Tanya, you like triangles.
These are interesting
a = sqrt(c^2-b^2);
this isn't a calculation!
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.
The Goldbach Conjecture, Part 2
Try 1.5.4: Celsius to Fahrenheit
Back to basics - mean of corner elements of a matrix
Compare two input matrices
Area of a circle
Area of an equilateral triangle
Is this triangle right-angled?
Dimensions of a rectangle
Find a Pythagorean triple
Area of an Isoceles Triangle
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