Learn on PengienVision, Mathematics, Grade 8Chapter 7: Understand and Apply the Pythagorean Theorem

Lesson 3: Apply the Pythagorean Theorem to Solve Problems

Grade 8 students learn how to apply the Pythagorean Theorem and its converse to solve real-world problems, including finding diagonal lengths in rectangles and using a two-stage approach to find the space diagonal of rectangular prisms. This lesson from Chapter 7 of enVision Mathematics Grade 8 covers practical scenarios such as determining whether a triangular shelf forms a right angle using the converse of the Pythagorean Theorem. Students practice setting up and solving equations of the form a² + b² = c² in both two-dimensional and three-dimensional contexts.

Section 1

Applying the Pythagorean Theorem in Two Dimensions

Property

The Pythagorean theorem states that for a right triangle with legs of length aa and bb and a hypotenuse of length cc, the relationship is a2+b2=c2a^2 + b^2 = c^2.

This theorem is used to find an unknown side length in any right triangle when the other two side lengths are known. It is widely applied in real-world problems involving distances and heights.

Examples

  • What is the length of the diagonal of a rectangular TV screen that is 12 inches tall and 16 inches wide? The diagonal is the hypotenuse, so its length is 122+162=144+256=400=20\sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 inches.
  • A 17-foot ladder is placed against a wall. The base of the ladder is 8 feet away from the wall. How high up the wall does the ladder reach? Let the height be hh. Then h2+82=172h^2 + 8^2 = 17^2, so h2=28964=225h^2 = 289 - 64 = 225. The ladder reaches a height of h=225=15h = \sqrt{225} = 15 feet.
  • A ship sails 9 miles north and then 12 miles east. How far is the ship from its starting point? The path forms a right triangle. The distance dd is 92+122=81+144=225=15\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 miles.

Section 2

Applying the Pythagorean Theorem in Three Dimensions

Property

To find the length of the diagonal dd that passes through the interior of a rectangular prism (a box) with length ll, width ww, and height hh, we can use a three-dimensional version of the Pythagorean theorem. The formula is:

d=l2+w2+h2d = \sqrt{l^2 + w^2 + h^2}

This is derived by applying the Pythagorean theorem twice: first for the diagonal of the base, and then for the triangle formed by the base diagonal, the height, and the interior diagonal.

Examples

  • What is the length of the longest fishing pole that can fit in a box that is 3 feet wide, 4 feet long, and 12 feet tall? The length is 32+42+122=9+16+144=169=13\sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 feet.
  • What is the length of the diagonal inside a cube with a side length of 2 cm? The length is 22+22+22=4+4+4=12=23\sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} cm.
  • A room is 12 feet long, 16 feet wide, and 9 feet high. The diagonal on the floor is 122+162=144+256=400=20\sqrt{12^2+16^2} = \sqrt{144+256} = \sqrt{400} = 20 feet. The space diagonal is then 202+92=400+81=481\sqrt{20^2 + 9^2} = \sqrt{400+81} = \sqrt{481} feet.

Explanation

To find the longest straight line inside a box, you use the Pythagorean theorem in 3D. Imagine finding the diagonal on the floor, then using that line and the box's height to form a new right triangle that cuts through the box's center.

Section 3

Solving Real-World Problems

Property

The Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2, can be used to find unknown lengths in real-world scenarios that can be modeled by a right triangle. In the formula, aa and bb represent the lengths of the legs, and cc represents the length of the hypotenuse.

Examples

Book overview

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Chapter 7: Understand and Apply the Pythagorean Theorem

  1. Lesson 1

    Lesson 1: Understand the Pythagorean Theorem

  2. Lesson 2

    Lesson 2: Understand the Converse of the Pythagorean Theorem

  3. Lesson 3Current

    Lesson 3: Apply the Pythagorean Theorem to Solve Problems

  4. Lesson 4

    Lesson 4: Find Distance in the Coordinate Plane

Lesson overview

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Section 1

Applying the Pythagorean Theorem in Two Dimensions

Property

The Pythagorean theorem states that for a right triangle with legs of length aa and bb and a hypotenuse of length cc, the relationship is a2+b2=c2a^2 + b^2 = c^2.

This theorem is used to find an unknown side length in any right triangle when the other two side lengths are known. It is widely applied in real-world problems involving distances and heights.

Examples

  • What is the length of the diagonal of a rectangular TV screen that is 12 inches tall and 16 inches wide? The diagonal is the hypotenuse, so its length is 122+162=144+256=400=20\sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 inches.
  • A 17-foot ladder is placed against a wall. The base of the ladder is 8 feet away from the wall. How high up the wall does the ladder reach? Let the height be hh. Then h2+82=172h^2 + 8^2 = 17^2, so h2=28964=225h^2 = 289 - 64 = 225. The ladder reaches a height of h=225=15h = \sqrt{225} = 15 feet.
  • A ship sails 9 miles north and then 12 miles east. How far is the ship from its starting point? The path forms a right triangle. The distance dd is 92+122=81+144=225=15\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 miles.

Section 2

Applying the Pythagorean Theorem in Three Dimensions

Property

To find the length of the diagonal dd that passes through the interior of a rectangular prism (a box) with length ll, width ww, and height hh, we can use a three-dimensional version of the Pythagorean theorem. The formula is:

d=l2+w2+h2d = \sqrt{l^2 + w^2 + h^2}

This is derived by applying the Pythagorean theorem twice: first for the diagonal of the base, and then for the triangle formed by the base diagonal, the height, and the interior diagonal.

Examples

  • What is the length of the longest fishing pole that can fit in a box that is 3 feet wide, 4 feet long, and 12 feet tall? The length is 32+42+122=9+16+144=169=13\sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 feet.
  • What is the length of the diagonal inside a cube with a side length of 2 cm? The length is 22+22+22=4+4+4=12=23\sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} cm.
  • A room is 12 feet long, 16 feet wide, and 9 feet high. The diagonal on the floor is 122+162=144+256=400=20\sqrt{12^2+16^2} = \sqrt{144+256} = \sqrt{400} = 20 feet. The space diagonal is then 202+92=400+81=481\sqrt{20^2 + 9^2} = \sqrt{400+81} = \sqrt{481} feet.

Explanation

To find the longest straight line inside a box, you use the Pythagorean theorem in 3D. Imagine finding the diagonal on the floor, then using that line and the box's height to form a new right triangle that cuts through the box's center.

Section 3

Solving Real-World Problems

Property

The Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2, can be used to find unknown lengths in real-world scenarios that can be modeled by a right triangle. In the formula, aa and bb represent the lengths of the legs, and cc represents the length of the hypotenuse.

Examples

Book overview

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

Continue this chapter

Chapter 7: Understand and Apply the Pythagorean Theorem

  1. Lesson 1

    Lesson 1: Understand the Pythagorean Theorem

  2. Lesson 2

    Lesson 2: Understand the Converse of the Pythagorean Theorem

  3. Lesson 3Current

    Lesson 3: Apply the Pythagorean Theorem to Solve Problems

  4. Lesson 4

    Lesson 4: Find Distance in the Coordinate Plane