Question

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The radius of a circular garden is 100m. There is a road 10 m wide , running all around it. Find the area of the road and the cost of levelling it at ₹20 per m^2. ( Take root 3 = 1.73 )​

Answer 1

Step-by-step explanation:

Answer:-

Given:-

• The radius of a circular garden is 100m.
• The road is 10 m long.

To find :-

• Area of the road.
• the cost of leveling it at ₹20 per m²

Solution :-

Here we see in the figure there are two concentric circles so,

Here we see in the figure there are two concentric circles so,Area of road = Area of outer circle- Area of circular garden

Here we see in the figure there are two concentric circles so,Area of road = Area of outer circle- Area of circular gardenArea of circle = πR² (where R = radius of circle) → eqn1

Let the radius of inner circle = r = 100 m

Let the radius of inner circle = r = 100 mAlso, radius of outer circle = R = 110 m (R = r + 10)

Let the radius of inner circle = r = 100 mAlso, radius of outer circle = R = 110 m (R = r + 10)Area of outer circle = π(110)²→ eqn2 (putting R in eqn1)

→ eqn2 (putting R in eqn1)Area of inner circle = π(100)²→ eqn2 (putting r in eqn1)

→ eqn2 (putting r in eqn1)∴ Area of road = π(110)² – π(100)² (from eqn2 and eqn3)

⇒ Area of road = π(12100 – 10000)

⇒ Area of road = π(12100 – 10000)⇒ Area of road = 2100 π (put π = 3.14)

⇒ Area of road = π(12100 – 10000)⇒ Area of road = 2100 π (put π = 3.14)⇒ Area of road = 2100 × 3.14

3.14∴ Area of road = 6594 m²

Cost of leveling = Rate of leveling × Area of road

⇒ Cost of leveling = 20 × 6594

∴ Cost of leveling = Rs.131880

Area of road is 6594 m2 and cost of leveling is Rs.131880.

Answer 2

Answer:

Given:-

$\sf \: Area \: of \: Road= \pi r {}^{2} - \pi r {}^{2}$

$\pi((100 {}^{2} )- (100 ){}^{2} )$

$\frac{22}{7} \: \: (12100 - 10000)$

$\pink{ \frac{22}{7} \: \times \: (2100)}$

$\green{ \sf \: 22 \times (300)}$

$\red{ \huge \frak{6600cm {}^{2} }}$

Cost=22×6600=13200