+14905 Find the area of triangle ACE.
Hello Guest!
\(A=\int_{0}^{17} \! f(x) \, dx+\int_{17}^{30} \! g(x) \, dx-\int_{0}^{30} \! h(x) \, dx\)
\(m_f=-\frac{4}{17}\\ f(x)=m(x-x_1)+y_1\\ f(x)=-\frac{4}{17}(x-0)+17\\ \color{blue}f(x)= -\frac{4}{17}x+17 \\ m_g=-1\)
\(g(x)=-1(x-17)+13\\ \color{blue}g(x)=-x+30\\ m_h=-\frac{17}{30}\\ h(x)=-\frac{17}{30}(x-0)+17\\ \color{blue}h(x)=-\frac{17}{30}x+17\)
\({ \color{blue}A_f}=\int_{0}^{17} \! ( -\frac{4}{17}x+17) \, dx=\ _0^{17}| -\frac{4}{2\cdot 17}x^2+17x|\color{blue}=255\)
\({\color{blue}A_g}=\int_{17}^{30} \! (-x+30) \, dx=\ _{17}^{30}|-\frac{x^2}{2}+30x|=450-365.5\color{blue}=84.5\)
\({\color{blue}A_h}=\int_{0}^{30} \! (-\frac{17}{30}x+17) \, dx =\ _0^{30}|-\frac{17x^2}{2\cdot 30}+17x| \color{blue}=255\)
\(A=A_f+A_g+A_h=255+84.5-255\)
\(A=84.5\)
The area of triangle ACE.is 84.5
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+14905 Find the area of triangle ACE.
Hello Guest!
This is how it works easier:
\(\Delta_{ACE}= ◻ _{ABFG+CDEF}-\Delta_{ABC}-\Delta_{CDE}-\Delta_{AEG}\)
\({\color{blue}\Delta _{ACE}}=17^2+13^2-\frac{17(17-13)}{2}-\frac{13^2}{2}-\frac{17(17+13)}{2}\color{blue}=84.5\)
The area of triangle ACE is 84.5 !
