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=(x2−x1)2+(y2−y1)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 ∣x2−x1∣|x_2 - x_1| and ∣y2−y1∣|y_2 - y_1|.

Examples

  • To find the distance between (1,3)(1, 3) and (5,6)(5, 6), we calculate d=(5−1)2+(6−3)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+(−5−7)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=(−5−4)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.

Lesson overview

Expand to review the lesson summary and core properties.

Expand

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=(x2−x1)2+(y2−y1)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 ∣x2−x1∣|x_2 - x_1| and ∣y2−y1∣|y_2 - y_1|.

Examples

  • To find the distance between (1,3)(1, 3) and (5,6)(5, 6), we calculate d=(5−1)2+(6−3)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+(−5−7)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=(−5−4)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.