Question

triangle rst, with vertices r(-6,-8), s(-3,-2), and t(-8,-4), is drawn inside a rectangle, as shown below. what is the area, in square units, of triangle rst?

Explanation:

Step1: Find the area of the rectangle

The length of the rectangle along the x - axis is the difference between the x - coordinates of the right - most and left - most points. The x - coordinates of the points are \(x_1=-8\) and \(x_2=-3\), so the length \(l = |-3-(-8)|=5\). The length of the rectangle along the y - axis is the difference between the y - coordinates of the top - most and bottom - most points. The y - coordinates are \(y_1=-2\) and \(y_2=-8\), so the width \(w=|-2 - (-8)| = 6\). The area of the rectangle \(A_{rect}=l\times w=5\times6 = 30\).

Step2: Find the areas of the three right - triangles outside \(\triangle RST\)

For the first right - triangle with vertices \(R(-6,-8)\), \(S(-3,-2)\) and the point \((-3,-8)\):
The base \(b_1=|-3-(-6)| = 3\) and the height \(h_1=|-2-(-8)|=6\). The area \(A_1=\frac{1}{2}\times b_1\times h_1=\frac{1}{2}\times3\times6 = 9\).
For the second right - triangle with vertices \(R(-6,-8)\), \(T(-8,-4)\) and the point \((-6,-4)\):
The base \(b_2=|-6-(-8)| = 2\) and the height \(h_2=|-4-(-8)| = 4\). The area \(A_2=\frac{1}{2}\times b_2\times h_2=\frac{1}{2}\times2\times4=4\).
For the third right - triangle with vertices \(T(-8,-4)\), \(S(-3,-2)\) and the point \((-8,-2)\):
The base \(b_3=|-3-(-8)| = 5\) and the height \(h_3=|-2-(-4)|=2\). The area \(A_3=\frac{1}{2}\times b_3\times h_3=\frac{1}{2}\times5\times2 = 5\).

Step3: Calculate the area of \(\triangle RST\)

The area of \(\triangle RST\) is \(A = A_{rect}-A_1 - A_2 - A_3\).
\(A=30-(9 + 4+5)=30 - 18=12\).

Answer:

12