Question

quadratic relationships
student activity sheet 1; overview
page 1 of 1

1. complete the following table to show possible dog kennels with a perimeter of

30 meters.

p(m)	l(m)	w(m)	a(m²)
30	13	2	26
30	12	3	36
30	11	4	44
30	10	5	50
30	9	6	54

1. if the width of the kennel is x meters, then what is the length in terms of x?
2. write an expression for the area of the rectangle in terms of x. then write an equation

you could solve to find the dimensions of a rectangle with an area of 50.

1. what is a possible solution to the equation you wrote in question 3? use the table you

completed in question 1 to answer this question.
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Explanation:

Question 1: Complete the table

Step1: Recall perimeter formula for rectangle

The perimeter \( P \) of a rectangle is given by \( P = 2(l + w) \). Given \( P = 30 \), we can solve for \( l \): \( 30 = 2(l + w) \implies l + w = 15 \implies l = 15 - w \).

Step2: Calculate length (\( l \)) and area (\( A \))

• For \( w = 1 \): \( l = 15 - 1 = 14 \), \( A = l \times w = 14 \times 1 = 14 \)
• For \( w = 2 \): \( l = 15 - 2 = 13 \), \( A = 13 \times 2 = 26 \)
• For \( w = 3 \): \( l = 15 - 3 = 12 \), \( A = 12 \times 3 = 36 \)
• For \( w = 4 \): \( l = 15 - 4 = 11 \), \( A = 11 \times 4 = 44 \)
• For \( w = 5 \): \( l = 15 - 5 = 10 \), \( A = 10 \times 5 = 50 \)
• For \( w = 6 \): \( l = 15 - 6 = 9 \), \( A = 9 \times 6 = 54 \)

Question 2: Length in terms of \( x \) (width)

Step1: Use perimeter formula

Perimeter of rectangle: \( P = 2(l + w) \). Given \( P = 30 \) and \( w = x \), substitute:
\( 30 = 2(l + x) \)

Step2: Solve for \( l \)

Divide both sides by 2: \( 15 = l + x \implies l = 15 - x \)

Question 3: Area expression and equation

Step1: Area formula for rectangle

Area \( A = l \times w \). From Question 2, \( l = 15 - x \) and \( w = x \), so \( A = x(15 - x) \).

Step2: Equation for area = 50

Substitute \( A = 50 \) into the area expression: \( x(15 - x) = 50 \). Expand: \( 15x - x^2 = 50 \), or \( -x^2 + 15x - 50 = 0 \) (or \( x^2 - 15x + 50 = 0 \) after multiplying by -1).

Question 4: Possible solution from table

Answer:

Step1: Check table for \( A = 50 \)

From the table in Question 1, when \( A = 50 \), \( w = 5 \) and \( l = 10 \) (or \( w = 10 \) and \( l = 5 \), but width is usually smaller). So \( x = 5 \) (width) or \( x = 10 \) (length, but if \( x \) is width, \( x = 5 \)).

Final Answers (Summarized):

1. Table completed as shown (lengths: 14,13,12,11,10,9; areas:14,26,36,44,50,54).
2. Length: \( \boldsymbol{l = 15 - x} \)
3. Area expression: \( \boldsymbol{A = x(15 - x)} \); Equation: \( \boldsymbol{x(15 - x) = 50} \) (or \( -x^2 + 15x - 50 = 0 \))
4. Possible solution: \( \boldsymbol{x = 5} \) (width) or \( \boldsymbol{x = 10} \) (length), from table \( w = 5, l = 10 \).