An isosceles triangle’s altitude will bisect its base. Which expression could be used to find the area of the isosceles triangle above?

Answer:[tex]\dfrac{\sqrt{40}\cdot \sqrt{40}}{2}[/tex]Step-by-step explanation:The length of the base is the distance between the points 4+2i and 10+4i, so[tex]\text{Base}=|10+4i-(4+2i)|=|10+4i-4-2i|=|6+2i|=\sqrt{6^2+2^2}=\\ \\=\sqrt{36+4}=\sqrt{40}[/tex]The middle point of the base is placed at point[tex]\dfrac{4+2i+10+4i}{2}=\dfrac{6i+14}{2}=7+3i[/tex]The length of the height is the distance between the points 5+9i and 7+3i[tex]\text{Height}=|5+9i-(7+3i)|=|5+9i-7-3i|=|-2+6i|=\sqrt{(-2)^2+6^2}=\\ \\=\sqrt{4+36}=\sqrt{40}[/tex]So, the area of the triangle is[tex]A=\dfrac{1}{2}\cdot \text{Base}\cdot \text{Height}=\dfrac{\sqrt{40}\cdot \sqrt{40}}{2}[/tex]