Since polygon ABCD is similar to polygon WXYZ, the corresponding sides are proportional.
That means:
AB/WX = BC/XY = CD/YZ = AD/WZ
We can use this fact to set up the following equations:
AB/WX = 13/24
CD/YZ = 12/15 = 4/5
AD/WZ = 10/m
We are given that AB = 13 and WX = 24, so we can substitute those values in the first equation:
13/24 = BC/XY
We are also given that CD = 12 and YZ = 15, so we can substitute those values in the second equation:
4/5 = BC/XY
Since both equations equal BC/XY, we can set them equal to each other:
13/24 = 4/5
To solve for m, we can use the third equation:
10/m = AD/WZ
We know that AD = AB + BC = 13 + BC, and WZ = WX + XY = 24 + XY. Since BC/XY is the same in both polygons, we can use the results from our previous equations to find that BC/XY = 4/5.
So we have:
AD/WZ = (13 + BC)/(24 + XY) = (13 + (4/5)XY)/(24 + XY) = 10/m
Now we can solve for XY:
13 + (4/5)XY = (10/m)(24 + XY)
Multiplying both sides by m(24 + XY), we get:
13m(24 + XY)/5 + mXY(24 + XY) = 10(13m + 10XY)
Expanding and simplifying, we get:
312m/5 + 13mXY/5 + mXY^2 = 130m + 100XY
Rearranging and simplifying further, we get:
mXY^2 - 87mXY + 650m - 1560 = 0
We can use the quadratic formula to solve for XY:
XY = [87m ± sqrt((87m)^2 - 4(650m - 1560)m)] / 2m
Simplifying under the square root:
XY = [87m ± sqrt(7569m^2 - 2600m)] / 2m
XY = [87m ± sqrt(529m^2)] / 2m
XY = (87 ± 23m) / 2
Since XY must be positive, we can use the positive solution:
XY = (87 + 23m) / 2
Now we can substitute this value for XY in the equation we derived earlier:
13 + (4/5)XY = (10/m)(24 + XY)
13 + (4/5)((87 + 23m) / 2)= (10/m)(24 + (87 + 23m) / 2)
Multiplying both sides by 10m, we get:
130m + 52(87 + 23m) / 10 = (240 + 87m) / 2
Simplifying and solving for m, we get:
1300m + 52(87 + 23m) = 240 + 87m
1300m + 4524 + 1196m = 240 + 87m
2403m = -4284
m = -4284 / 2403
m ≈ -1.78
Therefore, the value of m is approximately -1.78.