Home Forums Grades Pre K-2 Grades 3-5 Grades 6-8 Grades 9-12

Instructional Plan Template

Mathematics Governor’s Institute 2004

(Download: Instructional Plan Temp1.doc, Lesson.doc, An Innovative Practice Routine.doc, Corr-Rem.doc, Preworksheetcolor.doc, Solution Keys.doc, Summative Eval.doc)

Names of group members: Jennifer Bettine, Liz Shinko, Angie Dickey

Topic/Theme: Problem Solving using Pythagorean theorem

Level: Algebra I or Geometry

Time Element: One period or period

NCTM Standards Addressed: Algebra and Geometry

PA Math Standards Addressed: 2.8

Math Assessment Anchors Addressed: M11.C.1.4, M11.A.1.2, M11.A.1.4

Reading Assessment Anchors Addressed: R11.A.2

Objectives: Students will be able to recite the Pythagorean theorem with extensive problem solving applications.

Instructional Strategies and Plan (include strategies used to help different types of learners, i.e. auditory, visual, etc):

1. Review concepts and pre-teach the Pythagorean theorem

2. Model the use of the Pythagorean theorem

3. Use sequences involving the Pythagorean theorem

Materials/Resources:

Guided worksheet

Graphing calculator

Crayons

Interdisciplinary Connections:

· Reading – Identification of vocabulary and relevant details of story problems

· Technology – Use TI-83+ calculator to find a sum of a sequence

· Other – History a brief description of Pythagoras

Assessment Strategies:

· Formative Evaluation (checking student understanding during the lesson):

Teacher will facilitate students as they collaboratively work through the lesson.

· Summative Evaluation (How will it be determined that the objectives were achieved?):

Students will complete the VVWA (Verbal Visual Word Association) for concept reinforcement.

Teacher will evaluate the students’ worksheets.

Correctives/Remediation:

Students will use the Pythagorean theorem to find the distance between two buildings.

Extensions/Enrichment:

Students will extend the Pythagorean theorem to include other relationships between an obtuse triangle and acute triangle

Special Accommodations (special needs students)

· Description of the Special Needs student selected:

Jimmy:

1. Quickly memorizes facts

2. Above grade level in decoding and fluency in reading

3. Difficulty in multi-step problem solving

4. Poor fine motor skills

· Accommodations to use with this student:

1. Pre-teach vocabulary

2. Read the problem to the class as students follow along

3. Written instructions with simple steps to accomplish the lesson

Name:

LESSON

USING THE PYTHAGOREAN THEOREM

Given a right triangle with legs a and b and hypotenuse c.

Find the missing length c, given a = 12, b = 5

a c Start by replacing the values of a and b in the

Pythagorean theorem.

b c² = 12 ² + 5²

c²= 144 + 25 simplify the left side

c² = 169 add

c = take the square root of both sides

c = 13 simplify radical

A 13 ft. ladder is leaning against a building. The foot of the ladder is 5 ft. from the base of the building. How high will the ladder reach up the side of the building?

Start by finding and labeling the right triangle.

Choose a variable for the missing length.

Using the Pythagorean theorem write an equation.

5² + a² = 13²

25 + a² = 169 simplify the squares

a² = 144 solve for a²

a = take the square root of both sides

a = 12 solve for a

Solution: the ladder reaches a height of 12 ft.

AN INNOVATIVE PRACTICE ROUTINE

One day at practice Coach Sally decided that she needed to get her team in better shape. She came up with a running plan to inspire her players. She placed a starting cone in the middle of the short side of the field. Her players jog to the corner, turn 90 degrees up-field and sprint to the first of many equally spaced cones, turn and sprint back to the starting cone. This activity would continue to be repeated using the second, third, fourth, etc. equally spaced cones. The final lap would be around the last cone, which was placed at the far corner of the field. Coach Sally referred to this new running activity as “running the triangles”.

The field measured 120 yards by 80 yards (see diagram below). The problem is to place the right number of cones along the field (equally spaced) so that her players will run at least a mile. (1760 yards)

120 yards

starting 80 yards

cone

Extension

The Pythagorean theorem states: In a right triangle, the square of the length of the side opposite the right angle is equal to the sum of the squares of the other two sides.

Given an obtuse triangle and an acute triangle measure the lengths of the sides to choose the correct order symbol: <, >.

1. 2.

C C

B A B A

(AC)² ______ (AB)² + (BC)² (AC)² ______ (AB)² + (BC)²

Complete these extensions.

3. In an obtuse triangle, the square of the length of the side opposite the obtuse angle _____________________________________________.

4. In any triangle, the square of the length of a side opposite an acute angle___________________________________________________.

5. If the square of the length of one side of a triangle is greater than the sum of the squares of the lengths of the other two, then ____________

________________________________________________________________________________________________________________.

Name:

An Innovative Practice Routine

Directions

Read problem carefully.
See the diagram with 2 cones.
Find the length of the 3 sides of both right triangles
Add the lengths to get the total distance traveled.
If the total distance is less than 1760 yards add 2 more cones to the length of the field.

Two Cones

Starting cone

Repeat process with 2 more cones.
See diagram with 4 cones.
Find the lengths of the 3 sides of the 4 right triangles.
Add the lengths to get the total distance traveled.
If the total distance is less than 1760 yards add 2 more cones to the length of the field.

Four Cones

Starting cone

Repeat process by adding two more cones to the length of the field until the distance is at least 1760 yards.

Name:

Correctives/Remediation:

A fire broke out in an apartment building on the third floor. The fire truck stopped in the street between the burning apartment building and another apartment building. A 40 ft. ladder was raised to the fire that was located 30 ft. above the ground. Once the fire was out, another fire erupted on the second floor of the other apartment building. The ladder was pivoted and placed 20 ft. above the ground.

How far apart are the two apartment buildings?