# Height of triangle

by cupcakes
Tags: height, triangle
 P: 18 1. The problem statement, all variables and given/known data Please see picture attached. 2. Relevant equations 3. The attempt at a solution If I use the pythagoras formula with the ratio I get the height as 3 yet it is longer than the side that is 4. Am I doing something wrong? Thanks. Attached Thumbnails
Mentor
P: 21,304
 Quote by cupcakes 1. The problem statement, all variables and given/known data Please see picture attached. 2. Relevant equations 3. The attempt at a solution If I use the pythagoras formula with the ratio I get the height as 3 yet it is longer than the side that is 4. Am I doing something wrong? Thanks.
Show us what you're doing. Where did you get one side being 4 in length?
P: 19
 Quote by Mark44 Show us what you're doing. Where did you get one side being 4 in length?
He is saying that he halved line AC in order to get a half-length of 4, and he used the Pythagorean theorem with side AB (or BC) in order to get length BD in terms of the other sides' lengths.

The problem is that the illustration is not drawn to scale. Side BC should be shorter than AC, as can be seen by the ratio. What to do next, I don't know.

Mentor
P: 21,304
Height of triangle

 Quote by Acala He is saying that he halved line AC in order to get a half-length of 4
That reasoning is flawed. It's given that the ratio of BC to AC is 5/8, but that doesn't mean that BC is 5 and AC is 8. For example, BC could be 1, and AC could be 1.6, or BC could be 2, with AC being 3.2. There are literally an infinite number of pairs of values of BC and AC that would give this ratio; you can't get the lengths from the ratio.
 Quote by Acala , and he used the Pythagorean theorem with side AB (or BC) in order to get length BD in terms of the other sides' lengths. The problem is that the illustration is not drawn to scale. Side BC should be shorter than AC, as can be seen by the ratio. What to do next, I don't know.
P: 19
 Quote by Mark44 That reasoning is flawed. It's given that the ratio of BC to AC is 5/8, but that doesn't mean that BC is 5 and AC is 8. For example, BC could be 1, and AC could be 1.6, or BC could be 2, with AC being 3.2. There are literally an infinite number of pairs of values of BC and AC that would give this ratio; you can't get the lengths from the ratio.
That is certainly correct. From this calculation, though, he obtained the ratio of the height to the other sides, which would be 3:5:8. I don't think it does any harm, either, to think of the triangle in terms of these lengths, since the answer has to be in terms of the radius of the circle anyway. Any of those pairs you named would maintain the necessary ratio. I'm still not getting any ideas on how to solve it, though.
 P: 19 I believe I've come up with something. The height in terms of the radius is going to be 2r for the portion of the height enclosed by the circle plus some remaining height x. If we solve for the ratio of the height to the other sides as mentioned previously, we get that the height is h, the side DC is (4/3)h and the side BC is (5/3)h. We know that the angles caused by the radius intersecting the side BC are each 90 degrees, because the side is tangent to the circle. We also know that the angle between the height and side DC is 90 degrees. Therefore length b (in attachment) is equal to side DC, which is (4/3)h. We know that the total length of side BC is (5/3)h, so length a (in attachment) is therefore (1/3)h. We know by Pythagorean's theorem, then, that a^2 + r^2 = (h-r)^2. If we substitute (1/3)h for a and then substitute (2r + x) for h, we get ((1/3)(2r+x))^2+r^2 = (x+r)^2. You should be able to solve for x and plug it into h = 2r + x to get the height in terms of r. I have never worked with inscribed circles before, so there might be an error in this. That is the way I would go about it, though. Note: attachment is NOT drawn to scale, whatsoever! Attached Thumbnails
 P: 18 thank you :)
 P: 339 I have a solution but it's not a very elegant one, nor is it even satisfactory. See attachment to refer to the variables. After naming the variables in the pic, you can write the following three equations. s2 - 6s + h2 - 8h - p2 + 10p = 0 s2 + h2 - p2 = 0 d2 = (3 - s - d)2 - h2 Attached Thumbnails
 P: 21 I have yet another idea of how to approach this: All lines from the circles centre is going to bisect each angle at the triangle's corners. So lets add one more point to the traingle where that extra line, r is touching line BC - we'll call that point E. Line OE is hitting BC at a right angle. Line OC bisects the angle BCA. Due to the ration of sides given we know that BC = 5x, CD = 4x, and BD = sqrt of (5x)^2 - (3x)^2 = 4x Now let's find some similar triangles. We can see that triangle OCD is the same as OCE so therefore CD =CE = 4x This means that the length of line BE is going to be 5x-4x = 1x Looking at triangle OBE we want to figure out the length of line OB (we already know that OD is = to r) The length of line OB is equal to r + some value we will call Z So looking at triangle OBE we see that: (x)^2 + (r)^2 = (r+Z)^2 So Z = sqrt((x)^2 + (r)^2) - r Substitute x for , say CD/4, we get the length of BD as: BD = sqrt((CD/4)^2 + (r)^2) - r + 2r BD = sqrt((CD/4)^2 + (r)^2) + r

 Related Discussions Special & General Relativity 31 Precalculus Mathematics Homework 6 Precalculus Mathematics Homework 5 Introductory Physics Homework 0