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The radius of a circular garden is 100 m. There is a road 10 m wide, running all around it. Find the area of the road and the cost of levelling it at Rs. 20 per m2. [Use π = 3.14.]
Here we will first find out the area of the road running around the circular garden and then multiplying it with rate per square meter to calculate the cost of leveling.
Here we see in the figure there are two concentric circles so,
Area of road = Area of outer circle- Area of circular garden
Area of circle = πR2 (where R = radius of circle) → eqn1
Let the radius of inner circle = r = 100 m
Also, radius of outer circle = R = 110 m (R = r + 10)
Area of outer circle = π(110)2→ eqn2 (putting R in eqn1)
Area of inner circle = π(100)2→ eqn2 (putting r in eqn1)
∴ Area of road = π(110)2 – π(100)2 (from eqn2 and 3)
⇒ Area of road = π(12100 – 10000)
⇒ Area of road = 2100 π (put π = 3.14)
⇒ Area of road = 2100×3.14
∴ 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.
ABCD is a field in the shape of a trapezium, AD || BC, ∠ABC = 90° and ∠ADC = 60°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m.
Find the following:
(i) total area of the four sectors,
(ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.
A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP ⊥ CD, HQ ⊥ FI and EL ⊥ DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure. [Take π = 3.14.]