Which parallelogram will have the smallest area?
2 answers:
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Answer:
41°
Step-by-step explanation:
I could not think of an easy way to solve this, apart from having a graphing calculator do it. In the end, I found I could solve it analytically using a combination of the law of sines and the law of cosines.
Let x represent the length of the shortest side, and θ the smallest angle. Then the <em>law of sines</em> tells you ...
sin(θ)/x = sin(2θ)/(x+2)
Cross-multiplying and using the trig identity for sin(2θ), we have ...
(x +2)sin(θ) = 2x·sin(θ)cos(θ)
Dividing out sin(θ), we see that ...
cos(θ) = (x+2)/(2x)
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The law of cosines for the shortest side and smallest angle tells you ...
x^2 = (x+1)^2 + (x+2)^2 - 2(x+1)(x+2)·cos(θ)
Substituting the above expression for cos(θ), this can be rewritten as ...
0 = (x^2 +2x +1) +(x^2 +4x +4) -x^2 -(x+1)(x+2)^2/x
0 = x^2 +6x +5 -(x+1)(x+2)^2/x . . . . . . collect terms outside the fraction
0 = x(x+5)(x+1) -(x+1)(x+2)^2 . . . . . . . . factor and multiply by x
We know that x=-1 is not a solution, so we can divide by that factor:
0 = x^2 +5x -(x^2 +4x +4) . . . . . multiply it all out
0 = x -4 . . . . . . . . . . . . . . . . . . . . collect terms
4 = x
so, cos(θ) = (4+2)/(2·4) = 6/8 = 3/4
and the angle of interest is ...
θ = arccos(3/4) ≈ 41.40962° ≈ 41°
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The attachment shows a triangle-solver's result using the consecutive integers for side lengths. It confirms the answer we have here.
