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Published on: 2024-12-12 13:57:56
Author: James Miller

Pythagorean Theorem Calculator

Place any two of a,b,c value

Pythagorean Theorem Calculator

Use the Pythagorean theorem to determine the side lengths of a right triangle. A crucial tool in physics, building, and geometry, you can enter any two sides to find the third.

Pythagorean Theorem Formulas

Theorem Statement

The hypotenuse's square (c) in a right triangle is equal to the sum of the squares of the other two sides (a and b):

c² = a² + b²

Formulas to Find Missing Sides

  • To find the hypotenuse (c):

    c = √(a² + b²)

  • To find side a:

    a = √(c² - b²)

  • To find side b:

    b = √(c² - a²)

Applications of the Pythagorean Theorem

The Pythagorean Theorem is used in various fields such as:

  • Geometry to calculate distances.
  • Trigonometry for solving triangles.
  • Engineering and architecture for design and construction.
  • Physics to calculate vector magnitudes.

Pythagorean Theorem Problems and Solutions

Problem 1: Calculate the Hypotenuse

A right triangle has legs of lengths 6 and 8. What is the length of the hypotenuse?

Using the Pythagorean Theorem: \( c^2 = a^2 + b^2 \)

\( c^2 = 6^2 + 8^2 = 36 + 64 = 100 \)

\( c = \sqrt{100} = 10 \)

Answer: The hypotenuse is 10.
Problem 2: Finding a Leg

A right triangle has a hypotenuse of 13 and one leg of length 12. Find the other leg.

Using the Pythagorean Theorem: \( c^2 = a^2 + b^2 \)

\( 13^2 = 12^2 + b^2 \)

\( 169 = 144 + b^2 \)

\( b^2 = 169 - 144 = 25 \)

\( b = \sqrt{25} = 5 \)

Answer: The other leg is 5.
Problem 3: Is It a Right Triangle?

A triangle has sides of lengths 9, 12, and 15. Is it a right triangle?

Using the Pythagorean Theorem: \( c^2 = a^2 + b^2 \)

\( 15^2 = 9^2 + 12^2 \)

\( 225 = 81 + 144 \)

\( 225 = 225 \)

Answer: Yes, it is a right triangle.
Problem 4: Diagonal of a Rectangle

A rectangle has a length of 9 and a width of 12. Find the length of its diagonal.

Using the Pythagorean Theorem: \( c^2 = a^2 + b^2 \)

\( c^2 = 9^2 + 12^2 = 81 + 144 = 225 \)

\( c = \sqrt{225} = 15 \)

Answer: The diagonal is 15.
Problem 5: Distance Between Two Points

Find the distance between the points (1, 1) and (4, 5) on a coordinate plane.

The distance formula is derived from the Pythagorean Theorem:

\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

\( d = \sqrt{(4 - 1)^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

Answer: The distance is 5 units.
Problem 6: Triangle Height from Hypotenuse

A right triangle has a hypotenuse of 17 and one leg of length 8. Find the height (other leg) of the triangle.

Using the Pythagorean Theorem: \( c^2 = a^2 + b^2 \)

\( 17^2 = 8^2 + b^2 \)

\( 289 = 64 + b^2 \)

\( b^2 = 289 - 64 = 225 \)

\( b = \sqrt{225} = 15 \)

Answer: The other leg is 15.
Problem 7: Diagonal of a Square

A square has sides of length 7. Find the length of its diagonal.

For a square, diagonal \( c \) can be found using \( c^2 = a^2 + a^2 = 2a^2 \):

\( c^2 = 2(7^2) = 2(49) = 98 \)

\( c = \sqrt{98} \approx 9.90 \)

Answer: The diagonal is approximately 9.90.
Problem 8: Ladder Against a Wall

A ladder is 25 feet long and is placed against a wall. The base of the ladder is 7 feet away from the wall. How high up the wall does the ladder reach?

Using the Pythagorean Theorem: \( c^2 = a^2 + b^2 \)

\( 25^2 = 7^2 + b^2 \)

\( 625 = 49 + b^2 \)

\( b^2 = 625 - 49 = 576 \)

\( b = \sqrt{576} = 24 \)

Answer: The ladder reaches 24 feet up the wall.
Problem 9: Distance Between Two Points (3D)

Find the distance between the points (1, 2, 3) and (4, 6, 7) in 3D space.

The distance formula in 3D is derived from the Pythagorean Theorem:

\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \)

\( d = \sqrt{(4 - 1)^2 + (6 - 2)^2 + (7 - 3)^2} \)

\( d = \sqrt{3^2 + 4^2 + 4^2} = \sqrt{9 + 16 + 16} = \sqrt{41} \approx 6.40 \)

Answer: The distance is approximately 6.40 units.
Problem 10: Area of a Triangle Using Pythagoras

A right triangle has legs of lengths 9 and 12. Find its area.

The area of a right triangle is given by:

\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

\( \text{Area} = \frac{1}{2} \times 9 \times 12 = \frac{1}{2} \times 108 = 54 \)

Answer: The area is 54 square units.

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