Question

I just need the area in square units of triangle rst

Answer 1

A = 5 square units

Step-by-step explanation:

Vertices: R(-5, 5), S (-3, 8) and T (-7, 7)

We need to find the three sides of the triangle using the distance formula:

`dis\tan ce\text{ = }\sqrt[]{(y_2-y_1)^2+(x_2-x_1)^2}`
`\begin{gathered} \text{RS: }x_1=-5,y_1=5,x_2=-3,y_2\text{ = }8 \\ \text{Distance RS = }\sqrt[]{(8-5)^2+(-3-(-5))^2}\text{ = }\sqrt[]{3^2+2^2} \\ \text{Distance RS = }\sqrt[]{13} \\ \\ ST\colon\text{ }x_1=-3,y_1=8,x_2=-7,y_2\text{ = }7 \\ \text{Distance ST = }\sqrt[]{(7-8)^2+(-7-(-3))^2}\text{ = }\sqrt[]{-1^2+(-4)^2} \\ \text{Distance ST = }\sqrt[]{1\text{ + 16}}\text{ = }\sqrt[]{17} \\ \\ RT\colon\text{ }x_1=-5,y_1=5,x_2=-7,y_2\text{ = }7 \\ \text{Distance RT = }\sqrt[]{(7-5)^2+(-7-(-5))^2}\text{ = }\sqrt[]{2^2+(-2)^2} \\ \text{Distance RT = }\sqrt[]{4+4}\text{ = }\sqrt[]{8} \end{gathered}`

We have the 3 sides, we will use Heron's formula for the area:

`\begin{gathered} \text{Area = }\sqrt[]{s(s-RS)(s-ST)(s-RT} \\ s\text{ = }\frac{RS\text{ + ST + RT}}{2} \\ s\text{ = }\frac{\sqrt[]{13}\text{ + }\sqrt[]{17}\text{ + }\sqrt[]{8}}{2}\text{ = }\frac{\text{10.56}}{2}\text{ = 5.28} \end{gathered}`
`\begin{gathered} \text{Area = }\sqrt[]{5.28(5.28-\sqrt[]{13})(5.28-\sqrt[]{17})(5.28-\sqrt[]{8})} \\ \text{Area = }\sqrt[]{25.0752} \\ \text{Area = 5.0075 } \\ \text{Area is appro x i}mately5units^2 \end{gathered}`