Rectangle: scaling sides if area reduced by 10%

[TheJim]

Hi everyone,

Slowly losing my mind here. Logic tells me there's a simple solution that I'm just missing but I can't work it out.

My issue is I need to know the lengths of two sides of a rectangle if it's area is reduced by 10%. The lengths will remain in proportion.

eg.

5m x 3m rectangle. This gives us an area of 15m².

I then reduce the area by 10% (15*0.9) to give 13.5m².

Now I need to know what the two sides are. They will be proportional to the former side lengths, but smaller obviously.

I can't work out a way of doing this. Reducing them by 10% obviously won't work. I can't figure it out.


(Context: I'm trying to figure out how long and wide some underfloor heating mats need to be cut for our electricians to install at our customers' houses, where they need to be the exact same shape as the room but 10% smaller in area)

Any help would be greatly appreciated before I tear all of my hair out. I just need a formula that will relate the two.

Thanks very much.

 

[Deleted member 4993]

Hi everyone,

Slowly losing my mind here. Logic tells me there's a simple solution that I'm just missing but I can't work it out.

My issue is I need to know the lengths of two sides of a rectangle if it's area is reduced by 10%. The lengths will remain in proportion.

eg.

5m x 3m rectangle. This gives us an area of 15m².

I then reduce the area by 10% (15*0.9) to give 13.5m².

Now I need to know what the two sides are. They will be proportional to the former side lengths, but smaller obviously.

I can't work out a way of doing this. Reducing them by 10% obviously won't work. I can't figure it out.


(Context: I'm trying to figure out how long and wide some underfloor heating mats need to be cut for our electricians to install at our customers' houses, where they need to be the exact same shape as the room but 10% smaller in area)

Any help would be greatly appreciated before I tear all of my hair out. I just need a formula that will relate the two.

Thanks very much.

Short answer:

For the given problem the length and the width (each) must be reduced by factor of [= √(0.9) = 0.948683 =] 95% (depending on how accurate you can get in cutting (or want to get)

Proof: Let

the original rectangle have a length = L

the original rectangle have a width = W = k * L (where k remains constant after area-reduction)

original area = A = L * W = k * L²

reduced area = A_R= A * p (where p is the proportion by which the area was reduced)

the reduced rectangle have a length = L_R

the reduced rectangle have a length = W_R = k * L_R

So;

A_R= L_R * W_R = k * L_R² = A * p = p * k * L² →

L_R/L = √p ..... Answer

 

[stapel]

My issue is I need to know the lengths of two sides of a rectangle if it's area is reduced by 10%. The lengths will remain in proportion; eg, 5m x 3m rectangle, which gives us an area of 15m².

I then reduce the area by 10% (15*0.9) to give 13.5m².

Now I need to know what the two sides are. They will be proportional to the former side lengths, but smaller obviously.

So, as is customary, you want to "resize" and also "lock aspect ratio". To do this, try using the algebra you learned before starting geometry/trig:

. . . . .original length: (old L)

. . . . .original width: (old h)

What then is the formula for the original area A?

You want ninety percent of this area. What expression stands for this new area?

You want the ratio between the lengths and widths to be preserved, so you want (old L)/(old h) = (new L)/(new h).

Also, you know that the new area is equal to (new L)*(new h).

You have the values of (old A), (old L), (old h), and (new A). Solve the proportion for one of the "new" values in terms of the other. Then plug-n-chug into the equation for (new A). This will allow you to solve exactly for the one "new" value. Then back-solve for the other "new" value.

(Context: I'm trying to figure out how long and wide some underfloor heating mats need to be cut for our electricians to install at our customers' houses, where they need to be the exact same shape as the room but 10% smaller in area)

...and students wonder "when I'll need this for my job" or "for real life"!

 

[mmm4444bot]

If the room were square, the new dimension squared would need to be 9/10ths the original dimension squared.

