# How to Solve Pythagorean Theorem Problems: Geometry and Algebra Examples

The Pythagorean Theorem is one of the oldest known algebraic and geometric identities. Although the first formal proof of the equation is credited to the ancient Greek mathematician Pythagoras and his students, it was widely applied in the ancient world before his time, before anyone thought to officially prove that it worked in general. The theorem states that if a right triangle has sides of length A and B, and a hypotenuse of length C, then the three measures are related by the equation

A^2 + B^2 = C^2.

What it means in practical terms is that if you know two of the measurements and the third is unknown, you can solve for it.

A unknown, B & C known:

A = sqrt(C^2 - B^2)

B unknown, A & C known:

B = sqrt(C^2 - A^2)

C unknown, A & B known:

C = sqrt(A^2 + B^2)

For an arbitrary right triangle, A, B, and C may not all by integers or rational numbers. But there are some special right triangles for which all three sides are whole numbers. These are called Pythagorean triples, written (A, B, C). A few of them are

(3, 4, 5), 3^2 + 4^2 = 9 + 16 = 25 = 5^2

(5, 12, 13), 5^2 + 12^2 = 25 + 144 = 169 = 13^2

(8, 15, 17), 8^2 + 15^2 = 64 + 225 = 289 = 17^2

Here are some examples of using the Pythagorean Theorem to solve a variety of geometry problems.

## Example 1: Side Lengths of 22 and 37

Two sides of a right triangle are 22 and 37, with no information given about whether these refer to leg or hypotenuse measurements. What are the possible values for the missing side length?

In a right triangle, the hypotenuse is always the longest side. Since 22 is less than 37, we know that 22 has to be a leg. But 37 can be either a leg or a hypotenuse. There are two cases to consider which yield two possible values for the third side.

**Case 1:** If both 22 and 37 are leg mesurements, then we set A = 22 and B = 37 and solve for C using the equation C = sqrt(22^2 + 37^2), which gives us

sqrt(484 + 1369)

= sqrt(1853)

= 43.0465

**Case 2:**If 22 is one leg and 37 is the hypotenuse, we have A = 22 and C = 37. We then solve for B with the equation B = sqrt(37^2 - 22^2), which produces

sqrt(1369 - 484)

= sqrt(885)

= 29.7489

Therefore the third side could either be 43.0465 or 29.7489.

## Example 2: Calculate an Angle w/o a Protractor

You need to know the angle and length of an inclined ramp that you're going to build, but the only tool you have is a calculator that's missing some keys -- the only trig function button that works on the calculator is the inverse sine button (arcsin). If the ramp spans a vertical distance of 26 inches, and spans 67 inches horizontally, what are its length and angle of inclination?

As viewed from the side, the length of the ramp is the hypotenuse of a right triangle with legs 26 and 67. Since A = 26 and B = 67, we can then solve the the length of the ramp C:

C = sqrt(A^2 + B^2)

= sqrt(26^2 + 67^2)

= sqrt(676 + 4489)

= sqrt(5165)

= **71.868 inches**

To find the angle of the ramp, we note that the sine of the angle is the ratio opposite/hypotenuse. The opposite length is 26 inches and the hypotenuse length is 71.868 inches, therefore

sin(angle) = 26/71.868 = 0.361774

Taking the inverse sine of both sides allows you to find the angle, since sin and arcsin cancel each other out as functions.

arcsin[ sin(angle) ] = arcsin[0.361774]

angle = arcsin[0.361774]**angle = 21.21 degrees**

## Example 3: Height of a Triangle

A triangle has a base of 134 centimeters and two shorter sides, rising up from the base, whose lengths are 83 cm and 95 cm. What is the height of the triangle from the apex to the base?

As shown in the diagram, if you draw a line from the apex to the base, you partition the base into two segments. Say the length of the segment on the left is x and the length of the segment on the right is 134 - x. If the height of the triangle is denoted by y, then we can use the Pythagorean Theorem to construct a system of two equations in two unknowns.

The triangle on the left has a hypotenuse of 83 and legs of x and y. This produces the equation

x^2 + y^2 = 83^2

The triangle on the right has a hypotenuse of 95 and legs of 134-x and y. This gives us the equation

(134-x)^2 + y^2 = 95^2

If we subtract the first equation from the second, we get

(134-x)^2 + y^2 - x^2 - y^2 = 95^2 - 83^2

-268x + 17956 = 9025 - 6889

268x = 15820**x = 15820/268**

Now we just have to plug this x value into the first equation to solve for the height, y:

(15820/268)^2 + y^2 = 6889

y^2 = 6889 - (15820/268)^2

y = sqrt(6889 - (15820/268)^2)**y = 58.3479 cm**

## Example 4: Pythagorean Algebra

A right triangle has side lengths given by the expressions

5x + 3

x + 20

x + 17

for an unknown value of x, but no information is given about which side is the hypotenuse and which sides are the legs. What are all the possible values of x that would make (5x+3, x+20, x+17) a Pythagorean triple, in some order?

## More Geometry Help

- List of Geometry Formulas for Angle, Length, Area, and Volume
- Derivation of Heron's Formula for the Area of a Triangle
- How to Find the Area of a Parallelogram: 2 Different Methods
- How to Find the Area of a Triangle with 3 Coordinate Points
- Geometry and Math Facts About Regular Pentagons
- Volume of a Barrel: Formula and Examples
- Volume of a Spherical Cap: Formula and Examples

First, we know that the side corresponding to x + 17 cannot be the hypotenuse since for every value of x, the side x + 20 is longer. Therefore, either x + 20 or 5x + 3 can be the hypotenuse. We can find all the corresponding values of x examining these two cases.

**Case 1:**

If 5x + 3 is the hypotenuse and the sides x + 17 and x + 20 are the legs, then these sides satisfy the right triangle relation

(x+17)^2 + (x+20)^2 = (5x+3)^2

After simplifying this quadratic equation, you get the standard form

23x^2 - 39x - 680 = 0

Using the quadratic formula gives you the two solutions x = -4.6553 and x = 6.3509. The first solution can be discarded because when you plug x = -4.6553 into the expression 5x + 3, you get a negative side length. Therefore, Case I yields the single solution x = 6.3509.

**Case 2:** If x + 20 is the hypotenuse and 5x + 3 and x + 17 are the legs, the the Pythagorean Theorem gives us the quadratic equation

(5x+3)^2 + (x+17)^2 = (x+20)^2

25x^2 + 24x - 102 = 0

which yields the two solutions x = -2.5562 and x = 1.5962. The first solution yields a negative length when you plug it into the expression 5x + 3, therefore the only solution from this case is x = 1.5962.

The two solutions are **x = 6.3509** and **x = 1.5962**.