ACDBEF if
AL = 80 m, AM = 100 m, AN = 160 m, AB=240 m
CL = 50 m, DN = 60 m, AF = 30 m, EM = 40 m​

300L is the right answer of this question

chu***

hey mate here your answer

Step-by-step explanation:

Length of inner field=100-4=96m

•Breadth of inner field=60-4=56m

•Area of outer field=100×60=6000m^2

•Area of inner field=96×56=5376m^2

•Now, Area of the boundary=6000-5376=624m^2

•So, the cost of gravelling=624×30=Rs18720.

Hope! it helped you!

Mark me as brainliest

hey mate your answer here

Step-by-step explanation:

Length of inner field=100-4=96m

•Breadth of inner field=60-4=56m

•Area of outer field=100×60=6000m^2

•Area of inner field=96×56=5376m^2

•Now, Area of the boundary=6000-5376=624m^2

•So, the cost of gravelling=624×30=Rs18720.

Hope! it helped you!

mark me as brainliest

I don't solve this problem

please follow me i followed you

write your question properly

★ Area of Path = 624 m² ★

★ Cost of gravelling path = Rs 18,720★

Step-by-step explanation:

Given:

To Find:

Solution: Let ABCD be a rectangular park .

★ Area of ABCD = Length x Breadth ★

Area of ABCD = AB X BC

Area of ABCD = 100 x 60

Area of ABCD = 6000m²

★ In rectangle EFGH ★

∴ Area of EFGH = EF x FG

Area of EFGH = 96 x 56

Area of EFGH = 5376m²

Now we have to find the area of path:

Area of Path = ( 6000–5376)m²

Area of Path = 624m²

Hence, Area of Path is 624m²

† Cost of gravelling 1 m² = Rs 30 †

∴ Cost of gravelling 624m² = 624 x 30

→ Rs 18,720

Hence, Cost of gravelling is Rs 18,720