Math Problem Solver

The area of a triangle is $\frac{209}{5} = 41.8\, \text{cm}^2$.

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

Given: $$\begin{aligned}s_{1}&=b=19\\h&=b=\frac{22}{5} = 4.4\end{aligned}$$

Area

The area of a triangle is given by the equation: $$\begin{aligned}A&=\frac{1}{2} b h\end{aligned}$$ where $A$ is area, $b$ is base, and $h$ is height.

Solution:$$\begin{aligned}A&=\frac{1}{2} b h\\A&=\frac{1}{2} \left(19\right) \left(\frac{22}{5}\right)\\A&=\frac{209}{5}\end{aligned}$$

A triangle is a two dimensional polygon with three sides, three angles, and three vertices. Any side can be called the base b, and the corresponding height h, the distance between a point on the base and the vertex opposite to the base, is perpendicular to the base.

The perimeter P of a triangle is the sum of all the sides. $$P = s_{1} + s_{2} + s_{3}$$

The area A of a triangle is the region enclosed inside it. $$A = \frac{1}{2} b h$$

Heron's Formula

The area A of a triangle where all three sides are known is $$A = \sqrt{p (p-s_{1}) (p-s_{2}) (p-s_{3})}$$ where p is half the perimeter.

Note: Triangle Inequality Theorem

The sum of the length of any two sides of a triangle is greater than the length of the third side.

Classification: Polygon