SOLUTION: two squares are joined together. the bigger one has an area of 75m² and the smaller one has an area of 27m². find the combined perimter of them. express answer in radical form

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Question 1201827: two squares are joined together. the bigger one has an area of 75m² and the smaller one has an area of 27m². find the combined perimter of them. express answer in radical form
Found 3 solutions by mananth, greenestamps, math_tutor2020:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
Bigger square Area = 75cm^2
Side ^2 = area
side =
=
Similarly small square side =
Observe fig and get perimeter of the fig

3*3sqrt(3)+ 3*5sqrt(3) +2sqrt(3)
Add it up



Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The statement of the problem is deficient; it is not possible to come up with a definitive answer.

Given the areas of the two squares, we know the side lengths of the two squares are 5*sqrt(3) and 3*sqrt(3).

The trouble is that "joined together" is not well defined. It makes sense to assume that part of one edge of one square is part of one edge of the other square; but that can happen in different ways that give different combined perimeters.

The other tutor showed the two squares joined like the following figure:
+----------+
|          |
+          +------+
|          |      |
|          |      |
+----------+------+

That tutor then showed how to find the combined perimeter by finding the length of each piece of the perimeter:

5*sqrt(3)+5*sqrt(3)+5*sqrt(3)+2*sqrt(3)+3*sqrt(3)+3*sqrt(3)+3*sqrt(3) = 26*sqrt(3)

Note that the calculation is easier by noting that the perimeter of the figure above is the same as the perimeter of this figure:
+----------+------+
|          |      |
+          +------+
|          |      |
|          |      |
+----------+------+

The perimeter of that figure is simply

5*sqrt(3)+8*sqrt(3)+5*sqrt(3)+8*sqrt(3) = 26*sqrt(3)

And the perimeter of the combined figure is always 26*sqrt(3), AS LONG AS the edge of the smaller square does not extend past the edge of the larger square.

But the following figure -- in which the two squares are still "joined together" has a perimeter greater than 26*sqrt(3):
+----------+
|          |
|          |
|          +------+
|          |      |
+----------+      |
           +------+

So in order to come to a single answer to the problem, we need to make some assumptions -- and that makes the problem not well defined.
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

I agree with the tutor @greenestamps
The problem seems to lack instructions as to exactly how the squares are connected.

Here is an interactive GeoGebra diagram where you can slide the smaller square up and down.
Move the red point to do so.
https://www.geogebra.org/m/krtatgey
Click the curved arrow if the diagram doesn't show up. That should refresh the page.

The GeoGebra app will compute the perimeter to give approximate values as a result.
The diagram is to scale.

If the smaller square is fully attached to the larger square, i.e. no "spillover", then the perimeter is as the other tutors mention (the tutor @mananth doesn't mention it directly, but you can arrive at this figure after adding up the terms they wrote).

However, if some piece of the smaller square is not attached to the larger, then the perimeter will exceed
It will be within the interval of

Here's where I'm getting the
First compute the perimeter of each square separately
perimeter of larger square =
perimeter of smaller square =
Then add up the perimeters
This is the largest possible perimeter of the complex shape.
If we slide the red point up as far as you can go, we'll have two squares as separate "islands" so to speak, and it should help visualize why I just added the perimeters.

Also, the red shared wall is subtracted twice from that total sum to arrive at the perimeter of the complex shape.
This because this shared wall is included in the calculation of perimeterA+perimeterB, but clearly the shared wall is interior (not exterior). We subtract off 2 copies because there are 2 squares.

Footnotes:





Each decimal value is approximate.
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