Math Problem Statement

The perimeter of a square and a rectangle is the same. The width of the rectangle is 6cm and its area is 16c𝑚2 less than the area of the square. Find the area of the square

Solution

Let the side length of the square be ss, and the length of the rectangle be ll with a width of 6 cm.

Step 1: Express the perimeter relationship

The perimeter of the square is: Psquare=4sP_{\text{square}} = 4s The perimeter of the rectangle is: Prectangle=2(l+6)P_{\text{rectangle}} = 2(l + 6) Since the perimeters are equal: 4s=2(l+6)4s = 2(l + 6) Dividing both sides by 2: 2s=l+62s = l + 6 l=2s6l = 2s - 6

Step 2: Express the area relationship

The area of the square is: Asquare=s2A_{\text{square}} = s^2 The area of the rectangle is: Arectangle=l×6=6lA_{\text{rectangle}} = l \times 6 = 6l We are given that the area of the rectangle is 16 cm² less than the area of the square, so: Asquare=Arectangle+16A_{\text{square}} = A_{\text{rectangle}} + 16 s2=6l+16s^2 = 6l + 16

Step 3: Substitute ll from Step 1 into the area equation

Substitute l=2s6l = 2s - 6 into the area equation: s2=6(2s6)+16s^2 = 6(2s - 6) + 16 Simplifying: s2=12s36+16s^2 = 12s - 36 + 16 s2=12s20s^2 = 12s - 20 Rearranging the equation: s212s+20=0s^2 - 12s + 20 = 0

Step 4: Solve the quadratic equation

We can solve this quadratic equation using the quadratic formula: s=b±b24ac2as = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=1a = 1, b=12b = -12, and c=20c = 20.

Substituting these values into the formula: s=(12)±(12)24(1)(20)2(1)s = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(20)}}{2(1)} s=12±144802s = \frac{12 \pm \sqrt{144 - 80}}{2} s=12±642s = \frac{12 \pm \sqrt{64}}{2} s=12±82s = \frac{12 \pm 8}{2}

Thus, the two possible solutions for ss are: s=12+82=10ors=1282=2s = \frac{12 + 8}{2} = 10 \quad \text{or} \quad s = \frac{12 - 8}{2} = 2

Step 5: Determine the valid solution

If s=2s = 2, the side length would be too small because the area of the rectangle would be negative. Therefore, the valid solution is s=10s = 10.

Step 6: Find the area of the square

The area of the square is: Asquare=s2=102=100cm2A_{\text{square}} = s^2 = 10^2 = 100 \, \text{cm}^2

Thus, the area of the square is 100 cm².

Would you like more details on any part of this process or have any further questions?

Here are 5 related questions to explore:

  1. How do you solve quadratic equations using factoring?
  2. What is the formula for the perimeter of different shapes?
  3. How can you derive the quadratic formula step by step?
  4. Can you calculate the area of a rectangle using different dimensions?
  5. How do you check if a quadratic equation has real solutions?

Tip: Always verify your solutions by plugging them back into the original conditions of the problem to ensure consistency.

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Quadratic Equations

Formulas

Perimeter of square: P = 4s
Perimeter of rectangle: P = 2(l + w)
Area of square: A = s²
Area of rectangle: A = l × w
Quadratic equation formula: ax² + bx + c = 0

Theorems

Quadratic formula

Suitable Grade Level

Grades 8-10