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triangle of area 0.5 on a lattice grid

I'm trying to find the area of this triangle using the $\frac{1}{2} \times b \times h$ formula, but for some reason, it isn't quite working out. My workings:

My working:

working

$$\alpha = \sqrt{1^{2}+1^{2}} = \sqrt{2}$$ $$\beta = \sqrt{2^{2} + 2^{2}} = 2\sqrt{2}$$

$$\frac{1}{2} \times \sqrt{2} \times 2\sqrt{2} = 2?$$

I know the area is supposed to be 0.5 units^2, so what am I doing wrong here?

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    $\begingroup$ $\beta$ is clearly not the height if you take $\alpha$ to be your base. $\endgroup$
    – daruma
    Commented Jun 15, 2021 at 2:42
  • $\begingroup$ @Shane You can use Pick's theorem $\endgroup$
    – Asher2211
    Commented Jun 15, 2021 at 2:55

3 Answers 3

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The altitude is measured perpendicular to the (extended) base line. In your case, $h=\dfrac{\sqrt2}{2}$

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    $\begingroup$ If we call alpha the base then, by running the third point parallel to the base down to the bottom row, the height is $\frac{ \sqrt{2}} 2$ $\endgroup$
    – robert timmer-arends
    Commented Jun 15, 2021 at 4:19
  • $\begingroup$ i.sstatic.net/O7An3.png Like this? $\endgroup$
    – Shane Sarosh ℼℼℼ
    Commented Jun 15, 2021 at 5:38
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In your diagram, why did you take $\beta$ to be the diagonal of a $(2,2)$ square instead of the diagonal of a $(1,1)$ square? What it should be is the distance from the upper right end of $\alpha$ to the other side of the triangle. You can evaluate that by choosing one of the dots to be the origin of a coordinate system and writing the equations for $\beta$ and the base of the triangle, then finding the intersection. You should then find the distance from the upper right of $\alpha$ to that point to get the height of the triangle.

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You have misunderstood. $\alpha$ an $\beta$ must be at $90$ degree. BUt the angle between your $\alpha$ and $\beta$ is not $90$. Instead you could have taken this figure

enter image description here

Here $\alpha$ and $\beta$ form 90 degree angle. Just now use the formula $$ \frac{1}{2} \cdot \beta \cdot (\alpha + \text{dotted line length}) $$

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  • $\begingroup$ Dotted line length? That's not needed, is it? $\endgroup$
    – Shane Sarosh ℼℼℼ
    Commented Jun 16, 2021 at 4:16
  • $\begingroup$ See if you inclue in $\alpha$ length the dotted part then dotted part isn't needed however if you take alpha only as the connected line not the dotted line then it is not needed $\endgroup$
    – user876009
    Commented Jun 16, 2021 at 5:01

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