query: why a^2 + b^2 = c^2 in right-angled triangle
positive_passages.docid: 143775
positive_passages.text: What's the intuition behind Pythagoras' theorem?
negative_passages.docid: 150320
negative_passages.text: Why does $a^{b}·a^{c}=a^{b+c}$?
negative_passages.docid: 37744
negative_passages.text: Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$
negative_passages.docid: 92916
negative_passages.text: Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle
negative_passages.docid: 111003
negative_passages.text: Prove that if $A\triangle B = C\triangle B$, then $A = C$
negative_passages.docid: 11711
negative_passages.text: Prove trigonometry identity for $\cos A+\cos B+\cos C$
negative_passages.docid: 61580
negative_passages.text: Calculating the determinant gives $(a^2+b^2+c^2+d^2)^2$?
negative_passages.docid: 55436
negative_passages.text: Why does $A\sin{k(x+c)}=a\sin{kx}+b\cos{kx}$ imply that $A=\sqrt{a^2+b^2}$ and $\tan{c}=-b/a$?
negative_passages.docid: 87206
negative_passages.text: Prove that a triangle can be inscribed in the hyperbola $xy=c^2$ whose sides touch the parabola $y^2=4ax$.
negative_passages.docid: 69679
negative_passages.text: Geometric interpretation for sum of fourth powers
negative_passages.docid: 104427
negative_passages.text: Sines and cosines of angles in arithmetic progression
negative_passages.docid: 89038
negative_passages.text: Why does the magnitude of the cross-product of a and b give the area of a parallelogram spanned by a and b?
negative_passages.docid: 115929
negative_passages.text: In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$
negative_passages.docid: 152494
negative_passages.text: How is  $Ax + By = C$ the equation of a straight line?
negative_passages.docid: 116223
negative_passages.text: Show that the angles satisfy $x+y=z$
negative_passages.docid: 126040
negative_passages.text: Geometric explanation of $\sqrt 2 + \sqrt 3 \approx \pi$
negative_passages.docid: 158356
negative_passages.text: Need a proofreading why all the units are satisfied $a^2-2b^2 =\pm1$ for $\mathbf{Z}[\sqrt{2}]$