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Let $ABCD$ be a square with area $24$ ${cm}^2$. Point $M$ is the centroid of $\triangle ABC$. What is the area of $\triangle AMC$?

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If the side of the square is $a$, we have $$S_{ABCD}=a^2=24, a>0\Rightarrow a=\sqrt{24}=\sqrt{4\times6}=2\sqrt{6}.$$ Now I am not really good with areas, so I don't know what I am supposed to search for. Since $M$ is the centroid of $\triangle ABC$, the ratio $BM:MO=2:1$. The intersection $O$ of the diagonals is actually the midpoint of $AC$ and $BD$. Also the diagonal of every parallelogram bisects the area of the parallelogram. What else? Thank you in advance!

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2 Answers 2

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HINT

One way to find the area of a triangle is to measure its base and its height.

  • Base is $AC$. To find it, think about the following: if the side of a square is $s$, you can find the length $d$ of the diagonal from the Pythagorean Theorem.
  • Since you know $d = \overline{BD}$, you also know $\overline{OB}$ and since you are given the proportions, you can find $\overline{OM}$.

Now you know both base and height...

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    $\begingroup$ Thank you! That isn't very hard to come up with, to be honest. I don't know why I didn't try the easies way. We have $d=a\sqrt{2}=4\sqrt{3}$. I found that $MO=\dfrac{2\sqrt{3}}{3}$. The area is $4$ $cm^2$. I still think there's something better, e.g. without calculating the side of the square. $\endgroup$ Commented May 19, 2021 at 13:51
  • $\begingroup$ @Medi yeah I agree with gt6989b, I just calculated it, and OB is not equal to 3, it's 2sqrt(3) $\endgroup$ Commented May 19, 2021 at 13:54
  • $\begingroup$ @Medi now better :) $\endgroup$
    – gt6989b
    Commented May 19, 2021 at 13:55
  • $\begingroup$ @Medi ok I edited my solution so that now it has the correct answer :) $\endgroup$ Commented May 19, 2021 at 13:58
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Using the fact that $OB \perp AC$ (due to the diagonals of squares being perpendicular to each other), we have that $\angle{AOB} = 90$ degrees.

Hence, we can use Pythagorean Theorem to find out the length of $AO$. $AO^2 + OB^2 = AB^2$. Since $OB = \frac{DB}{2} = 2\sqrt{3}$, and $AB = 2\sqrt{6}$, $AO^2 + 2\sqrt{3}^2 = 2\sqrt{6}^2$.

Simplifying, we get $AO = 2\sqrt{3}$.

Since you mentioned $O$ is the center of $AC, AC = 2*AO = 4\sqrt{3}$.

The area of a triangle is $\frac{1}{2}*b*h$ where $b$ is the base and $h$ is the height.

$AC$ is the base and $OM$ is the height.

Hence, the area of the triangle = $\frac{1}{2}*4\sqrt{3}*\frac{2\sqrt{3}}{3} = \boxed{4}$.

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    $\begingroup$ I do not like you doing all the work for the OP. I don't think this is the way to teach people, just to encourage them to copy other people's work without thinking for themselves... $\endgroup$
    – gt6989b
    Commented May 19, 2021 at 13:46
  • $\begingroup$ @gt6989b I just answered the question... $\endgroup$ Commented May 19, 2021 at 13:47
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    $\begingroup$ exactly... doing parts of it and leaving the others for the OP to do himself would be (in my mind) preferable way to go -- OP then will do at least some work on his question... (for the record, I did not downvote) $\endgroup$
    – gt6989b
    Commented May 19, 2021 at 13:49
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    $\begingroup$ @gt6989b I didn't think about it that way, well thanks for telling me! I'll keep that in my mind for the future. $\endgroup$ Commented May 19, 2021 at 13:50
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    $\begingroup$ thanks for thinking about it -- I hope you stay around and contribute to the site. All of us are working to make this a better resource for people to learn and grow. Thank you for being a part of the effort :) $\endgroup$
    – gt6989b
    Commented May 19, 2021 at 13:51

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