A triangle has coordinates (0,0), (1,2), and (3,1). Find the perimeter and the area of the triangle.
+14913 A triangle has coordinates A(0,0), C(1,2), and B(3,1). Find the perimeter and the area of the triangle.
Hello Guest!
The triangle ABC has the sides a, b, c.
a, b and c are sections of linear functions in the rectangular coordinate system.
Then:
\(A=\int_0^1f_b(x)dx+\int_1^3 f_a(x)dx-\int_0^3 f_c(x)dx\)
\(f_b(x)=2x\\ f_c(x)= \frac{1}{3}x\)
\(f_a(x)=\frac{y_B-y_C}{x_B-x_C}(x-x_C)+y_C\\ f_a(x)=\frac{-1}{2}(x-1)+2=- \frac{1}{2}x+\frac{5}{2}\\ \color{blue} f_a(x)=-\frac{1}{2}x+\frac{5}{2}\\ \)
\({\color{blue}\int_0^1f_b(x)dx}=\int_0^1(2x)dx=| x^2|_0^1=1-0\color{blue}=1\)
\({\color{blue}\int_1^3 f_a(x)dx}=\int_1^3(-\frac{1}{2}x+\frac{5}{2})dx=|-\frac{1}{4}x^2+\frac{5}{2}x|_1^3\\ =(-\frac{9}{4}+\frac{15}{2})-(-\frac{1}{4}+\frac{5}{2})\color{blue}=3\)
\({\color{blue}\int_0^3 f_c(x)dx}=\int_0^3( \frac{1}{3}x)dx=|\frac{1}{6}x^2 |_0^3=\frac{9}{6}-0\color{blue}=\frac{3}{2}\)
\(A=\int_0^1f_b(x)dx+\int_1^3 f_a(x)dx-\int_0^3 f_c(x)dx=1+3-\frac{3}{2}\)
\(\large A=\frac{5}{2}\)
\(P=a+b+c\)
\(a=\sqrt{(2-1)^2+(3-1)^2}=\sqrt{1+4}=\sqrt{5}\\ b=\sqrt{1^2+2^2}=\sqrt{5}\\ c=\sqrt{1^2+3^2}=\sqrt{10}=\sqrt{2}\cdot\sqrt{5}\)
\(\large {\color{blue} P=a+b+c}=\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}=\color{blue}\sqrt{5}\cdot(\sqrt{2}+2)\)
\(\large P=7.6344\)
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