle at centre of the green park = 360° - central angle of pond = 360° - 60° = 300°

→ Radius = r = 420 m

So,

→ Circumference of the green part of the park = (θ/360°)•2πr + 2r

→ Circumference of the green part of the park = (300°/360°) × 2 × (22/7) × 420 + 2 × 420

→ Circumference of the green part of the park = (5/6) × 2 × 22 × 60 + 840

→ Circumference of the green part of the park = 10 × 22 × 10 + 840

→ Circumference of the green part of the park = 2200 + 840

→ Circumference of the green part of the park = 3040 m (a) (Ans.)

5)

→ central angle of the pond = 60°

→ r = 420 m

So,

→ Total length of the pond = Circumference of the pond = (θ/360°)•2πr + 2r

→ Total length of the pond = (60°/360°) × 2 × (22/7) × 420 + 2 × 420

→ Total length of the pond = (1/6) × 2 × 22 × 60 + 840

→ Total length of the pond = 2 × 22 × 10 + 840

→ Total length of the pond = 440 + 840

→ Total length of the pond = 1280 m

now,

→ 1000 m = 1 km

→ 1 m = (1/1000) km

→ 1280 m = (1/1000) × 1280 = 1.28 km (c) (Ans.)

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