Question

The length and breadth of a park are in the ratio 2:1 and its perimeter is 240 m. A path
of 2 m wide runs inside it, along boundary. Find the cost of paving the path at 5 per m².​

Answer 1

let the length and breadth of a park be 2x and x respectively.

so, area of the park

$2(l + b) \\ = 2(2x + x) \\ = 2 \times 3x \\ = 6x = 240 \: m \\ = > x = \frac{240}{6} m \\ = > x = 40m \\ = > l = 2x = 2 \times 40m \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 80m \\ = > b = x = 40 \: m$

therefore the length and the breadth of rectangle are are 80 metre and 40 metre respectively.

Area of the park = 40 x 80 = 3200 m²

Area of the inner park ¥

Length = 80 - 2 - 2 = 76 m

Breadth = 40 - 2 - 2 = 36 m

Area = 76 x 36 = 2736 m²

area of the path:

3200 - 2736 = 464 m²

cost of paving the 1 m² = rs. 5

cost of paving the path(464 m²) = rs. 5×464

= rs. 2320

hence , the cost of paving the path at 5 per m²

is rs. 2320

I hope it will helps you.

please mark this answer as brainliest.

Answer 2

Given: Length:Breadth = 2:1

Perimeter = 240 cm

Rate of paving = 5₹/m^2

To Find: Cost of paving path

STEP 1:

Let the common multiple be x

i.e, Lenght = 2x

Breadth = x

STEP 2:

Perimeter of rectangle =2(l+b)=240

=> 2(2x+x)=240

=> 6x=240

=> x=40 m => Breadth

=> 2x=80 m => Lenght

Step 2: Width=2m (in both sides)

i.e, new lenght = 80+2+2 =84 cm

new breadth = 40+2+2 =44 cm

Step 3: Total area for pavement=

=> Area(ABQP) + Area(PEHS) + Area(FQRG) + Area(SRCD)

=> 2×(44×2) + 2×(80×2)

=> 496 m^2

Step 4: Cost of pavement=496×5=2480₹

Hence, Cost of pavement is 2480₹