Math Problem Solver

The perimeter of a square is $24\, \text{cm}$.

Note: Approximate solutions are rounded to the nearest hundredths place.

Given: $$\begin{aligned}l&=6\end{aligned}$$

Perimeter

The perimeter of a square is given by the equation: $$\begin{aligned}P&=4 s\end{aligned}$$ where $P$ is perimeter and $s$ is side.

Need:$$\begin{aligned}l&=s\end{aligned} \quad \Rightarrow \quad \begin{aligned}6&=s\end{aligned} \quad \Rightarrow \quad \begin{aligned}s&=6\end{aligned}$$

Solution:$$\begin{aligned}P&=4 s\\P&=4 \left(6\right)\\P&=24\end{aligned}$$

A square is a two dimensional polygon with four equal sides, four right angles, and four vertices. Opposite sides are parallel to each other. Each side has a length of s. Less commonly, one pair of opposite sides is the length l, and the other pair is the width w, or base b and height h, respectively. The diagonal d is the distance between any two non-adjacent vertices, dividing the square into two congruent right triangles.

The diagonal d of a square is the hypotenuse of the right triangle it forms. $$\begin{aligned} d &= s \sqrt{2} = \sqrt{s^{2} + s^{2}} \\ &= \sqrt{l^{2} + w^{2}} \\ &= \sqrt{b^{2} + h^{2}} \end{aligned}$$

The perimeter P of a square is the sum of all the sides. $$\begin{aligned} P &= 4 s \\ &= 2 l + 2 w = 2 (l + w) \\ &= 2 b + 2 h = 2 (b + h) \\ &= 2 \sqrt{2} d \end{aligned}$$

The area A of a square is the region enclosed inside it. $$\begin{aligned} A &= s^{2} \\ &= l w \\ &= b h \\ &= \frac{1}{2} d^{2} \end{aligned}$$

Classification: Polygon, Quadrilateral, Trapezoid, Parallelogram, Rectangle, Rhombus