Question

if the length of a rectangle is increased by 30% and the width of the same rectangle is decreased by 30% what is the effect on the area of the rectangle?

It is increased by 60%
it is unchanged
it is decreased by 15%
it is decreased by 9%

Answer 1

@Pixel

Answer 2

Im thinking its c so far or b

Answer 3

@Shadow

Answer 4

@Hero

Answer 5

@RealLifeJuul for questions like these, it is useful to create a real world scenario, i.e. create the conditions given in the problem "A rectangle of length \(a\)and width \(b\). Then increase \(a\) by \(30\%\) and decrease \( b\) by \(30\%\) as the problem suggests and then assess the area of the rectangle under these conditions compared to the original rectangle.

Answer 6

oh ok , so if im applying it the way you say to. Then Wouldnt the area of the rectangle remain th same, or am I wrong in assuming that if you increase the length and decrease the width by the same amount technically the rectangle would change at all regarding the area

Answer 7

??

Answer 8

Create the conditions stated in the problem, then assess. Otherwise you're just guessing.

Answer 9

ok so I made a rectangle with the measurements of 30L and 20W, and went on to multiply those by .30 and added 9 because .30 x 30 = 9 I just added that onto the L and then did the same to the width but this time subtracted and got 14W so my new rectangle taken into account the measurements from before has a new area of 546 and the old area is 600. So now what am I supposed to do here, btw sorry for the explanation being all over the place.

Answer 10

if I multiply my old area by the answer given and compare it to the new area do you think I can find my answer?

Answer 11

To decrease \(a\) by \(30\%\) you compute \(a - 0.30a\).

Answer 12

I think I got it what I did was 30a x .30 +9 got the L and then I did 20b x .30 -6 and got 39a and 14b
I went on to get the area of both subtracted 600 - 546 and then divided 54 over 600 and got a 9% decrease is this right?

Answer 13

Nope that's not it either. You're confusing the operations.

Answer 14

@RealLifeJuul are you still here?

Answer 15

yee

Answer 16

Suppose the length of a rectangle is \(l = 12\) units and the width is \(w = 6\) units. The area of the rectangle is \(A = 72\) square units If we increase the length by \(30\%\) and decrease the width by \(30\%\) then the length and width of the altered rectangle is
\( l = 12 + .30(12) = 15.6\) units and \(w = 6 - .30(6) = 4.2\) units The area of the altered rectangle is \(A = (15.6)(4.2) = 65.52\) square units. Notice that \(0.09(72) = 65.52\). Which means that the Area of the altered rectangle has decreased by \(9\%\) of the original.

Answer 17

Yeah I see that, so it's not unchanged it's decreased by 9% which I arrived at to with my rectangle.

Answer 18

or did I do something wrong in my equation?

Answer 19

Try posting your work again. I did not follow what you wrote the first time and the operations seemed off.

Answer 20

This time focus on typing the operations you computed correctly.

Answer 21

Ok, so first I got my original rectangle with L being 30 and W being 20 then I multiplied them both by .30 so .30 x 30 = 9 and then did 30+9=39 and got the new Length for the width I did the same thing except I subtracted so .30 x 20 = 6 and then 20-6=14 then I went and got the new area of the altered rectangle that being so LxW 39 x 14 = 546 I also had the area of the unaltered rectangle too. So I have unaltered rectangle A=600 and altered rectangle A= 546. I then went on to subtract the two and then divide with the old rectangles area ontop. so 600-546= 54 and then divided by 54/600 and got .9.

Answer 22

Correct.

Answer 23

Well, except, you should have gotten \(0.09\) which is \(9\%\)

Answer 24

Hey thanks for all the help :D

Answer 25

You're welcome