Respuesta :
Answer:
Length = 6, Breadth = 2
Step-by-step explanation:
Given:
A rectangle is 3 times as long as it is wide.
If the length is increased by 6 and the width is increased by 8, its area is increased by 108.
Question asked:
Find the original dimensions.
Solution:
Let width of rectangle = [tex]x[/tex]
As given that a rectangle is 3 times as long as it is wide.
Length of rectangle = [tex]3x[/tex]
[tex]Area\ of\ rectangle=length\times breadth[/tex]
[tex]=x\times3x=3x^{2}[/tex]
Now, as given that length is increased by 6 and the width is increased by 8,
New length = [tex]3x+6[/tex]
New breadth = [tex]x+8[/tex]
New area = [tex](3x+6)(x+8)[/tex]
[tex]=3x(x+8)+6(x+8)\\\\=3x^{2} +24x+6x+48\\=3x^{2} +30x+48[/tex]
As new area increased by 108, we can say:-
New area - old area = 108
[tex]3x^{2} +30x+48-(3x^{2} )=108\\3x^{2} +30x+48-3x^{2} =108\\\\30x+48=108\\[/tex]
Subtracting both sides by 48
[tex]30x+48-48=108-48\\30x=60[/tex]
Dividing both sides by 30
[tex]x=2[/tex]
Width of rectangle = [tex]x[/tex] = 2
Length of rectangle = [tex]3x[/tex] = [tex]3\times2=6[/tex]
Therefore, original length of rectangle was 6 and original width of rectangle was 2.
