geometrytriangles
The area of a right triangle is equal to $2 \sqrt{3}$. Determine its height projected to the hypotenuse if it divides the right angle in a ratio of 1:2.
I don't really understand how to obtain the height projected to the hypotenuse here. I've tried using the formulas $S = \frac12ab$ and $S= \frac12ch_c$ where a and b are the legs, c is the hypotenuse and $h_c$ is the height projected to the hypotenuse. Answer given is $\sqrt3$.
Can anybody help me here?
From the Wikipedia article on Thales' theorem:
If you know the hypotenuse, you know the base of this diagram; if you also know the area, then you can compute the height (from $A=\frac12 bh$). Then you can draw a horizontal line at the required height; where it intersects the circle is the vertex with the right angle.
Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.
However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.
In this case, $8^2 + 8^2 = (8\sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $\frac 12 (8)(8) = 32$.
(In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8\sqrt 2)$, you can apply the converse to simplify your work a bit).
Best Answer
The altitude to the hypotenuse of a right triangle divides the triangle into two similar copies of itself. So knowing that the altitude divides the right angle in the ratio $1:2$, we conclude that the right triangle's acute angles are $30^\circ$ and $60^\circ$.
Let $x$ and $x\sqrt3$ represent the lengths of the legs of the right triangle. We know that the area of the triangle is $\frac12\cdot x\cdot x\sqrt 3=2\sqrt3$, so $\frac12 x^2=2$ and thus $x=2$. The hypotenuse of the triangle is therefore $4$ and the height to the hypotenuse is given by $\frac12\cdot 4\cdot h=2\sqrt 3$ or $h=\sqrt3$.