Three equal squares are superimposed as indicated in the figure. The areas of the overlapping zones are 2 cm², 5 cm², and 8 cm², respectively. The total area of the non-overlapping area of the squares is 117 cm². What is the length of the side of each square?

  The square side lengths are  1 cm, 5 cm, and 11 cm.

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Explanation:

Refer to the diagram below.

Define the following areas

It doesn't really matter how you set up the labels, or what letters you pick, as long as you stay consistent.

We're told that "The areas of the overlapping zones are 2 cm², 5 cm², and 8 cm²", so let's make

Furthermore, we are told that "The total area of the non-overlapping area of the squares is 117 cm²" which means A+B+C = 117

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Refer to the diagram. For now, focus on the upper right square. It has regions A, E, and F. The area of this square is A+E+F

Similarly, the area of the left-most square is D+B+E

The area of the bottom-most square is C+F+D

Add up the expressions we found and we get

(A+E+F)+(D+B+E)+(C+F+D) = A+B+C+2D+2E+2F

This represents the sum of all three square areas even if we peeled the squares apart to not overlap.

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Let's plug in A+B+C = 117

We get

A+B+C+2D+2E+2F

117+2D+2E+2F

Now plug in D = 2, E = 5, and F = 8

117+2(2)+2(5)+2(8)

117+4+10+16

117+30

147

The sum of the three square areas is 147 square cm.

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Now we have to do a bit of trial and error. We're looking for ways to break up 147 such that it is a sum of three squares.

In other words, we want to find integers m,n,p such that m^2+n^2+p^2 = 147

Let's try subtracting off 121

147-121 = 26

I picked 121 since it's the largest perfect square just smaller than 147

Repeat this process, but do so for 25 since 25 is smaller than 26

26-25 = 1

We can see that

1 + 25 + 121 = 147

1^2 + 5^2 + 11^2 = 147

One solution is m = 1, n = 5, p = 11. There may be other solutions. Keep in mind that m,n,p are positive whole numbers.

So we got a bit lucky with our trial and error process. If we didn't find such a triple, we'd have to try other possible combos of squares until we get the right match.

Because 1^2 + 5^2 + 11^2 = 147, this means the squares' side lengths are 1, 5, and 11.

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step-by-step explanation:

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step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v