Abc Is A Right-Angled Triangle Calculate The Length Of Ab? The 49 Latest Answer

Lesson #3: Triangles

By Rebecca Adcock

right triangles

A right triangle is any triangle that has an angle equal to 90 degrees. In the triangle below, the right angle is BCA. The longest side of a right triangle is called the hypotenuse and is the side opposite the right angle. AB (also called c) is the hypotenuse. BC (also called a) and AC (also called b) are legs.

Pythagorean theorem

You have already encountered the Pythagorean theorem somewhere in your mathematics studies. Recall that the Pythagorean theorem states

“In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the leg lengths.”

If you don’t remember these words exactly, you probably remember the following:

Special right triangles. 30-60-90 triangle

We create a 30-60-90 triangle by bisecting one angle of an equilateral triangle. The two resulting triangles are congruent, but we’ll focus on just one of them. In triangle ABC (left), all sides are equal. AB=BC=AC. BD bisects angle ABC, so BD also bisects side AC. So AD=DC and AD=1/2(AB).

According to the Pythagorean theorem,

Let the length of side AB be 1. So BC=1, AC=1 and AD=1/2.

Special right triangles. 45-45-90 triangle

Check your understanding. See Lesson #3 in Lesson Assessments.

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You’re often asked in math and trigonometry how to find the length of a triangle. Maybe you need to find the missing side of a right triangle, maybe you know both side b and side c, or maybe you just know the opposite angle of the length of a side you’re trying to find. In any case, we have formulas that will help.

Case #1: When you know the area of ​​a triangle

If you know the area of ​​a triangle and either the base or the height, you can easily find the length using the area formula:

Let’s use the formula to find the base of a triangle with area 20 and height 5:

This also works for equilateral triangles and isosceles triangles!

Case #2: When you find the length of a right triangle

To find the hypotenuse of a right triangle, use the Pythagorean theorem. Start with the two known sides and use the famous formula of the Greek mathematician Pythagoras, which states that the sum of the squares of the sides is equal to the square of the length of the third side:

As an example, find the length of the third side for a triangle with two other side lengths of 5 and 12:

From there, you square the sides of the triangle, add them up, and compare them to the square root (sometimes abbreviated as sqrt) of the unknown side. The best? This works for all triangles that have right angles. Just remember that c always refers to the hypotenuse, or the longest side of the triangle.

Case #3: When you use the Law of Sines and the Law of Cosines

The Law of Sines states that for all angles in a triangle, the ratio of the sine of that angle to its opposite side is always the same.

Here is an example of the Law of Sines in action:

The length of side c is 2.98.

The law of cosines states that you can determine the length of each side of a triangle if you know the angle opposite and the lengths of the other two sides.

Here is an example of the law of cosines in action:

The best formula to find the length of a triangle

It all depends on what information you start with. They often know one or two sides of a triangle, missing angles, or other clues.

Check your formulas like the area formula, the Pythagorean theorem and the law of sines and the law of cosines, and you’ll be good to find the length of any triangle!

More math homework help

Perpendicular line segment from the side of a triangle to the opposite vertex

“Orthocenter” and “Orthocenter” redirect here. For the orthocentric system, see Orthocentric system

The three altitudes of a triangle intersect at the orthocenter, which is inside the triangle for an acute triangle.

In geometry, an altitude of a triangle is a line segment through a vertex and perpendicular to (i.e., forming a right angle) a line containing the base (the side opposite the vertex). This line containing the opposite side is called the extended base of elevation. The intersection of the extended base and the elevation is called the foot of elevation. The length of the height, often referred to simply as “the height”, is the distance between the extended base and the apex. The process of drawing the elevation from the apex to the root is called lowering the elevation at that apex. It is a special case of the orthogonal projection.

Heights can be used to calculate the area of ​​a triangle: half the product of the length of a height and the length of its base is equal to the area of ​​the triangle. So the longest height is perpendicular to the shortest side of the triangle. The altitudes are also related to the sides of the triangle through the trigonometric functions.

In an isosceles triangle (a triangle with two congruent sides), the altitude with the incongruent side as its base has the midpoint of that side as its foot. Also the height with the incongruent side as base is the bisector of the vertex angle.

