Answer:
The percentage change is 140%
Step-by-step explanation:
Given
[tex]L= 3W_1[/tex] ---- initial dimension
[tex]W_2 = 6m[/tex] --- new width
[tex]A_2 = 45m^2[/tex] --- new dimension
Required
The percentage increment
The length remains constant because only the width is extended.
The new area is:
[tex]Area =Length * Width[/tex]
[tex]A_2=L * W_2[/tex]
Make L the subject
[tex]L = \frac{A_2}{ W_2}[/tex]
Substitute values for A and W
[tex]L = \frac{45m^2}{6m}[/tex]
[tex]L = \frac{45m}{6}[/tex]
[tex]L = 7.5m[/tex] --- this is the length of the garden
Calculate the initial width:
[tex]L= 3W_1[/tex]
Make W1 the subject
[tex]W_1 = \frac{1}{3} * L[/tex]
[tex]W_1 = \frac{1}{3} * 7.5[/tex]
[tex]W_1 = 2.5[/tex]
So, the initial area is:
[tex]A_1 = L_1 * W_1[/tex]
[tex]A_1 = 2.5 * 7.5[/tex]
[tex]A_1 = 18.75[/tex]
The percentage change in area is:
[tex]\%A = \frac{A_2 - A_1}{A_1}[/tex]
[tex]\%A = \frac{45 - 18.75}{18.75}[/tex]
[tex]\%A = \frac{26.25}{18.75}[/tex]
[tex]\%A = 1.4[/tex]
Express as percentage
[tex]\%A = 1.4*100\%[/tex]
[tex]\%A = 140\%[/tex]
