Figure With Area And Perimeter Of 64

A figure with areaand perimeter of 64 presents a captivating geometry puzzle that merges algebraic manipulation with spatial reasoning. This problem invites learners to explore how a single numeric value can simultaneously describe both the interior space and the outer boundary of a shape, encouraging deeper comprehension of mathematical relationships and fostering problem‑solving confidence.

Understanding the Problem

When we speak of a figure with area and perimeter of 64, we are looking for a geometric shape whose measured interior space equals 64 square units and whose total edge length also equals 64 units. The most straightforward candidate is a rectangle, because its area and perimeter can be expressed with two simple formulas:

• Area = length × width - Perimeter = 2 × (length + width)

Setting both expressions equal to 64 creates a system of equations that can be solved to reveal the possible dimensions. This approach not only demonstrates the power of algebra in geometry but also highlights the interplay between different measurement concepts.

Solving for Dimensions Let the rectangle’s length be (L) and its width be (W). The conditions translate to:

1. (L \times W = 64)  (area condition)
2. (2(L + W) = 64)  (perimeter condition)

From the perimeter equation we isolate one variable: [ L + W = 32 \quad\Rightarrow\quad W = 32 - L ]

For more on this topic, read our article on white black shirt mens or check out why do we dress up for halloween.

Substituting this expression for (W) into the area equation yields: [ L(32 - L) = 64 \

• L^{2} + 32L - 64 = 0 \ L^{2} - 32L + 64 = 0 ]

Solving the quadratic equation using the quadratic formula (\displaystyle L = \frac{32 \pm \sqrt{32^{2} - 4 \cdot 1 \cdot 64}}{2}) gives:

[ L = \frac{32 \pm \sqrt{1024 - 256}}{2} = \frac{3

Continuing from the quadratic solution, we compute the discriminant:

[ \sqrt{768} = \sqrt{256 \times 3} = 16\sqrt{3} ]

Substituting back, the solutions for (L) are:

[ L = \frac{32 \pm 16\sqrt{3}}{2} = 16 \pm 8\sqrt{3} ]

This yields two pairs of dimensions:

• Length (16 + 8\sqrt{3}) and width (16 - 8\sqrt{3})
• Length (16 - 8\sqrt{3}) and width (16 + 8\sqrt{3})

Numerically, these approximate to (29.Now, 856) units by (2. 144) units (or vice versa).