It is customary to mark the elevation with the letter h (as in elevation), often prefixed with the name of the side on which the elevation is drawn.

(p + q, r, s ) , (r, p, h ) and (s, h, q ) ,

( p + q ) 2 = r 2 + s 2 p 2 + 2 p q + q 2 = p 2 + h 2 ⏞ + h 2 + q 2 ⏞ 2 p q = 2 h 2 ∴ h = p q {\displaystyle {\begin {aligned}(p+q)^{2}\;\;&=\quad r^{2}\;\;\,+\quad s^{2}\\p^{2}\!\! +\!2pq\!+\!q^{2}&=\overbrace {p^{2}\!\!+\!h^{2}} +\overbrace {h^{2}\!\! +\!q^{2}} \\2pq\quad \;\;\;&=2h^{2}\;\;\therefore h\!=\!{\sqrt {pq}}\\\end{aligned }}} The height of a right triangle from its right angle to its hypotenuse is the geometric mean of the lengths of the segments into which the hypotenuse is divided. Application of the Pythagorean theorem to the 3 triangles of sides and

In a right triangle, the elevation drawn to the hypotenuse c divides the hypotenuse into two segments of lengths p and q. If we denote the length of the height by h c, then we have the relationship

h c = p q {\displaystyle h_{c}={\sqrt {pq}}} geometric mean theorem)

In a right triangle, the height of each acute angle coincides with one leg and intersects the opposite side at the right-angled vertex (has its foot on) the orthocenter.

The altitudes of each of the acute angles of an obtuse triangle lie entirely outside the triangle, as does the orthocenter H.

For acute triangles, the feet of the altitudes all fall on the sides of the triangle (not extended). In an obtuse triangle (one with an obtuse angle), the foot of elevation to the obtuse vertex falls on the interior of the opposite side, but the feet of elevation to the acute vertex fall on the opposite extended side, outside of the triangle. This is illustrated in the adjacent diagram: In this obtuse triangle, a height falling perpendicularly from the apex, which has an acute angle, intersects the extended horizontal side outside the triangle.

Orthocenter[ edit ]

Three heights that intersect at the orthocenter

The three (possibly extended) altitudes intersect at a single point called the orthocenter of the triangle, usually denoted H. [1] [2] The orthocenter is within the triangle if and only if the triangle is acute (i.e. has no angle greater than or equal to a right angle). When an angle is a right angle, the orthocenter coincides with the vertex at the right angle.[2]

Let A, B, C be the corners and also the angles of the triangle, and let a = |BC|, b = |CA|, c = |AB| be the side lengths. The orthocenter has trilinear coordinates[3]

Sek ⁡ A : Sek ⁡ B : Sek ⁡ C = Kos ⁡ A – Sin ⁡ B Sin ⁡ C : Kos ⁡ B – Sin ⁡ C Sin ⁡ A : Kos ⁡ C – Sin ⁡ A Sin ⁡ B , {\displaystyle \sec A:\sec B:\sec C=\cos A-\sin B\sin C:\cos B-\sin C\sin A:\cos C-\sin A\sin B,}

and barycentric coordinates

( a 2 + b 2 − c 2 ) ( a 2 – b 2 + c 2 ) : ( a 2 + b 2 – c 2 ) ( – a 2 + b 2 + c 2 ) : ( a 2 – b 2 + c 2 ) ( − a 2 + b 2 + c 2 ) {\displaystyle \displaystyle (a^{2}+b^{2}-c^{2})(a^{2}-b^{2} +c^{2}):(a^{2}+b^{2}-c^{2})(-a^{2}+b^{2}+c^{2}):(a ^{2}-b^{2}+c^{2})(-a^{2}+b^{2}+c^{2})} = tan ⁡ A : tan ⁡ B : tan ⁡ C . {\displaystyle =\tan A:\tan B:\tan C.}

Since the barycentric coordinates for a point inside a triangle are all positive, but at least one is negative for a point outside, and two of the barycentric coordinates for a vertex are zero, the barycentric coordinates given for the orthocenter show the orthocenter is inside of an acute triangle, on the right apex of a right triangle, and outside of an obtuse triangle.

