$ABCD$ is a square, $M$ is the centroid of $\triangle ABC$. What is the area of $\triangle AMC$?

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

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!

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...

• 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. Commented May 19, 2021 at 13:51
• @Medi yeah I agree with gt6989b, I just calculated it, and OB is not equal to 3, it's 2sqrt(3) Commented May 19, 2021 at 13:54
• @Medi now better :) Commented May 19, 2021 at 13:55
• @Medi ok I edited my solution so that now it has the correct answer :) Commented May 19, 2021 at 13:58

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 = $$\frac{1}{2}*4\sqrt{3}*\frac{2\sqrt{3}}{3} = \boxed{4}$$.

• 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... Commented May 19, 2021 at 13:46
• @gt6989b I just answered the question... Commented May 19, 2021 at 13:47
• 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) Commented May 19, 2021 at 13:49
• @gt6989b I didn't think about it that way, well thanks for telling me! I'll keep that in my mind for the future. Commented May 19, 2021 at 13:50
• 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 :) Commented May 19, 2021 at 13:51