# Which Triangle has a larger altitude?

1. Nov 14, 2009

### um0123

1. The problem statement, all variables and given/known data

Which of these two triangles have a higher altitude (height)?

NOTE: BOTH TRIANGLES ARE EQUILATERAL!

Triangle 1 (all of the legnths going to the center are 665)

Triangle 2 (lengths of each line going to center writen)

2. Relevant equations

none

3. The attempt at a solution

i thought that the height of the first one is 1330 (twince of 665) but that cant be right, and i dont understand how to solve the second.

2. Nov 14, 2009

### SeanGillespie

I'm confused by your question, as an altitude on a triangle tends to be the line from one point and perpendicular to the opposite side. In your first triangle it appears to meet that condition, but in the second it does not. If the triangles are equilateral then the perpendicular line should be in the centre of each side. The second triangle does not appear to meet this criteria.

3. Nov 14, 2009

### um0123

all im asking is for the height of the triangle given those measurements.

4. Nov 15, 2009

### CarlAK

The center of a triangle is 1/3 the distance along each altitude. So since the center of each triangle is 665 from the base, the vertical(?) altitude of each is 3 times that, 1995.

But the second triangle is not equilateral. It has sides of 2303.6, 2307.1, and 2300.2
First triangle has all sides 2303.6.

5. Nov 15, 2009

### sreeram

how did you find the sides of second triangle?

6. Nov 15, 2009

### um0123

im confused also.

7. Nov 15, 2009

### CarlAK

Actually I drew it in a CAD program. Given a "center" point 665 above a base (side); draw circles centered on that point of radius 2* 664 & 2*666 to intersect the base at the vertices, draw 2 altitudes of 3*664 & 3*666 from vertices through center point, then can draw the 2 sides.

8. Nov 15, 2009

### um0123

but that doesnt help me actually solve the problem....

9. Nov 15, 2009

### sreeram

Okay, once you have all the sides of the second triangle, apply cosine rule
a*a=b*b+c*c-2*b*cosA where a, b and c are the sides opposite to angles A, B and C of triangle ABC. Thus you can find an angle of the triangle. then apply trigonometry. altitude=side*sinA. This should solve your problem

10. Nov 15, 2009

### um0123

how do i get all the sides of the second triangle?

11. Nov 15, 2009

### CarlAK

If you draw it with all the known information I think the solution will come to you..
Draw all three altitudes, which then forms 6 smaller triangles.
you know the length of all 3 altitudes, as well as the distance between the vertex & the center of each (which is 2/3 of the altitude).
Using the pythagorean theorem, you find the length of all other sides.

12. Nov 15, 2009

### um0123

but when u draw it it doesnt form 6 smaller triangles it forms 3 smaller quadralaterals. im confused!!!!

13. Nov 15, 2009

### CarlAK

OK in looking closer I see that I used a wrong assumption, that an altitude passes through the center of the triangle. That's only for an equilateral triangle. But, it is true that the distance from a center to a side (perpendicular) is one-third the altitude. So, you can still conclude both triangles have the same altitude. But what the triangle dimensions are, I don't see the derivation.

14. Nov 15, 2009

### um0123

no, they are both equilateral but only the first one has an altitude of 3 times the center to the side. The second one has different legnth to the center but is somehow still equilateral.

Last edited: Nov 16, 2009
15. Nov 16, 2009

### CarlAK

Well I just have to disagree about the second triangle being equilateral, isn't possible if those dimensions are distances to the centroid. Every triangle has a perpendicular distance to the centroid of 1/3 the perpendicular height, for each side. So I still contend that both triangles have the same altitude.

I also redrew it more accurately in CAD, so that the centroid is 1/3 along each altitude, and each angle bisector passes through the centroid, and get the same triangle sides I gave in Post #4, just to more decimal places. I don't know how to solve it directly, in CAD I iterated on an accurate solution. The second triangle is "similar" to the first, with its top vertex displaced about 4.5 to the right.

**edit - attached dimensioned triangle**

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• ###### Triangle.pdf
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Last edited: Nov 16, 2009
16. Nov 16, 2009

### willem2

Suppose the points at the base of the perpendiculars are A, B and C, the middle of the triangle is M. The corner between A and B is called D, between B and C is called E and between C and A is called F
Known are lenghts AM, BM and CM which we'll call a, b and c

The tetrangle AMBD has two adjacent sides known and all the corners are known, so it should be possible to find the other 2 sides. I did this by using coordinates, setting the corner D at (0,0) calling AD = d, so point A is at (d,0) and point M is at (d,a). Now find the point of intersection of the lines BD: y = sqrt(3)x and MB: y = (1/3)sqrt(3)(d-x) + a.
the distance from the intersection point found and M should be equal to b. This equation allows you to find AD. The same argument allows you to find AF from the tetrangle AMCF, and AD+AF is a side of the triangle, since it's equilateral, the altitude is (1/2)sqrt(3) times the side.

17. Nov 16, 2009

### SeanGillespie

If both triangles are equilateral, it is safe to say that on the first the lines are serving as both the altitudes and the medians. On the second it seems that the lines are neither altitudes or medians.

My assumption is that the lines do not intercept with the corners in the second triangle, as this would change the dimensions of the sides and mean that the triangle is not equilateral.

18. Nov 16, 2009

### willem2

After a dense page of calculations the answer is surprisingly simple:

$$AD = \frac {1}{3}\sqrt{3}(a+2b)$$
$$AF = \frac {1}{3}\sqrt{3}(a+2c)$$
$$BD = \frac {1}{3}\sqrt{3}(b+2a)$$
$$BE = \frac {1}{3}\sqrt{3}(b+2c)$$
$$CE = \frac {1}{3}\sqrt{3}(c+2b)$$
$$CF = \frac {1}{3}\sqrt{3}(c+2a)$$

the length of al the sides is $$\frac {2}{3}\sqrt{3}(a+b+c)$$

so the altitude is $$\frac {1}{2}\sqrt{3}(a+b+c)$$

(a+b+c) is the same for both problems, so the altitude is the same