How do I use the right triangle altitude theorem according to these definitions, and what is an easy way to remember it? This is what it says in my textbook: Right Triangle Altitude Theorem - If...
How do I use the right triangle altitude theorem according to these definitions, and what is an easy way to remember it?
This is what it says in my textbook:
- Right Triangle Altitude Theorem - If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and each other.
- Corollary 1 of Right Triangle Altitude Theorem - When the altitude is drawn to the hypotenuse of a right triangle, the length of the altitude is the geometric mean between the segments of the hypotenuse.
- Corollary 2 of Right Triangle Altitude Theorem - When the altitude is drawn to the hypotenuse of a right triangle, each leg is the geometric mean between the hypotenuse and the segment of the hypotenuse that is adjacent to the leg. /|\ If two triangles are similar, then the ratio of any two corresponding segments (such as medians, altitudes, angle bisectors) equals the ratio of any two corresponding sides
I've never really understood this in geometry (Though this stuff is trigonometrical, but I'm learning it in geometry anyways). I need help! Thanks!
Much of this will refer to the attachment I showed. These may become somewhat more understandable if you break up the triangles. For instance, I drew a right triangle in red on MS Paint. I drew an altitude to the hypotenuse of it. That effectively makes 3 similar triangles; I will call them small, medium, and large. I separated them out in black.
In similar shapes, the ratio of sides are proportional. For instance, in the black triangles, in the smallest one, we could write a ratio of sides:
cd/db (ratio of small leg over large leg)
(Don't forget that cd and dc are the same; the labels are just backwards)
Also, on the large triangle, we can see that:
cb/ba (again, ratio of small leg over large leg)
In similar figures, these ratios (not the lengths but the ratios) would be equal, so:
cd/db = cb/ba
These can be used to solve a variety of problems. For instance, with this equation, if we are given 3 of the sides, then we can solve for the 4th part of the equation.
The first part means that the two triangles formed are similar to each other. This means that they are proportional to each other. There are many examples we can use to show this, one example based off your picture: `(JM)/(JL) = (LM)/(LK)`
This is the short side of each triangle divided by the hypotenuse.
The second part is just a rule, and you need to know what the geometric mean is for it to work. To find the geometric mean of two numbers you take the square root of them multiplied together. For example the geometric mean of 2 and 8 is `sqrt(2*8) = 4`
From your picture, the altitude LM equals the geometric mean of the parts of the hypotenuse JM and MK
I will do problem 23)
LM = 6, JM = 4, find MK.
We plug this into the formula `6 = sqrt(4*(MK))`
`36 = 4*MK`
`MK = 9`