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Given a triangle with sides of length a, b, and c, if a 2 + b 2 = c 2, then the angle between sides a and b is a right angle. For any three positive real numbers a , b , and c such that a 2 + b 2 = c 2 , there exists a triangle with sides a , b and c as a consequence of the converse of the triangle inequality .
The name is derived from the Pythagorean theorem, stating that every right triangle has side lengths satisfying the formula + =; thus, Pythagorean triples describe the three integer side lengths of a right triangle. However, right triangles with non-integer sides do not form Pythagorean triples. For instance, the triangle with sides = = and ...
The area of a triangle can be demonstrated, for example by means of the congruence of triangles, as half of the area of a parallelogram that has the same base length and height. A graphic derivation of the formula T = h 2 b {\displaystyle T={\frac {h}{2}}b} that avoids the usual procedure of doubling the area of the triangle and then halving it.
The area of a rational-sided right triangle is called a congruent number, so no congruent number can be square. A right triangle and a square with equal areas cannot have all sides commensurate with each other. There do not exist two integer-sided right triangles in which the two legs of one triangle are the leg and hypotenuse of the other ...
A right triangle ABC with its right angle at C, hypotenuse c, and legs a and b,. A right triangle or right-angled triangle, sometimes called an orthogonal triangle or rectangular triangle, is a triangle in which two sides are perpendicular, forming a right angle (1 ⁄ 4 turn or 90 degrees).
In this example, the triangle's side lengths and area are integers, making it a Heronian triangle. However, Heron's formula works equally well when the side lengths are real numbers. As long as they obey the strict triangle inequality, they define a triangle in the Euclidean plane whose area is a positive real number.
in pink, the areas a 2, b 2 on the left and the areas 2ab cos γ and c 2 on the right; in blue, the triangle ABC, on the left and on the right; in grey, auxiliary triangles, all congruent to ABC, an equal number (namely 2) both on the left and on the right. The equality of areas on the left and on the right gives
The apparent triangles formed from the figures are 13 units wide and 5 units tall, so it appears that the area should be S = 13×5 / 2 = 32.5 units. However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent.
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