Well this is an easy question. Take out all the factors of 150, so that they give an area of 150 on multiplication.
Length X width = Area
10 x 15= 150
25 x 6 = 150
50 x 3 = 150
30 x 5 = 150
75 x 2 = 150
Now chose the length and width amongst the options above that will result on the least amount of fencing.
2(10 + 15)= 50 {10 is the length and 15 is the width, both are multiplied into two as there are two lengths and two widths in a rectangle}
Like wise;
2(25 + 6) = 62
2(50 + 3) = 300
2(30 + 5) = 70
2(75 + 2) = 154
The least amount of fencing will occur if the rectangular field has the dimensions 10 and 15.
Length X width = Area
10 x 15= 150
25 x 6 = 150
50 x 3 = 150
30 x 5 = 150
75 x 2 = 150
Now chose the length and width amongst the options above that will result on the least amount of fencing.
2(10 + 15)= 50 {10 is the length and 15 is the width, both are multiplied into two as there are two lengths and two widths in a rectangle}
Like wise;
2(25 + 6) = 62
2(50 + 3) = 300
2(30 + 5) = 70
2(75 + 2) = 154
The least amount of fencing will occur if the rectangular field has the dimensions 10 and 15.