Let the points A, B and C in the complex plane be the numbers z A {\displaystyle z_{A}} , z B {\displaystyle z_{B}} and z C {\displaystyle z_{C} } and assume that the circumcenter of triangle ABC lies at the origin of the plane. Then the complex number

z H = z A + z B + z C {\displaystyle z_{H}=z_{A}+z_{B}+z_{C}}

is represented by the point H, namely the orthocenter of the triangle ABC.[4] From this, the following characterizations of the orthocenter H can be derived without further ado using free vectors:

OH → = ∑ c y c l i c O A → , 2 ⋅ H O → = ∑ c y c l i c H A → . {\displaystyle {\vec {OH}}=\sum \limits _{\scriptstyle {\rm {cyclic}}}{\vec {OA}},\qquad 2\cdot {\vec {HO}}=\sum \limits _{\scriptstyle {\rm {cyclic}}}{\vec {HA}}.}

The first of the previous vector identities is also known as the Sylvester problem proposed by James Joseph Sylvester.[5]

Properties[edit]

Let D, E, and F denote the feet of the altitudes of A, B, and C, respectively. Then:

The product of the lengths of the segments into which the orthocenter divides a height is the same for all three heights:[6][7]

A H ⋅ H D = B H ⋅ HE = C H ⋅ H F . {\displaystyle AH\cdot HD=BH\cdot HE=CH\cdot HF.} The circle centered at H with the radius of the square root of this constant is the arctic circle of the triangle.[8]

The sum of the ratios on the three heights of the distance of the orthocenter from the base to the length of the height is 1:[9] (This property and the next are applications of a more general property of any interior point and the three Cevians throughout.)

H D A D + H E B E + H F C F = 1. {\displaystyle {\frac {HD}{AD}}+{\frac {HE}{BE}}+{\frac {HF}{CF}}=1.}

The sum of the ratios on the three heights of the distance of the orthocenter from the vertex to the length of the height is 2:[9]

A H A D + B H B E + C H C F = 2. {\displaystyle {\frac {AH}{AD}}+{\frac {BH}{BE}}+{\frac {CH}{CF}}=2.}

The isogonal conjugate of the orthocenter is the circumcenter of the triangle.[10]

Four points in the plane, one of which is the orthocenter of the triangle formed by the other three, is called an orthocentric system or quadrilateral.

Relation to circles and conics[ edit ]

Denote the circumradius of the triangle by R. Then[12][13]

a2 + b2 + c2 + AH2 + BH2 + CH2 = 12R2 . {\displaystyle a^{2}+b^{2}+c^{2}+AH^{2}+BH^{2}+CH^{2}=12R^{2}.}

Furthermore, if r is denoted as the radius of the triangle’s incircle, r a , r b and r c as the radii of its excircles, and R in turn as the radius of its circumcircle, the following relationships hold with respect to the distances of the orthocenter from the vertices: [14]

r a + r b + r c + r = A H + B H + C H + 2 R , {\displaystyle r_{a}+r_{b}+r_{c}+r=AH+BH+CH+2R,}r a 2 + r b 2 + r c 2 + r 2 = AH 2 + BH 2 + CH 2 + ( 2 R ) 2 . {\displaystyle r_{a}^{2}+r_{b}^{2}+r_{c}^{2}+r^{2}=AH^{2}+BH^{2}+CH^ {2}+(2R)^{2}.}

If an arbitrary altitude, say AD, is extended to intersect the circumcircle at P such that AP is a chord of the circumcircle, then the foot D bisects the segment HP:[7]

H.D. = D.P. . {\displaystyle HD=DP.}

The directrixes of all parabolas externally tangent to one side of a triangle and tangent to the extensions of the other sides pass through the orthocenter.[15]

A circumconus passing through the orthocenter of a triangle is a rectangular hyperbola.[16]

Relationship to other centers, the nine-point circle [ edit ]

The orthocenter H, centroid G, circumcenter O, and center N of the nine-point circle all lie on a single line known as the Euler line.[17] The center of the nine-point circle is at the midpoint of the Euler line between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half that between the centroid and the orthocenter:[18]

