Right triangle

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A right triangle is a triangle in which one angle is a right angle.

The side opposite the right angle is called the hypotenuse (side [BC] in the figure below). In addition, the sides adjacent to the right angle are called legs or catheti (singular: cathetus). Side [AB] is called the side adjacent to angle α and opposed to angle β, while side [AC] is the side adjacent to angle β and opposed to angle α. Compared with angle α, side [AB] is said to be the adjacent cathetus and [AC] is the opposite cathetus.

Principal propertiesEdit

AreaEdit

As with any triangle, to calculate the area, multiply the base and the corresponding height, and divide it by two. If ABC is a right triangle in A, each of the sides [AB] and [AC] can be considered as the height; the base is then the other side of the right angle ([AC] and [AB], respectively). The area S of the triangle is equal to $S = \frac{AB\times AC}{2}$.

For example, we have a right triangle in A with [AB] = 4 cm, [AC] = 3 cm, and hypotenuse [BC] = 5 cm. Area S is calculated as $S = \frac{4 \times 3}{2} = 6$; the area of the triangle is therefore 6 cm2.

The area of the triangle could also be calculated by using the hypotenuse as the base. One would then have to calculate the height associated with the hypotenuse, as it would no longer be one of the sides.

Using trigonometry:

$A= s^2\frac{sin(\frac{360}{n})}{4}= \frac{s^2}{4csc(\frac{360}{n})}$

Pythagorean theoremEdit

The Pythagorean theorem states that:

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

This is stated in equation form as BC2 = AC2 + AB2.

This theorem is a consequence of the definition of the distance between two points from the inner square of the vector. Indeed

$BC^2 = \overrightarrow{BC}\cdot \overrightarrow{BC}=\left(\overrightarrow{BA}+ \overrightarrow{AC}\right)\cdot\left(\overrightarrow{BA}+ \overrightarrow{AC}\right)$
$BC^2=\overrightarrow{BA}\cdot\overrightarrow{BA}+2 \overrightarrow{BA}\cdot \overrightarrow{AC}+ \overrightarrow{AC}\overrightarrow{AC}=AB^2+AC^2$

since $\overrightarrow{BA}\cdot\overrightarrow{AC}=0$ as vectors $\overrightarrow{BA}$ and $\overrightarrow{AC}$ are orthogonal.

Median theoremEdit

For a right triangle, the median theorem reads:

If M is the midpoint of the hypotenuse, then AM = ½ BC. One can also say that point A is located on the circle with diameter [BC]. Conversely, if A is any point of the circle with diameter [BC] (except B or C themselves) then angle A in triangle ABC is a right angle.

There are several demonstrations of this theorem. The direct effect may be purely geometric: by definition M is the midpoint of [BC]. The triangle ABC is half of a rectangle ABCD. A rectangle is a parallelogram, so its diagonals bisect in the center; therefore, M, which is the midpoint of [BC], is also the midpoint of [AD]. The diagonals of a rectangle are of equal length, so AD = BC = AD and AM / 2 = BC / 2.

One can also use the vector:

$\overrightarrow{AM} = \overrightarrow{AB} + \frac{1}{2}\overrightarrow{BC}$ and $\overrightarrow{BC} = \overrightarrow{BA} + \overrightarrow{AC}$ where $\overrightarrow{AM} = \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC}).$

These last two vectors are orthogonal; thus, AM² = (AB² + AC²)/4.

On the other hand, applying the Pythagorean theorem to the triangle ABC yields the equation BC² = AB² + AC², and also AM = BC / 2.

One can also apply the theorem of the angle at the center, which also helps to demonstrate the converse. Consider the circle circumscribed about the triangle ABC and note its center, O. According to the theorem, the angle BOC is twice the angle BAC. Thus:

$\widehat{BOC} = 2\widehat{BAC} = \pi.$

The points B, O, and C are aligned. As BO = OC, O is the midpoint of [BC], so O = M.

Conversely, if we know that A is a point on a circle with diameter [BC], then according to the theorem of the angle at the center, angle BAC is half of angle BOC, so it is π / 2. Therefore, BAC is a right triangle in A.

Center of gravityEdit

With the previous notations (ABC with a right angle on A and M as the midpoint of [BC]), the center of gravity G verifies

$\overrightarrow{AG} = \frac{2}{3}\overrightarrow{AM}.$

M is also the projected midpoint of [AB] and [AC] (ABM and ACM are isosceles triangles). Point G is therefore predicted to be one-third of [AB] and [AC]:

$\overrightarrow{AG} = \frac{1}{3}(\overrightarrow{AB} + \overrightarrow{AC}).$

Other propertiesEdit

For any integer n greater than or equal to 3, one can always find a triangle with a length of one side of the right angle and n is the length of the other two sides are whole numbers. Indeed:

• If n is an even number, $n = 2k$.
Take the length of the other side of the right angle equal to $k^2 - 1$. The Pythagorean theorem then gives a hypotenuse of length $k^2+1$.
• If n is an odd number, $n = 2k + 1$.
Take the length of the other side of the right angle equal to $2(k^2 + k)$. The Pythagorean theorem then gives a hypotenuse of greater than or equal to $2(k^2 + k) + 1$.