Math Problem Solver

The area of a rhombus is $360\, \text{cm}^2$.

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

Given: $$\begin{aligned}P&=180\\h&=8\end{aligned}$$

Area

The area of a rhombus is given by the equation: $$\begin{aligned}A&=b h\end{aligned}$$ where $A$ is area, $b$ is base, and $h$ is height.

Need:$$\begin{aligned}P&=4 s\end{aligned} \quad \Rightarrow \quad \begin{aligned}180&=4 s\end{aligned} \quad \Rightarrow \quad \begin{aligned}s&=45\end{aligned}$$$$\begin{aligned}b&=s\end{aligned} \quad \Rightarrow \quad \begin{aligned}b&=\left(45\right)\end{aligned} \quad \Rightarrow \quad \begin{aligned}b&=45\end{aligned}$$

Solution:$$\begin{aligned}A&=b h\\A&=\left(45\right) \left(8\right)\\A&=360\end{aligned}$$

A rhombus is a two dimensional polygon with four equal sides, four angles, and four vertices. Opposite sides are parallel to each other. Opposite angles are of equal measurement. Each side has a length of s. Any side can be the base b.The height h is the perpendicular distance between the base and its parallel side. The diagonal is the distance between any two non-adjacent vertices, dividing the rhombus into two congruent triangles.

Let one angle be ∠A, and its adjacent angle be ∠B. The diagonal d₁ of a rhombus is the line segment opposite to ∠A. The diagonal d₂ of a rhombus is the line segment opposite to ∠B. $$d_{1} = s \sqrt{2 + 2 \cos A}=s \sqrt{2 - 2 \cos B}$$ $$d_{2} = s \sqrt{2 + 2 \cos B}=s \sqrt{2 - 2 \cos A}$$

The two diagonals of a rhombus are perpendicular to and bisect each other, forming four congruent right triangles. The hypotenuse of a right triangle is the side of the rhombus. $$s = \frac{\sqrt{d_{1}^{2} + d_{2}^{2}}}{2} = \sqrt{\left(\frac{d_{1}}{2}\right)^{2} + \left(\frac{d_{2}}{2}\right)^{2}}$$

The perimeter P of a rhombus is the sum of all the sides. $$\begin{aligned}P &= 4 s \\ &= 2 \sqrt{d_{1}^{2} + d_{2}^{2}} \end{aligned}$$

The area A of a rhombus is the region enclosed inside it. $$\begin{aligned} A &= b h \\ &= \frac{d_{1} d_{2}}{2} \\ &= \frac{1}{2} d_{1} \sqrt{4 s^{2} - d_{1}^{2}} = \frac{1}{2} d_{2} \sqrt{4 s^{2} - d_{2}^{2}} \end{aligned}$$

Classification: Polygon, Quadrilateral, Trapezoid, Parallelogram