Find the area

Given that the sides are

$a = 3 \\ b = 4 \\ c = 5$

Semiperimeter( $s$ ) = $\frac{a+b+c}{2}= \frac{3+4+5}{2} = 6 \ \text{units}$

Using Heron's Formula, we have

$\text{Area of triangle } = \sqrt{s(s-a)(s-b)(s-c)} \\ \Rightarrow \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} =\boxed{ 6 \text{ sq. units}}$

Since the triangle has side lengths of 3, 4, and 5, it is a right triangle with legs of length 3 and 4. Therefore, the area is $\frac{3\cdot 4}{2}=\boxed{6}$ .

The given triangle is a right triangle. The area of the triangle is given by $A=\dfrac{1}{2}bh$ where $b$ is the base and $h$ is the height of the triangle. So the area of the triangle is

$A=\dfrac{1}{2}(3)(4)=\boxed{6}$

Let the sides lengths be $a=3,b=4$ and $c=5$ . Let $C$ be the angle opposite side $c$ . By the law of cosines, we have

$5^2=3^2+4^2-2(3)(4)(\cos C)$

$C=90^\circ$

So the area is,

$A=\dfrac{1}{2}ab \sin C=\dfrac{1}{2}(3)(4)(\sin 90)=$ $\boxed{6}$