Math Problem Solver

The perimeter of a rectangle is $26\, \text{cm}$.

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

Given: $$\begin{aligned}l&=8\\w&=5\end{aligned}$$

Perimeter

The perimeter of a rectangle is given by the equation: $$\begin{aligned}P&=2 l + 2 w\end{aligned}$$ where $P$ is perimeter, $l$ is length, and $w$ is width.

Solution:$$\begin{aligned}P&=2 l + 2 w\\P&=2 \left(8\right) + 2 \left(5\right)\\P&=26\end{aligned}$$

A rectangle is a two dimensional polygon with four sides, four right angles, and four vertices. Opposite sides are parallel to each other and are of equal length. 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 rectangle into two congruent right triangles.

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

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

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

Classification: Polygon, Quadrilateral, Trapezoid, Parallelogram