Learn on PengiBig Ideas Math, Course 2, AcceleratedChapter 4: Real Numbers and the Pythagorean Theorem

Lesson 5: Using the Pythagorean Theorem

In this Grade 7 lesson from Big Ideas Math, Course 2 Accelerated, students learn to apply the converse of the Pythagorean Theorem to determine whether a triangle is a right triangle given its side lengths. The lesson also covers using the Pythagorean Theorem to find distances in a coordinate plane and solve real-life problems. It is part of Chapter 4: Real Numbers and the Pythagorean Theorem and addresses Common Core standards 8.G.6, 8.G.7, and 8.G.8.

Section 1

Defining the Converse of the Pythagorean Theorem

Property

For a triangle with side lengths a,b,a, b, and cc, if the sides satisfy the equation a2+b2=c2a^2 + b^2 = c^2, then the triangle is a right triangle. The right angle is always opposite the longest side, cc.

Examples

  • A triangular garden has sides measuring 8 meters, 15 meters, and 17 meters. Is it a right triangle? Check: 82+152=64+225=2898^2 + 15^2 = 64 + 225 = 289. The longest side squared is 172=28917^2 = 289. Yes, it's a right triangle.
  • A carpenter builds a frame with sides 5 ft, 12 ft, and a diagonal of 13 ft. Since 52+122=25+144=1695^2 + 12^2 = 25 + 144 = 169, and 132=16913^2 = 169, the frame must have a 90-degree corner.

Section 2

Distance Formula with Pythagorean Derivation

Property

The distance dd between points P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) is

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
This formula is derived from the Pythagorean theorem, where the distance dd is the hypotenuse of a right triangle whose legs have lengths x2x1|x_2 - x_1| and y2y1|y_2 - y_1|.

Examples

  • To find the distance between (1,3)(1, 3) and (5,6)(5, 6), we calculate d=(51)2+(63)2=42+32=16+9=25=5d = \sqrt{(5 - 1)^2 + (6 - 3)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.
  • The distance between (2,7)(-2, 7) and (3,5)(3, -5) is d=(3(2))2+(57)2=52+(12)2=25+144=169=13d = \sqrt{(3 - (-2))^2 + (-5 - 7)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.
  • The distance between (4,1)(4, -1) and (5,1)(-5, -1) is d=(54)2+(1(1))2=(9)2+02=81=9d = \sqrt{(-5 - 4)^2 + (-1 - (-1))^2} = \sqrt{(-9)^2 + 0^2} = \sqrt{81} = 9.

Explanation

Think of this as the Pythagorean theorem on a coordinate plane. The horizontal change (run) and vertical change (rise) between two points form the legs of a right triangle. The distance formula simply calculates the length of the hypotenuse.

Book overview

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Chapter 4: Real Numbers and the Pythagorean Theorem

  1. Lesson 1

    Lesson 1: Finding Square Roots

  2. Lesson 2

    Lesson 2: Finding Cube Roots

  3. Lesson 3

    Lesson 3: The Pythagorean Theorem

  4. Lesson 4

    Lesson 4: Approximating Square Roots

  5. Lesson 5Current

    Lesson 5: Using the Pythagorean Theorem

Lesson overview

Expand to review the lesson summary and core properties.

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Section 1

Defining the Converse of the Pythagorean Theorem

Property

For a triangle with side lengths a,b,a, b, and cc, if the sides satisfy the equation a2+b2=c2a^2 + b^2 = c^2, then the triangle is a right triangle. The right angle is always opposite the longest side, cc.

Examples

  • A triangular garden has sides measuring 8 meters, 15 meters, and 17 meters. Is it a right triangle? Check: 82+152=64+225=2898^2 + 15^2 = 64 + 225 = 289. The longest side squared is 172=28917^2 = 289. Yes, it's a right triangle.
  • A carpenter builds a frame with sides 5 ft, 12 ft, and a diagonal of 13 ft. Since 52+122=25+144=1695^2 + 12^2 = 25 + 144 = 169, and 132=16913^2 = 169, the frame must have a 90-degree corner.

Section 2

Distance Formula with Pythagorean Derivation

Property

The distance dd between points P1(x1,y1)P_1(x_1, y_1) and P2(x2,y2)P_2(x_2, y_2) is

d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
This formula is derived from the Pythagorean theorem, where the distance dd is the hypotenuse of a right triangle whose legs have lengths x2x1|x_2 - x_1| and y2y1|y_2 - y_1|.

Examples

  • To find the distance between (1,3)(1, 3) and (5,6)(5, 6), we calculate d=(51)2+(63)2=42+32=16+9=25=5d = \sqrt{(5 - 1)^2 + (6 - 3)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.
  • The distance between (2,7)(-2, 7) and (3,5)(3, -5) is d=(3(2))2+(57)2=52+(12)2=25+144=169=13d = \sqrt{(3 - (-2))^2 + (-5 - 7)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.
  • The distance between (4,1)(4, -1) and (5,1)(-5, -1) is d=(54)2+(1(1))2=(9)2+02=81=9d = \sqrt{(-5 - 4)^2 + (-1 - (-1))^2} = \sqrt{(-9)^2 + 0^2} = \sqrt{81} = 9.

Explanation

Think of this as the Pythagorean theorem on a coordinate plane. The horizontal change (run) and vertical change (rise) between two points form the legs of a right triangle. The distance formula simply calculates the length of the hypotenuse.

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 4: Real Numbers and the Pythagorean Theorem

  1. Lesson 1

    Lesson 1: Finding Square Roots

  2. Lesson 2

    Lesson 2: Finding Cube Roots

  3. Lesson 3

    Lesson 3: The Pythagorean Theorem

  4. Lesson 4

    Lesson 4: Approximating Square Roots

  5. Lesson 5Current

    Lesson 5: Using the Pythagorean Theorem