So, I'm thinking that, for any rectangle, the square of each new dimension must be 9/10ths the square of each original dimension (respectively).

(Reduced L)^2 = 9/10 * L^2
(Reduced W)^2 = 9/10 * W^2

or

Reduced L = sqrt(9/10) * L
Reduced W = sqrt(9/10) * W

This matches Subhotosh's general result.

Using your example:

L = 5
W = 3

Reduced L = sqrt(9/10) * 5 ≈ 4.74
Reduced W = sqrt(9/10) * 9 ≈ 2.85

Original area (5*3) is 15 square units.

Reduced area (4.74*2.85) is 13.5 square units (rounded).

And 90% of 15 is 13.5 :cool:

A decimal approximation for sqrt(9/10) is 0.9486833, so you may multiply the original dimensions by this value, to get the reduced dimensions.

 

[mmm4444bot]

To confirm my hunch, I worked symbolically. (The following is likely similar to what stapel posted.)

To simplify the notation, I'll use these symbols:

a = original length
b = original width

c = reduced length
d = reduced width

Then we have:

0.9 ab = cd [1]

and the proportion (a/b = c/d) gives:

ad = bc [2]

Solving [1] for c, we get:

c = 0.9 ab/d

Solving [2] for c, we get:

c = a/b * d

Therefore:

0.9 ab/d = a/b * d [3]

Multiplying each side of [3] by d, we get:

0.9 ab = a/b * d^2 [4]

Multiplying each side of [4] by b/a, we get:

d^2 = 0.9 * b^2

Taking the square root of each side, we get:

d = sqrt(0.9) * b

In other words, the reduced width is 0.9486833 times the original width.

A similar solve gives the same relationship for the lengths.

 

[Deleted member 4993]

Another approach...

r = percentage reduction (given)
larger *triangle*: u=width, v = length (those are givens)
smaller rectangle: w=width, x=length

Since same ratio:
w / x = u / v : x = vw / u

area smaller rectangle: uv(1 - r)

So:
w(vw / u) = uv(1 - r) → w² = u²(1-r) → w = u * √(1-r)

Solve for w, then x

That darn "easy looking" problem is sure
not as "easy" as it appears...right, Mark?!

Where, how, why are you getting triangle here?

Assuming you meant triangle, I think you are making it complicated (completely against your nature). It is as easy as it appears!! According to your convention, the result should be:

w = u * √(1-r)

and

x = v * √(1-r)

And it is..... as shown in post 2 (with different variables)

 

[Deleted member 4993]

[1]: was still on 1st coffee; meant rectangle, of course!

[2]: agree : RED star for you

I want GOLD star !!!!

 

[TheJim]

Hi guys,

Thanks for all of that!

While waiting for the responses I actually worked it out for myself as I realised where I'd gone wrong. Twice.

I took the example of a 5m x 3m room and tried to solve using quadratics (w = x, L = x+2) because I'm an idiot. Not only was this a stupid way of going about doing it, I also gave myself a whole bunch of really time consuming and difficult quadratic equations. Of course I ended up with the correct square area when checking for proof, but I realised that logically speaking the length of the sides must be wrong.

So then I went back and did it in a super simplified way using fractions.

If the original rectangle has length L and width W and area A, and the new one has length l, width w and area a, then:

lw = a

l = (L/W)w

So then I just substituted in the known values. Sticking with the original example, that would b L=5, W=3, A=15

So

A * 0.9 = 13.5

lw = 13.5

l = (5/3)w

w * (5/3)w = 13.5

x² = 13.5/(5/3)

x² = 8.1

x = 2.846

y = 5/3 * 2.846

y = 4.743

4.743 * 2.846 = 13.5


I tried it with some other examples and it seems to work, but it's interesting to see other methods too!

Thanks very much everyone. My job involves a lot of maths and while competent, I do sometimes get stuck and my brain can't do the lateral thinking required, so you may be seeing more of me on here!

Cheers

Jim