Is 43 a two digit odd number?

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payshencec21

05.02.2021

Check the picture below.as you know, the area of a triangle is (1/2)bh.now, the base for this one is the segment cd, and its altitude as we already know is ef. $\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ c(\stackrel{x_1}{3}~,~\stackrel{y_1}{-2})\qquad d(\stackrel{x_2}{-1}~,~\stackrel{y_2}{2})\qquad \qquad % distance value d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ cd=\sqrt{[-1-3]^2+)]^2}\implies cd=\sqrt{(-1-3)^2+(2+2)^2} \\\\\\ cd=\sqrt{(-4)^2+4^2}\implies cd=\sqrt{2(4^2)}\implies \boxed{cd=4\sqrt{2}}\\\\ /tex][tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ e(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{1}) \\\\\\ ef=\sqrt{(0-2)^2+(1-3)^2}\implies ef=\sqrt{(-2)^2+(-2)^2} \\\\\\ ef=\sqrt{2(2^2)}\implies \boxed{ef=2\sqrt{2}}\\\\ \stackrel{\textit{area of the triangle}}{\cfrac{1}{2}\cdot cd\cdot ef}\implies \cfrac{1}{2}(4\sqrt{2})(2\sqrt{2})\implies 4\sqrt{2}\cdot \sqrt{2} \\\\\\ 4(\sqrt{2})^2\implies 4\cdot 2\implies 8$

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