Question

a rectangle is 3 times as long as it is wide. if length is increased by 6 and the width by 8, the area is increased by 108cm. what are the original dimensions

Answer 1

The area of a rectangle is the length * width, or A = L*W. You also know that the length is 3*width. So, you can write one equation:

L = 3*W

You know that if you increase the length by 6 cm and the width by 8 cm, the area increases by 108 cm^2. Can you write a formula for the area of the increased size rectangle? If you let that be A2, then A2 - A = 108 cm^2.

Answer 2

plaese do a equation i dont get what you are saying

Answer 3

Let length=L, Width=W,
"3 times as long as it is wide" -->
3W=L
"if length is increased by 6 and the width by 8, the area is increased by 108cm"-->
(L+6)(W+8)=A+108
And we know this already:
A(Area)=W*L, so

3W=L, A=W*L --> A=3(W^2)

(L+6)(W+8)=A+108
(3W+6)(W+8)=3(W^2)+108
3(W^2)+30W+48=3(W^2)+108
30W=60
W=2
L=3W=6
so,
A(Area)=W*L=12

Answer 4

Always a good idea to check the solution to make sure it is correct. We have W = 2 and L = 6. The area of the original rectangle is therefore 2*6 = 12. The problem said that if we add 6 to the length and 8 to the width, the area is increased by 108. The new rectangle would have a length of 6+6 = 12 and a width of 2+8=10, giving it an area of 12*10 = 120. If we subtract the area of the original rectangle, we find that the increased area is 120 - 12 = 108, just as the problem stated. Our answer is correct.