Description : Find the area of the quadrilateral whose vertices are (3, 4), (0, 5), (2, –1) and (3, –2). -Maths 9th
Last Answer : Let A(-36, 7), B(20, 7) and C(0, -8) be the vertices of the given triangle.Then, a = BC = \(\sqrt{(0-20)^2+(-8-7)^2}\) = \(\sqrt{400+225}\) = \(\sqrt{625}\) = 25b = AC =\(\sqrt{(0-36)^2+(-8-7 ... (rac{-900+780}{120},rac{175+273-448}{120}\bigg)\) = \(\bigg(rac{-120}{120},rac{0}{120}\bigg)\) = (-1, 0)