Simple Unit Squares and Lines

Let $D$ be the point of intersection of the extension of $AC$ with the vertical through $B$ . Then $\Delta BCD$ and $\Delta ACB$ are similar, and since $|BD| = 1/2$ and $|AB| = 1$ the area of $\Delta ACB$ will be 4 tines that of $\Delta BCD$ . The area of $\Delta ABC$ will thus be $4/5$ the area of $\Delta ABD$ , which is $1/4$ , giving us an answer of $4/5 \times 1/4 = 1/5 = \boxed{0.2}$ .

The triangle sought is a right triangle with hypotenuse $AB=1$ . It is similar to the large right triangles formed, which have hypotenuse $\sqrt{1^{2}+2^{2}}=\sqrt{5}$ and area $\frac{1}{2}*1*2=1$ . The scale factor is thus $\frac{1}{\sqrt{5}}$ and so the area for the triangle is just the square of this. $\frac{1}{5}=\boxed{0.2}$

Since $\triangle BFG \cong \triangle ADE$ , $\angle ACB=90$ . Since $\triangle ACB \sim \triangle ADE$ , we have

$\dfrac{BC}{AC}=\dfrac{DE}{AD}$

$\dfrac{BC}{AC}=\dfrac{1}{2}$

$AC=2BC$

By pythagorean theorem on $\triangle ACB$ ,

$1^2=(BC)^2 + (AC)^2$

$1=(BC)^2+(2BC)^2$

$1=5(BC)^2$

$BC=\sqrt{\dfrac{1}{5}}$

It follows that, $AC=2\sqrt{\dfrac{1}{5}}$ .

The area of $\triangle ACB$ is

$A=\dfrac{1}{2}(CB)(AC)=\dfrac{1}{2} \left(\sqrt{\dfrac{1}{5}}\right) \left(2\sqrt{\dfrac{1}{5}}\right) = \dfrac{1}{5}=\boxed{0.2}$

Set up a coordinate system wih the origin at lower left.Call the point E.Let C be the intersection point,D lower right.E =(0,0), A= (0,3). B= (1,2), D = (2,1). Equation of line EB is y = 2x. Equation of line AD is y = (-1/2)x +2. The intersection point C = (.8,1.6). The product of slopes = -1, so the lines are perpendicular. CB = .447214, AC = .894417. Area of Area of ABC = (1/2) BC AC = .2 Ed Gray

Employ Cartersian coordinates - let the equations of the two diagonal lines be $y=2x-2$ and $y=-\dfrac{x}{2}$ where the topmost and leftmost corner is taken to be the origin. Now $x=-2y$ so we have:

$y=2(-2y)-2$

$y=-4y-2$

$(1+4)y=-2$

$5y=-2$

$y=-\dfrac{2}{5}$

So the height of the triangle is $\dfrac{2}{5}$ which means its area is $\dfrac{1}{2} \times 1 \times \dfrac{2}{5}=\dfrac{1}{5} =\boxed{0.2}$ .