Find the area of triangle OBC .
Hello Guest!
The triangle is symmetrical in the X-Y coordinate system with BC on the ordinate axis.
\(A(0,\sqrt{5^2-3.5^2})= A(0,\sqrt{12.75})\\ B(-3.5,\ 0)\\ C(3.5,\ 0)\\ M(\frac{7}{4},\frac{\sqrt{12.75}}{2})\ [center\ \overline{AC}]\\ m_{AC(x)}=-\frac{\sqrt{12.75} }{3.5}\\ m=\frac{3.5}{\sqrt{12.75}}\)
\(MO(x)=m_{}(x-x_M)+y_M\\ MO(x)=\frac{{3.5}}{\sqrt{12.75}}(x-\frac{7}{4})+\frac{3.5707}{2}\\ h=MO(0)=\frac{{3.5}}{\sqrt{12.75}}(0-\frac{7}{4})+\frac{\sqrt{12.75}}{2}\\ h=0.0700147\\ A_{OBC}=\dfrac{0,0700147\cdot 7}{2}\\ \color{blue}A_{OBC}=0.1225245\\ Finally\ solved!\\ (The\ circular\ function:\ f(x)=\sqrt{3.5^2+0.0700147^2-x^2}+0.0700147)\)
!