OH = 2NH , {\displaystyle OH=2NH,}

2 O G = G H . {\displaystyle 2OG=GH.}

The orthocenter is closer to the center I than the center of gravity, and the orthocenter is farther from the center of gravity than the center:

H i I G . {\displaystyle HG>IG.}

Referring to the sides a, b, c, the inradius r and the circumradius R,[19]

O H 2 = R 2 − 8 R 2 cos ⁡ A cos ⁡ B cos ⁡ C = 9 R 2 − ( a 2 + b 2 + c 2 ) , {\displaystyle OH^{2}=R^{2 } -8R ^{2}\cos A\cos B\cos C=9R^{2}-(a^{2}+b^{2}+c^{2}),} [20] : p. 449

H i 2 = 2 r 2 − 4 R 2 cos ⁡ A cos ⁡ B cos ⁡ C . {\displaystyle HI^{2}=2r^{2}-4R^{2}\cos A\cos B\cos C.}

Local triangle[ edit ]

abc (respectively in the text DEF) is the orthotic triangle of the triangle ABC Triangle (respectively in the text) is the orthotic triangle of the triangle

When triangle ABC is oblique (contains no right angle), the pedal triangle of the orthocenter of the original triangle is called the orthic triangle or triangle of elevation. That is, the bases of the altitudes of an oblique triangle form the orthotic triangle DEF. Also the midpoint (the center of the inscribed circle) of the orthotic triangle DEF is the orthomidpoint of the original triangle ABC.[21]

Trilinear coordinates for the vertices of the orthotic triangle are given by

D = 0 : sec B : sec C

E = sec A : 0 : sec C

F = sec A : sec B : 0 .

The extended sides of the orthotic triangle meet the opposite extended sides of its reference triangle at three collinear points.[22][23][21]

In any acute triangle, the inscribed triangle with the smallest perimeter is the orthotic triangle.[24] This is the solution to Fagnano’s problem posed in 1775.[25] The sides of the orthotic triangle are parallel to the tangents to the circumcircles at the vertices of the original triangle.

The orthotic triangle of an acute triangle gives a triangular light path.[27]

The lines tangent to the nine-point circle at the midpoints of the sides of ABC are parallel to the sides of the orthotic triangle, forming a triangle similar to the orthotic triangle.[28]

The orthotic triangle is closely related to the tangent triangle, which is constructed as follows: Let L A be the tangent to the circumcircle of triangle ABC at vertex A, and define L B and L C analogously. Let A” = L B ∩ L C , B” = L C ∩ L A , C” = L C ∩ L A . The tangent triangle is A”B”C”, whose sides are the tangents to the circumcircle of triangle ABC at its vertices; it is homothetic to the orthotic triangle. The circumcentre of the tangential triangle and the similarity center of the orthotic and tangential triangle lie on the Euler line.[20]: p. 447

Trilinear coordinates for the vertices of the tangent triangle are given by

A” = − a : b : c

B” = a : − b : c

C” = a : b : −c .

The reference triangle and its orthotic triangle are orthological triangles.

More information about the Orthotic Triangle can be found here.

Some extra height sets

Height in relation to the sides[edit]

For any triangle with sides a, b, c and semicircumference s = (a + b + c) / 2, the height from side a is given by

h a = 2 s ( s − a ) ( s − b ) ( s − c ) a . {\displaystyle h_{a}={\frac {2{\sqrt {s(s-a)(s-b)(s-c)}}}{a}}.}

This follows from combining Heron’s formula for the side area of ​​a triangle with the area formula (1/2)×base×height, where the base is taken as side a and the height is the height of A.

Inradius theorems[ edit ]

Consider any triangle with sides a, b, c and corresponding altitudes h a , h b , and h c . The heights and the incircle radius r hang together[29]: Lemma 1

1r = 1ha + 1hb + 1hc . {\displaystyle \displaystyle {\frac {1}{r}}={\frac {1}{h_{a}}}+{\frac {1}{h_{b}}}+{\frac {1} {h_{c}}}.}

Circumradius theorem [ edit ]

If the height of one side of a triangle is denoted as h a , the other two sides as b and c and the circumradius of the triangle (radius of the circumscribed circle of the triangle) as R, the height is given by [30]

h a = b c 2 R . {\displaystyle h_{a}={\frac {bc}{2R}}.}

Inner point[ edit ]

