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Lesson 7A Proof of the Pythagorean Theorem

Let’s prove the Pythagorean Theorem.

Learning Targets:

  • I can explain why the Pythagorean Theorem is true.

7.1 Notice and Wonder: A Square and Four Triangles

Two squares are shown with triangles connected to the side lenths. One square has right triangles of sides and the other does not.
What do you notice? What do you wonder?

7.2 Adding Up Areas

Both figures shown here are squares with a side length of a + b . Notice that the first figure is divided into two squares and two rectangles. The second figure is divided into a square and four right triangles with legs of lengths a and b . Let’s call the hypotenuse of these triangles c .

Two squares of the same area are labeled “F” and “G”. Square F is divided into the following: A square with side lengths “a”. Two rectangles with side lengths “a” and “b”. A square with side lengths “b”. Square G is divided into the following: Four identical triangles on each corner of the square with sides labeled “a” and “b”. A square in the center with unlabeled side lengths.
  1. What is the total area of each figure?
  2. Find the area of each of the 9 smaller regions shown the figures and label them.
  3. Add up the area of the four regions in Figure F and set this expression equal to the sum of the areas of the five regions in Figure G. If you rewrite this equation using as few terms as possible, what do you have?

Are you ready for more?

Take a 3-4-5 right triangle, add on the squares of the side lengths, and form a hexagon by connecting vertices of the squares as in the image. What is the area of this hexagon?

A 3-4-5 right triangle, with added on the squares of the side lengths. A hexagon is formed by connecting vertices of the squares

7.3 Let’s Take It for a Spin

Find the unknown side lengths in these right triangles.

7.4 A Transformational Proof

Use the applets to explore the relationship between areas.

  • Consider Squares A and B .

  • Check the box to see the area divided into five pieces with a pair of segments.

  • Check the box to see the pieces.

  • Arrange the five pieces to fit inside Square C .

  • Check the box to see the right triangle.

  • Arrange the figures so the squares are adjacent to the sides of the triangle.

  1. If the right triangle has legs a and b and hypotenuse c , what have you just demonstrated to be true?
open applet in presentation mode
  1. Try it again with different squares. Estimate the areas of the new Squares,  A , B , and C and explain what you observe.
open applet in presentation mode
  1. Estimate the areas of these new Squares, A , B , and C , and then explain what you observe as you complete the activity.
open applet in presentation mode

4. What do you think we may be able to conclude?

Lesson 7 Summary

The figures shown here can be used to see why the Pythagorean Theorem is true. Both large squares have the same area, but they are broken up in different ways. (Can you see where the triangles in Square G are located in Square F? What does that mean about the smaller squares in F and H?) When the sum of the four areas in Square F are set equal to the sum of the 5 areas in Square G, the result is  a^2 + b^2 = c^2 , where c is the hypotenuse of the triangles in Square G and also the side length of the square in the middle. Give it a try!

Two squares of the same area are labeled “F” and “G”. Square F is divided into the following: A square with side lengths “a”. Two rectangles with side lengths “a” and “b”. A square with side lengths “b”. Square G is divided into the following: Four identical triangles on each corner of the square with sides labeled “a” and “b”. A square in the center with unlabeled side lengths.

This is true for any right triangle. If the legs are a and b and the hypotenuse is c , then a^2+b^2=c^2 . This property can be used any time we can make a right triangle. For example, to find the length of this line segment:

A line segment slanted downward and to the right on a square grid. The bottom endpoint is 7 units down and 24 units to the right from the top endpoint.

The grid can be used to create a right triangle, where the line segment is the hypotenuse and the legs measure 24 units and 7 units:

A right triangle on a square grid. The horizontal side has a length of 24 and the vertical side has a length of 7. The hypotenuse is labeled c.

Since this is a right triangle,  24^2+7^2=c^2 . The solution to this equation (and the length of the line segment) is c=25 .

Lesson 7 Practice Problems

    1. Find the lengths of the unlabeled sides.
      A right triangle with a horizontal side on the top and a vertical side on the left. The top side is labeled 6 and the side on the left is labeled 2.
      A right triangle with a horizontal side on top and a vertical side on the left. The top side is labeled 8 and the left side is labeled 6.
    2. One segment is n units long and the other is p units long. Find the value of n and p . (Each small grid square is 1 square unit.)
      A line segment labeled “n” on a square grid. The line segment starts at an intersection point on the grid and slants downward and to the right to an end point that is 1 unit to the right and 3 units down.
      A line segment labeled “p” on a square grid. The line segment starts at an intersection point on the grid and slants upward and to the right to an end point that is 3 units to the right and 4 units up.

  2. Use the areas of the two identical squares to explain why  5^2+12^2=13^2 without doing any calculations.

    Two squares are shown. The first square is made up of two 12 by 5 rectangles, a 12 by 12 square, and a 5 by 5 square. The second box is made of 4 right triangles with side lengths of 5 and 12 and the hypotenuse of 13 and a 13 by 13 square.

  3. Each number is between which two consecutive integers?

    1. \sqrt{10}

    2. \sqrt{54}

    3. \sqrt{18}

    4. \sqrt{99}

    5. \sqrt{41}

    1. Give an example of a rational number, and explain how you know it is rational.

    2. Give three examples of irrational numbers.

  5. Write each expression as a single power of 10.

    1. 10^5 \boldcdot 10^0
    2. \frac{10^9}{10^0}

  6. Andre is ordering ribbon to make decorations for a school event. He needs a total of exactly 50.25 meters of blue and green ribbon. Andre needs 50% more blue ribbon than green ribbon for the basic design, plus an extra 6.5 meters of blue ribbon for accents. How much of each color of ribbon does Andre need to order?