geometrytrianglestrigonometry
Question: In triangle ABC, AC=6 and BC=8. It is also known that the median drawn from the vertices A and B are perpendicular to each other. Find the area of triangle ABC.
I realize that because the medians are perpendicular, right triangles are formed within triangle ABC. However, I cannot figure out how to calculate the length of the medians using the fact that there are right triangles.
The area is base times height.
Here is an only slightly dishonest proof. You will need scissors and paper. Produce a parallelogram $ABCD$. So that this parallelogram will look like the one I have in my head, the vertices $A$, $B$, $C$, $D$ are in counterclockwise order, the base $AB$ is at the bottom, and the point $A$ is on the left. Also, the parallelogram "leans" to the right, but does not lean by a ridiculous amount.
Let $P$ be the point where the perpendicular $D$ meets the line $AB$. If you are drawing the picture I have in mind, $P$ is between $A$ and $B$. Draw th perpendicular from $C$ to the extension of line segment $AB$. Suppose that this perpendicular meets $AB$ extended at $Q$.
OK, we are all set up. Note that triangles $APD$ and $BQC$ are congruent. This is easy, the angles match and $AD=BC$.
Now take a pair a scissors, cut out $\triangle APD$, and place the sliced off triangle so that it covers $\triangle BQC$. So we have sliced our parallelogram into two pieces, and reassembled the pieces to make rectangle $PQCD$. Thus the parallelogram and the rectangle have the same area. But the area of the rectangle is clearly $PQ$ times $DP$, which is $AB$ times $PD$.
Remark: My assertion that I cheated may be puzzling. But draw the parallelogram so that it leans ridiculously to the right, so that the perpendicular $DP$ meets the line $AB$ to the right of $B$. Then the argument just given breaks down. It can be fixed.
There is a lot of interesting mathematics connected with dissecting geometric figures, and rearranging the pieces to make another geometric figure. For example, there are very nice cut and paste proofs of the Pythagorean Theorem. You may also be interested in the Bolyai-Gerwien Theorem.
Let $X$, $Y$ and $Z$ be the side midpoints. Construct the parallelogram PYBZ and connect PC.
By construction, PY = BZ = AZ, PY || AZ and then PYZA is also a parallelogram, which leads to AP || ZY || XC and AP = XC, and in turn the parallelogram APCX. Thus, PC = AX and the sides of the triangle PCZ are the medians of ABC.
The parallelogram PYZA also yields AQ = QY = $\frac14$AC and then
$$\frac{Area_{PCZ}}{Area_{ABC}}=\frac{Area_{QCZ}}{Area_{AZC}}=\frac{QC}{AC} = \frac34 $$
Best Answer
Let $A'$ be the midpoint of $BC$; let $B'$ be the midpoint of $AC$. Let $M$ be the intersection point of the medians. We will find the lengths of the medians $AA'$ and $BB'$ from the fact that the medians are divided by the intersection point $M$ forming the ratio $2/1$, that is, $$ {AM\over A'M} = {BM\over B'M} = 2. $$ Suppose that $${1\over3}AA'=MA'=x, \qquad {1\over3}BB'=MB'=y.$$
By the Pythagorean theorem, from the right triangles $AMB'$ and $BMA'$ we have the system of equations $$ 4x^2 + y^2 = 9, $$ $$ x^2 + 4y^2 = 16. $$ Solving these equations we find $$ x = {2\over\sqrt3}, \qquad y = \sqrt{11\over3}. $$ Therefore, the medians are $$ AA' = 3x = {2\sqrt3}, \qquad BB' = 3y = \sqrt{33}. $$ We can find the third side, $AB$, from the right triangle $AMB$: $$ AB^2 = AM^2 +BM^2 = 4x^2 + 4y^2 = 20. $$ So we now know all tree sides of triangle $ABC$: $$ a = BC = 8, \qquad b=AC=6, \qquad c = AB = \sqrt{20}. $$ Finally, we can find the area of $ABC$ using an alternative form of Heron's formula: $$ \mbox{Area}(ABC)=\frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} =4\sqrt{11}\approx13.266. $$