Tanya, you like triangles.
a = sqrt(c^2-b^2);
test_1 a=sqrt(3) is longer than b=1
the good solution would be
a= min(sqrt(c^2 -b^2),b)
Kudos for finding this solution - it did indeed meet all the test cases, but wasn't quite what I had in mind! I will add a new test.
Back to basics 6 - Column Vector
Determine Whether an array is empty
Compute a dot product of two vectors x and y
Return 'on' or 'off'
ベクトル [1 2 3 4 5 6 7 8 9 10] の作成
Is this triangle right-angled?
Area of an equilateral triangle
Area of an Isoceles Triangle
Dimensions of a rectangle
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