Altitude of the shorter side of a parallelogram-base of a right pyramid is divided in half by the point of intersection between base and height?

[Ognjen]

The geometric problem I came across left me stunned. For some reason, to me it seems like it's assuming that altitude of the shorter side of a parallelogram is practically the same as a line that connects midpoints of parallel ( and equal ) sides of a given parallelogram. Confusingly, this seems to be the assumption required to solve the problem at all.

Here's the problem:

The base of a pyramid is a parallelogram whose sides are 10cm and 15cm, and the area of its base equal to 90cm^2. The height of the pyramid passes through the intersection of the diagonals of the base and equals 6cm. Calculate the area of the wrapper of the pyramid.

The official solution instructs assignment of values 10 and 15 to sides a and b respectively. Then it names a point O, representing the orthogonal projection of the top point of the pyramid ( that it names S ) on the base of the pyramid. Then, it names orthogonal projections of the point O upon sides a and b - M and N, respectively. HOWEVER, now the official solution assumes the thing I don't understand: If we indicate altitudes of the parallelogram adequate to sides a and b using ha and hb respectively, then OM = ha/2 and ON = hb/2. I attached the picture of the pyramid given in the solution.

I don't quite see how altitude of the shorter side of the parallelogram even necessarily crosses the center of the base ( point of intersection between the base and the height of the pyramid ), let alone ( and this is the most important point ) why that altitude divides the side it passes from and to in half.

I would greatly appreciate any help in understanding this conceptually, as the latter part of the solution to this, once the foregoing dilemma is dispelled, is quite trivial, for which reason I didn't bother showing it here ( In short, it consists of simply getting the values of ha and hb from the value of the base and the two sides, then applying Pythagorean theorem to deduce h1 and h2, ultimately using them alongside the values of sides given in the problem itself to calculate the areas of the 4 triangles.

Thanks in advance.

 

[Dr.Peterson]

The official solution instructs assignment of values 10 and 15 to sides a and b respectively. Then it names a point O, representing the orthogonal projection of the top point of the pyramid ( that it names S ) on the base of the pyramid. Then, it names orthogonal projections of the point O upon sides a and b - M and N, respectively. HOWEVER, now the official solution assumes the thing I don't understand: If we indicate altitudes of the parallelogram adequate to sides a and b using ha and hb respectively, then OM = ha/2 and ON = hb/2. I attached the picture of the pyramid given in the solution.

I don't quite see how altitude of the shorter side of the parallelogram even necessarily crosses the center of the base ( point of intersection between the base and the height of the pyramid ), let alone ( and this is the most important point ) why that altitude divides the side it passes from and to in half.

Here is a picture of the base alone:

​

I think your main error is in thinking that there is just one "altitude" of a parallelogram to a given side. Any segment parallel to two opposite sides can be called an altitude; they are just choosing the one passing through O. It has nothing to do with midpoints. (In fact, as you see in the picture, it doesn't have to intersect the side itself at all!)

Secondly, they don't say the altitudes divide the sides in half. The division by 2 is because OM and ON are half the altitudes h_a and h_b. (I'm assuming those are meant to be subscripts.)

Their picture, however, is horrendous. It doesn't show the base as being a parallelogram. It is very misleading.

 

[Dr.Peterson]

I should add that they are actually using the word "altitude" not of a segment at all, but of the distance between parallel sides. But, as I said, you can think of particular altitudes (segments) passing through O:

​

The altitude h_a is the length of MP, and h_b is the length of NQ. O bisects each of them.

 

[Ognjen]

Here is a picture of the base alone:

View attachment 32976​

I think your main error is in thinking that there is just one "altitude" of a parallelogram to a given side. Any segment parallel to two opposite sides can be called an altitude; they are just choosing the one passing through O. It has nothing to do with midpoints. (In fact, as you see in the picture, it doesn't have to intersect the side itself at all!)

Secondly, they don't say the altitudes divide the sides in half. The division by 2 is because OM and ON are half the altitudes h_a and h_b. (I'm assuming those are meant to be subscripts.)

Their picture, however, is horrendous. It doesn't show the base as being a parallelogram. It is very misleading.

The reason why I assumed the altitude divides the appropriate side in half is because the next part of the problem includes application of Pythagorean theorem: h1 = √(H^2 + (ha/2)^2). This assumes that the altitude of the wrapper triangle meets the altitude of the appropriate side which passes through O: this is why I assumed the necessity of the base altitude dividing the side in half ( as far as I know, these are all isosceles triangles as wrappers, so its altitude begins at half of its base ( base of the triangle ) ).

Even if I'm mistaken and the wrapper triangles aren't isosceles, how is it so certain that a or b's altitude that passes through O is EXACTLY the one that interscepts the altitude h1 / h2 ?

 

[Dr.Peterson]

The reason why I assumed the altitude divides the appropriate side in half is because the next part of the problem includes application of Pythagorean theorem: h1 = √(H^2 + (ha/2)^2). This assumes that the altitude of the wrapper triangle meets the altitude of the appropriate side which passes through O: this is why I assumed the necessity of the base altitude dividing the side in half ( as far as I know, these are all isosceles triangles as wrappers, so its altitude begins at half of its base ( base of the triangle ) ).

Even if I'm mistaken and the wrapper triangles aren't isosceles, how is it so certain that a or b's altitude that passes through O is EXACTLY the one that interscepts the altitude h1 / h2 ?

The term "wrapper" is not standard; I assume it means the lateral surface. The lateral triangles are not isosceles. That's the false assumption you are making from the picture. So only your last question is appropriate.

The particular formula you identify, h₁ = √(H^2 + (h_a/2)^2), gives the altitude of the lateral triangle; but it does not assume it is isosceles. Here is the triangle on side a corresponding to my picture (my vertices in both are labeled incorrectly):

​

Altitude h₁ is calculated from right triangle SOM, and has nothing to do with the shape of this triangle.

The picture they provided shows a rectangular pyramid (making M look like the midpoint, the lateral triangles isosceles, etc.), but the angle relationships shown are still applicable to the actual pyramid. The plane SOM is perpendicular to side a, so that angle SMB is also a right angle, making SM an altitude.

 

[Ognjen]

The term "wrapper" is not standard; I assume it means the lateral surface. The lateral triangles are not isosceles. That's the false assumption you are making from the picture. So only your last question is appropriate.

The particular formula you identify, h₁ = √(H^2 + (h_a/2)^2), gives the altitude of the lateral triangle; but it does not assume it is isosceles. Here is the triangle on side a corresponding to my picture (my vertices in both are labeled incorrectly):

View attachment 32989​

Altitude h₁ is calculated from right triangle SOM, and has nothing to do with the shape of this triangle.

The picture they provided shows a rectangular pyramid (making M look like the midpoint, the lateral triangles isosceles, etc.), but the angle relationships shown are still applicable to the actual pyramid. The plane SOM is perpendicular to side a, so that angle SMB is also a right angle, making SM an altitude.

Thank you so much, it is all clear now !