3.7: Special Triangles

Definitions and Theorems

A special right triangle is a right triangle whose angles force fixed ratios among its side lengths. Two of these occur often enough that you should recognize them on sight: knowing their side ratios lets you find any missing side exactly, without applying the Pythagorean theorem each time.

Definition: 30-60-90 Triangle

A right triangle in which one angle is \( 30^{ \circ } \) and another angle is \( 60^{ \circ } \) is called a \( 30^{ \circ } \)-\( 60^{ \circ } \)-\( 90^{ \circ } \) triangle.

Theorem: Side Relationships for a 30-60-90 Triangle

In any right triangle in which the two acute angles are \( 30^{ \circ } \) and \( 60^{ \circ } \), the hypotenuse is always twice the length of the shortest side (the side opposite the \( 30^{ \circ } \) angle), and the remaining side (opposite the \( 60^{ \circ } \) angle) is always \( \sqrt{3} \) times the shortest side.

Proof

Consider an equilateral triangle with side lengths \( 2a \), as shown in the figure below.

Since this triangle is equilateral, each angle is \( 60^{ \circ } \). Drawing an altitude divides the triangle into two smaller triangles. Since the altitude is perpendicular to the base, each of these smaller triangles is a right triangle. Moreover, since one angle in each of these smaller right triangles is \( 60^{ \circ } \), the remaining angle must be \( 30^{ \circ } \). Furthermore, the side opposite the \( 30^{ \circ } \) angle must have length \( a \) (in the original equilateral triangle, the side opposite \( 60^{ \circ } \) had length \( 2a \), so it should make sense that the side opposite \( 30^{ \circ } \) is half of this). We now consider one of these smaller right triangles (see the figure below).

The missing side length, which we will call \( x \) for now, can be found using the Pythagorean Theorem.\[\begin{array}{rrcl}
& x^2 + a^2 & = & (2a)^2 \\
\implies & x^2 + a^2 & = & 4a^2 \\
\implies & x^2 & = & 3a^2 \\
\implies & x & = & a \sqrt{3} \\
\end{array} \nonumber \]

Definition: 45-45-90 Triangle

Since an isosceles right triangle has two angles that are \( 45^{ \circ } \), it is commonly called a \( 45^{ \circ } \)-\( 45^{ \circ } \)-\( 90^{ \circ } \) triangle.

Theorem: Side Relationships for a 45-45-90 Triangle

In any right triangle in which the two acute angles are \( 45^{ \circ } \), the hypotenuse is always \( \sqrt{2} \) times the side length.

Proof

The sides opposite the \( 45^{ \circ } \) angles in a \( 45^{ \circ } \)-\( 45^{ \circ } \)-\( 90^{ \circ } \) triangle must have the same length. Let \( a \) be the lengths of these sides and \( x \) be the length of the hypotenuse, as shown in the figure below.

Since this is a right triangle, we can use the Pythagorean Theorem to find the length of the hypotenuse in terms of \( a \).\[\begin{array}{rrcl}
& a^2 + a^2 & = & x^2 \\
\implies & 2 a^2 & = & x^2 \\
\implies & a \sqrt{2} & = & x \\
\end{array} \nonumber \]

Caution: Match Each Side to Its Opposite Angle

In a 30-60-90 triangle, the factor of \(\sqrt{3}\) belongs to the longer leg—the one opposite the \(60^{\circ}\) angle—not the shorter leg opposite the \(30^{\circ}\) angle. When a problem gives you the longer leg or the hypotenuse, first solve for the shorter leg \(x\), then build the remaining sides from it.

Examples 

Example \(\PageIndex{1}\): Leg to Hypotenuse

The legs of a 45-45-90 triangle each measure \(7\). Find the length of the hypotenuse.

Solution

The two legs are congruent, so \(x=7\). By the 45-45-90 ratio, the hypotenuse has length \(x\sqrt{2}=7\sqrt{2}\). Thus the hypotenuse measures \(7\sqrt{2}\).

Example \(\PageIndex{2}\): Hypotenuse to Leg

The hypotenuse of a 45-45-90 triangle measures \(10\). Find the length of each leg.

Solution

Let each leg have length \(x\). By the 45-45-90 ratio, the hypotenuse equals \(x\sqrt{2}\), so \(x\sqrt{2}=10\). Solving for \(x\) and rationalizing the denominator,\[x=\dfrac{10}{\sqrt{2}}=\dfrac{10}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{10\sqrt{2}}{2}=5\sqrt{2}.\nonumber\]Each leg measures \(5\sqrt{2}\).

Example \(\PageIndex{3}\): Shorter Leg Given

In a 30-60-90 triangle, the leg opposite the \(30^{\circ}\) angle measures \(6\). Find the lengths of the remaining leg and the hypotenuse.

Solution

The given side is the shorter leg, so \(x=6\). The longer leg, opposite the \(60^{\circ}\) angle, has length \(x\sqrt{3}=6\sqrt{3}\), and the hypotenuse has length \(2x=12\).

Example \(\PageIndex{4}\): Hypotenuse Given

The hypotenuse of a 30-60-90 triangle measures \(14\). Find the lengths of the two legs.

Solution

The hypotenuse equals \(2x\), so \(2x=14\), giving \(x=7\). The shorter leg, opposite the \(30^{\circ}\) angle, has length \(x=7\), and the longer leg, opposite the \(60^{\circ}\) angle, has length \(x\sqrt{3}=7\sqrt{3}\).

Example \(\PageIndex{5}\): Longer Leg Given

In a 30-60-90 triangle, the leg opposite the \(60^{\circ}\) angle measures \(9\). Find the lengths of the shorter leg and the hypotenuse.

Solution

The given side is the longer leg, so \(x\sqrt{3}=9\). Solving for \(x\) and rationalizing the denominator,\[x=\dfrac{9}{\sqrt{3}}=\dfrac{9}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{9\sqrt{3}}{3}=3\sqrt{3}.\nonumber\]The shorter leg measures \(3\sqrt{3}\), and the hypotenuse has length \(2x=6\sqrt{3}\).

Example \(\PageIndex{6}\): Altitude of an Equilateral Triangle

An equilateral triangle has side length \(10\). Find the length of its altitude.

Solution

An altitude of an equilateral triangle bisects the base and splits the figure into two congruent 30-60-90 triangles. In each, the hypotenuse is a full side of length \(10\), and the shorter leg, opposite the \(30^{\circ}\) angle, is half the base. Since the hypotenuse equals \(2x\), we have \(2x=10\), so \(x=5\). The altitude is the longer leg, opposite the \(60^{\circ}\) angle, with length \(x\sqrt{3}=5\sqrt{3}\).

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Sources

Several parts of this text use modifications from the following source:

• Wikipedia article: "Special right triangle"

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