a field in the shape of a rhombus have the distance between pairs of opposite vertices are as 14 M and 48 M what is the cost of fencing the field at rupees 20 per metre
Answers
Answered by
2
Length of one side=√{(48÷2)^2+(14×2)^2} M
=√{576+49} M
=√625 M
=25 M
Total perimeter of the field=25×4 M =100 M
Therefore total cost of fencing will be=(100×20)/- =2000/-
=√{576+49} M
=√625 M
=25 M
Total perimeter of the field=25×4 M =100 M
Therefore total cost of fencing will be=(100×20)/- =2000/-
Similar questions