Let the width of the rectangle be 'w'
According to the question length of the rectangle (l) is 4 more than 3 times it's width;
[tex] \longrightarrow [/tex] l = 3w + 4
Perimeter of rectangle = 152
Formula of perimeter of rectangle = 2(Length + Width)
So,
[tex]\rm \implies 2(Length + Width) = 152 \\ \\ \rm \implies 2(3w + 4+ w) = 152 \\ \\ \rm \implies 2(4w + 4) = 152 \\ \\ \rm \implies 2 \times 4(w + 1) = 152 \\ \\ \rm \implies 8(w + 1) = 152 \\ \\ \rm \implies w + 1 = \dfrac{152}{8} \\ \\ \rm \implies w + 1 = 19 \\ \\ \rm \implies w = 19 - 1 \\ \\ \rm \implies w = 18[/tex]
[tex] \therefore [/tex] Width of the rectangle (w) = 18
Length of rectangle (l) = 3 × 18 + 4 = 58
