What is the perimeter and area of a triangle? J(-5,6). K(3,4) L(-2,1)

Answer:Perimeter of triangle JKL: [tex]2 \sqrt{17} + 2\sqrt{34}[/tex].Area of triangle JKL: 17.Step-by-step explanation:None of the three sides of triangle JKL is parallel to either the x-axis or the y-axis. Apply the Pythagorean Theorem to find the length of each side.[tex]\rm JK = \sqrt{(3 - (-5))^{2} + (4- 6)^{2}} = \sqrt{8^{2} + (-2)^{2}} = \sqrt{68} = 2\sqrt{17}[/tex].[tex]\rm JL = \sqrt{(-2 - (-5))^{2} + (1- 6)^{2}} = \sqrt{3^{2} + (-5)^{2}} = \sqrt{34}[/tex].[tex]\rm KL = \sqrt{(-2 - 3)^{2} + (1-4)^{2}} = \sqrt{(-5)^{2} + (-3)^{2}} = \sqrt{34}[/tex].The perimeter of triangle JKL will be:[tex]\rm JK + JL + KL = 2\sqrt{17} + \sqrt{34} + \sqrt{34} = 2 \sqrt{17} + 2\sqrt{34}[/tex].Finding the Area of JKL:Method OneIn case you realized that [tex]\rm JK : JL : KL = \sqrt{2} : 1 : 1[/tex], which makes JKL an isosceles right triangle:Area of a right triangle: [tex]\begin{aligned}\displaystyle \rm Area &= \frac{1}{2} \times \text{First Leg} \times \text{Second Leg}\\ &=\frac{1}{2} \times \sqrt{34}\times\sqrt{34}\\&= 17\end{aligned}[/tex].Method TwoAlternatively, apply the Law of Cosines to find the cosine of any of the three internal angles. This method works even if the triangle does not contain a right angle.Taking the cosine of angle K as an example:[tex]\displaystyle\begin{aligned}\rm \cos{K}&=\frac{(\text{First Adjacent Side})^{2} + (\text{Second Adjacent Side})^{2}-(\text{Opposite Side})^{2}}{2\times (\text{First Adjacent Side})\times(\text{Second Adjacent Side})}\\&\rm =\frac{(JK)^{2} + (JL)^{2} -(KL)^{2}}{2\times JK \times JL}\\&=\frac{(2\sqrt{17})^{2}+(\sqrt{34})^{2}-(\sqrt{34})^{2}}{2\times\sqrt{34} \times(2\sqrt{17})}\\ &=\frac{2^{2}\times 17}{2\times \sqrt{2}\times\sqrt{17}\times 2\times \sqrt{17}}\\&=\frac{1}{\sqrt{2}}\end{aligned}[/tex].Apply the Pythagorean Theorem to find the sine of angle K:[tex]\displaystyle \rm \sin{K} = \sqrt{1 - (\cos{K})^{2}} = \sqrt{1 - \left(\frac{1}{\sqrt{2}}\right)^{2} } = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}[/tex].The height of JKL on the side JK will be:[tex]\displaystyle \rm KL \cdot \sin{K} = \sqrt{34} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{68}}{2} = \frac{2\sqrt{17}}{2} = \sqrt{17}[/tex].What will be the area of JKL given its height [tex]\sqrt{17}[/tex] on a base of length [tex]2\sqrt{17}[/tex]?[tex]\displaystyle \rm Area = \frac{1}{2} \times Base\times Height = \frac{1}{2}\times (2\sqrt{17})\times \sqrt{17} = 17[/tex].