Pyramid Problem

**empireruler** · June 9th 2010, 03:29 PM

The base edges of a triangular pyramid are 25, 39, and 50 inches long, and all three lateral faces have a slant height of 24. What is the height of the pyramid?

I am really stuck on this problem. I don't think I did it right.

To start out with I found the circumcenter of the triangle (I figure the height projected onto the triangle would be its circumcenter. I then found the distance from the circumcenter to the sides of the triangle and was planning on using pythagorean theorum to find the height. However, the distance from the circumcenter to the side is greater than 24. Did I do something wrong?

**earboth** · June 10th 2010, 12:05 AM

Originally Posted by empireruler

The base edges of a triangular pyramid are 25, 39, and 50 inches long, and all three lateral faces have a slant height of 24. What is the height of the pyramid?

I am really stuck on this problem. I don't think I did it right.

To start out with I found the circumcenter of the triangle (I figure the height projected onto the triangle would be its circumcenter. I then found the distance from the circumcenter to the sides of the triangle and was planning on using pythagorean theorum to find the height. However, the distance from the circumcenter to the side is greater than 24. Did I do something wrong?

1. All slanted sides are isosceles triangles. The heights and the medians of these triangles are equal. (blue) Use Pythagorean theorem to calculate it's length.

2. The line segment from a vertex of the base triangle to footpoint of the median of the opposite slanted side must be a median too. (green) Use Cosine law to calculate the interior angles of the base triangle and consequently the length of the base median.

3. You now know the dimensions of the greyed triangle which contains the height H of the pyramid.

**empireruler** · June 10th 2010, 06:34 PM

Originally Posted by earboth

1. All slanted sides are isosceles triangles. The heights and the medians of these triangles are equal. (blue) Use Pythagorean theorem to calculate it's length.

2. The line segment from a vertex of the base triangle to footpoint of the median of the opposite slanted side must be a median too. (green) Use Cosine law to calculate the interior angles of the base triangle and consequently the length of the base median.

3. You now know the dimensions of the greyed triangle which contains the height H of the pyramid.

Thank you so much for your reply. Isn't the slant height what you marked as h? When I did as you said, I looked at the answer posted, and it was different. I got a height of around 7, but the answer says 22.48. Can you show how to get this? Again, thanks.

**earboth** · June 10th 2010, 10:56 PM

I've modified my first statement and I hope it is now better understandable what I meant:

Originally Posted by earboth

1. All slanted sides are isosceles triangles. Heights and medians of these triangles are the same. (blue) Use Pythagorean theorem to calculate it's length.

...

Originally Posted by empireruler

Thank you so much for your reply. Isn't the slant height what you marked as h? <<<<<< correct
When I did as you said, I looked at the answer posted, and it was different. I got a height of around 7, but the answer says 22.48. Can you show how to get this? Again, thanks.

1. Since the base lines of the isosceles triangles are all different the slant heights in these triangles must be different too.

2. I've attached a completed sketch which I use for the following calculations:

$h=\sqrt{24^2-12.5^2}=\sqrt{\frac{1679}4}\approx 20.4878$

$\cos(S)=\dfrac{50^2-39^2-25^2}{-2 \cdot 39 \cdot 25}~\implies~S\approx 100.4594^\circ$

$m=\sqrt{39^2+12.5^2-2\cdot 12.5 \cdot 39 \cdot \cos(S)}=\sqrt{1854.25}\approx 43.0610$

$\cos(T)=\dfrac{h^2-m^2-24^2}{-2 \cdot m \cdot 24}~\implies~T\approx 13.4189^\circ$

$\boxed{H=24 \cdot \sin(T)\approx 5.5696}$

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