The Area of a Right Triangle
The area of any triangle of base B and height H is:
[tex]A=\frac{B\cdot H}{2}[/tex]
The base and the height must be perpendicular, i.e, the angle between them must be 90°.
The trick here is to prove the triangle is right at the point (15, 5).
If two lines are perpendicular, the product of their slopes is -1.
Calculate the slope of the line that joins the vertices at (5, 15) and (15, 5):
[tex]\begin{gathered} m_1=\frac{5-15}{15-5} \\ m_1=-\frac{10}{10} \\ m_1=-1 \end{gathered}[/tex]
Calculate the slope of the line that joins the vertices at (20, 10) and (15, 5);
[tex]\begin{gathered} m_2=\frac{5-10}{15-20} \\ m_2=\frac{-5}{-5} \\ m_2=1 \end{gathered}[/tex]
It can be verified that m1 * m2 = -1, thus the lines are perpendicular and we can use the formula given above to compute the area.
Find the length of both lines with the formula of the distance:
[tex]\begin{gathered} L_1=\sqrt[]{(5-15)^2+(15-5)^2} \\ \text{Calculating:} \\ L_1=\sqrt[]{200} \end{gathered}[/tex][tex]\begin{gathered} L_2=\sqrt[]{(5-10)^2+(15-20)^2} \\ L_2=\sqrt[]{50} \end{gathered}[/tex]
Apply the formula of the area:
[tex]\begin{gathered} A=\frac{\sqrt[]{200}\cdot\sqrt[]{50}}{2} \\ A=\frac{\sqrt[]{10000}}{2} \\ A=\frac{100}{2}=50 \end{gathered}[/tex]
The area is 50 square units
