Unit 8: Problem Solving · Lesson 7
Problem Solving - Lesson 7
What this lesson is for
- Students will understand and practice Problem Solving - Lesson 7
- Students will solve problems related to this topic
Train up a child in the way he should go: and when he is old, he will not depart from it.
Proverbs 22:6
Geometry Applications
Using Algebra to Solve Geometry Problems Geometry formulas become powerful tools when combined with algebra. When a shape's dimensions are unknown, we write expressions for them, set up an equation using a geometry formula, and solve.
Review of Key Formulas
- Perimeter of rectangle: P = 2l + 2w
- Area of rectangle: A = l × w
- Area of triangle: A = ½ × b × h
- Pythagorean theorem: a² + b² = c² (right triangles only, c = hypotenuse)
- Area of circle: A = πr²
- Circumference: C = 2πr
Worked Example 1 — Perimeter: A rectangle has a perimeter of 48 ft. The length is twice the width. Find both dimensions.
- Let w = width; then l = 2w
- Equation: 2(2w) + 2w = 48 → 4w + 2w = 48 → 6w = 48 → w = 8 ft
- Length = 2(8) = 16 ft
- Check: 2(16) + 2(8) = 32 + 16 = 48 ✓
Worked Example 2 — Area/Cost: A rectangular room is 15 ft × 12 ft. Carpet costs $3.50 per square foot. What is the total carpet cost?
- Area = 15 × 12 = 180 sq ft
- Cost = 180 × $3.50 = $630
Worked Example 3 — Pythagorean Theorem: A 10-foot ladder leans against a wall. The base is 6 ft from the wall. How high on the wall does the ladder reach?
- a² + b² = c² → 6² + h² = 10²
- 36 + h² = 100 → h² = 64 → h = 8 ft
Worked Example 4 — Composite Figures: An L-shaped room can be divided into two rectangles. Rectangle A: 10 ft × 6 ft = 60 sq ft. Rectangle B: 8 ft × 4 ft = 32 sq ft. Total area = 60 + 32 = 92 sq ft.
Alternatively, find the area of the large enclosing rectangle and subtract the missing corner.
Worked Example 5 — Triangle Area Word Problem: A triangular garden has a base of 14 ft and a height of 9 ft. How many bags of mulch are needed if each bag covers 21 sq ft?
- Area = ½ × 14 × 9 = 63 sq ft
- Bags needed = 63 ÷ 21 = 3 bags
Problem-Solving Checklist
- Draw and label a diagram
- Write expressions for unknowns
- Choose the correct formula
- Set up and solve the equation
- Check your answer and include units
"The heavens declare the glory of God; and the firmament sheweth his handywork." — Psalm 19:1
Guided practice
A rectangular garden has perimeter 40 meters and length is 3 times the width. Find the width.
Show the steps and the answer
Step 1: Let w = width, then l = 3w. Step 2: Perimeter: 2(3w) + 2w = 40. Step 3: 8w = 40. Step 4: w = 5 meters.
Answer: 5
A 10-foot ladder leans against a wall with its base 6 feet from the wall. How high does it reach?
Show the steps and the answer
Step 1: Use Pythagorean theorem: 6² + h² = 10². Step 2: 36 + h² = 100. Step 3: h² = 64. Step 4: h = 8 feet.
Answer: 8
A triangular garden has base 12 feet and height 8 feet. What is its area?
Show the steps and the answer
Step 1: Area of triangle = (1/2) × base × height. Step 2: (1/2) × 12 × 8 = 48 square feet.
Answer: 48
Homework
Complete the practice problems for Problem Solving - Lesson 7.
Note for the parent
Today's lesson covers Problem Solving - Lesson 7. Estimated time: 50 minutes. Encourage independent work but be available for questions.
The first three of this lesson's 10 questions
The perimeter formula for a rectangle is:
Show the answer
Answer: P = 2l + 2w
A rectangle has perimeter 48 ft and length = twice the width. What is the width?
Show the answer
Answer: 8 ft
Using the same rectangle, what is the length?
Show the answer
Answer: 16 ft
Another 7 questions follow in the lesson itself, each one graded the moment it is answered.