Question

(d) None of the above Direction Answer the questions from 36-40 based on the following case. Shivam has his hostel located at P and his school located at Q. Shivam drives his motorbike three day in a week and rides his car in the remaining 3 days, to go to his school and back to hostel. POQ is a sector of a circle with centre 0, central angle 60° and radius 4.2 km. Path POQ is the route for driving by motorbike and path PRQ is for car only. if the angle of sector changed from 60° to 90°,then the total length of the available paths is​

Answer 1

Answer:

1, 50.4 km

2, 26.4 km

3, 12.06 km sq.

4, Rs. 480

5, 17.14 km

Answer 2

1. The total distance travelled by Shivam through the motorbike in a week to go to school is equal to 50.4 km .
2. The total distance travelled by Shivam through the car in a week to go to school is equal to 26.4 km .
3. The area of sector POQ is equal to 9.24 km² .
4. The total cost of fuel used in a week ingoing to school by motorbike is equal to ₹ 1008 .
5. The total length of the available new paths is equal to 15 km .

Given :-

• Motorbike path for 3 days = POQ
• Car path for 3 days = PRQ
• ∠POQ = 60°
• OP = OQ = 4.2 km

To Find :-

1) The total distance travelled by Shivam through the motorbike in a week to go to school ?

2) The total distance travelled by Shivam through the car in a week to go to school ?

3) The area of sector POQ ?

4) If the cost of fuel for the motorbike is ₹ 20 per km, then find the total cost of fuel used in a week ingoing to school ?

5) If the angle of sector changed from 60° to 90°, then find the total length of the available path ?

Formula used :-

• Length of sector = (θ/360°)•2πr { where θ is angle at centre }
• Area of sector = (θ/360°)•πr²

Solution :-

1)

→ Motorbike path = POQ

So,

→ Distance travelled while going to school = PO + OQ = 4.2 km + 4.2 = 8.4 km

→ Distance travelled while coming to hostel = QO + OP = 4.2 km + 4.2 = 8.4 km

So,

→ Distance covered by motorbike in 1 day = 8.4 + 8.4 = 16.8 km

then,

→ Total distance travelled by Shivam through the motorbike in 3 days to go to school = 3 × 16.8 = 50.4 km (Ans.)

2)

→ Central angle = θ = 60°

→ Radius = r = 4.2 km

So,

→ Distance travelled by shivam while going to school in his car = (θ/360°)•2πr = (60°/360°) × 2 × (22/7) × 4.2 = (1/6) × 2 × 22 × 0.6 = 44 × 0.1 = 4.4 km

and,

→ Distance travelled by shivam while coming back to hostel in his car = 4.4 km

then,

→ Total distance covered by shivam in car in 1 day = 4.4 km + 4.4 km = 8.8 km

therefore,

→ The total distance travelled by Shivam through the car in 3 days to go to school = 8.8 × 3 = 26.4 km (Ans.)

3)

we have,

→ ∠POQ = θ = 60°

→ r = 4.2 km

So,

→ Area of sector POQ = (θ/360°)•πr²

→ Area of sector POQ = (60°/360°) × (22/7) × 4.2 × 4.2

→ Area of sector POQ = (1/6) × 22 × 0.6 × 4.2

→ Area of sector POQ = 22 × 0.1 × 4.2

→ Area of sector POQ = 9.24 km² (Ans.)

4)

→ The cost of fuel for the motorbike = ₹20 / km

→ Total distance covered by motorbike in a week = 50.4 km

So,

→ Total cost of fuel for the motorbike in a week = 20 × 50.4 = ₹ 1008 (Ans.)

5)

→ Central angle = θ = 90°

→ Radius = r = 4.2 km

So,

→ Length of distance covered by motorbike = 4.2 km + 4.2 km = 8.4 km

and,

→ Length of distance covered by car = (θ/360°)•2πr = (90°/360°) × 2 × (22/7) × 4.2 = (1/4) × 2 × 22 × 0.6 = 11 × 0.6 = 6.6 km

then,

→ Total length of available paths = 8.4 km + 6.6 km = 15 km (Ans.)

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