If p 1 , p 2 , and p 3 are the perpendicular distances from each point P to the sides, and h 1 , h 2 , and h 3 are the heights to the respective sides, then[31]

p 1 h 1 + p 2 h 2 + p 3 h 3 = 1. {\displaystyle {\frac {p_{1}}{h_{1}}}+{\frac {p_{2}}{h_{2 }}}+{\frac {p_{3}}{h_{3}}}=1.}

Face set [ edit ]

Denotes the heights of any triangle from sides a, b, and c as h a {\displaystyle h_{a}} , h b {\displaystyle h_{b}} and h c {\displaystyle h_{c}} and denotes the midpoint of the Reciprocals of heights as H = ( h a − 1 + h b − 1 + h c − 1 ) / 2 {\displaystyle H=(h_{a}^{-1}+h_{b}^{- 1}+h_{c }^{-1})/2} we have[32]

A re a − 1 = 4 H ( H − h a − 1 ) ( H − h b − 1 ) ( H − h c − 1 ) . {\displaystyle \mathrm {area} ^{-1}=4{\sqrt {H(H-h_{a}^{-1})(H-h_{b}^{-1})(H-h_ {c}^{-1})}}.}

General point to a height [ edit ]

If E is any point at an altitude AD of any triangle ABC, then[33]: 77–78

A C 2 + E B 2 = A B 2 + C E 2 . {\displaystyle AC^{2}+EB^{2}=AB^{2}+CE^{2}.}

Special case triangles[edit]

Equilateral triangle[edit]

For any point P within an equilateral triangle, the sum of the perpendiculars to the three sides is equal to the triangle’s height. This is Viviani’s theorem.

Right triangle[edit]

Comparison of the inverse Pythagorean theorem with the Pythagorean theorem

In a right triangle, the three heights h a , h b and h c (the first two of which are equal to the leg lengths b and a respectively) are according to [34][35]

1 h a 2 + 1 h b 2 = 1 h c 2 . {\displaystyle {\frac {1}{h_{a}^{2}}}+{\frac {1}{h_{b}^{2}}}={\frac {1}{h_{c} ^{2}}}.}

This is also known as the inverse Pythagorean theorem.

history [edit]

The theorem that the three altitudes of a triangle meet at a single point, the orthocenter, was first proved in a 1749 paper by William Chapple.[36]

See also[edit]

Notes [edit]

References[edit]

Use the Pythagorean theorem for right triangles…

Question:

If I have a right triangle and two of its sides, how can I find the length of the third side? Can I do this if it’s not a right triangle?

answers

Finding the missing side of a right triangle is fairly easy when two sides are known. One of the more well-known mathematical formulas is \(a^2+b^2=c^2\), which is known as the Pythagorean theorem. The theorem states that the hypotenuse of a right triangle can be easily calculated from the side lengths. The hypotenuse is the longest side of a right triangle.

Given the lengths of the two sides, finding the hypotenuse is easy. Just square the sides, add, and then take the square root. Here is an example:

Since we know the two legs of the triangle are 3 and 4, plug these into the Pythagorean equation and solve for the hypotenuse:

$$ a^2+b^2=c^2 $$ $$ 3^2+4^2=c^2 $$ $$ 25 = c^2 $$ $$ c = \sqrt{25} $$ $$c = 5$$

When you’re given the hypotenuse and one of the legs, it gets a little more complicated, but only because you have to do some algebra first. Suppose you know that one leg is 5 and the hypotenuse (longest side) is 13. Plug these into the appropriate places in the Pythagorean equation:

$$ a^2+b^2=c^2 $$ $$ 5^2+b^2=13^2 $$ $$ 25+b^2=169 $$ $$ b^2=144 $$ $$b = 12$$

As you can see, using the Pythagorean theorem to find the missing side length of a right triangle is fairly easy. But – what if it’s not a right triangle? Obviously, if you change this angle in the triangle, there can be a number of possibilities for the hypotenuse! Therefore, you need more information to solve the problem. You can try using the Law of Sines or the Law of Cosines to find side lengths in other triangles.

Try the “Triangle Calculator” below: