Answer:
Let the length and breadth of the plot be \[11x\,m\] and \[4x\,m\] respectively. Then, perimeter of the plot \[=2\times \] (Length + Breadth) \[=2\times (11x+4x)m\] \[=2\times (15x)m\] \[=30x\,m\] \[\therefore \] Cost of fencing the plot \[=Rs.\,(30x)\times 100\] \[=Rs.\,\,3000\,x\] According to the question, \[3000x=75000\] \[\Rightarrow \] \[x=\frac{75000}{3000}=25\] | Dividing both sides by 3000 \[\therefore \,\,\] Length of the plot \[=11x=11\times 25=275\,m\] and, breadth of the plot \[=4x=4\times 25=100m\] Hence, the dimensions of the plot are 275 m and 100 m. Check: \[275:\,100=\frac{275}{100}\] \[=\frac{275\div 25}{100\div 25}\] \[=\frac{11}{4}\] = 11 : 4 \[2(275+100)\times 100=2(375)\times 100\] \[=750\times 100\] = 75000 Hence, the result is verified.
You need to login to perform this action.
You will be redirected in
3 sec