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Home , Base (geometry)

Tetrahedron A presentation by Shreya Majumdar, Talia Smith, and Caitlyn Sonn

Monday, September 14 Cohort 2 Introduction to the Problem The sides of ABC measure 11 cm, 20 cm, and 21 cm. Let P, Q, and R be the of the sides. Fold along PQ, QR, and RP until vertices A, B, and C coincide. Calculate the volume of the resulting . Finding of the base (pt. 1)

A , by definition splits a segment into two congruent line segments.

A The Triangle Midpoint Theorem: The line segment R connecting the midpoints of two sides of a triangle is parallel P to the third side and is congruent to one half of the third side.

C B Q ❖ So now we know that RQ has to be 5.5cm (½ of AB=11)

P R We are now left with a with a pair of opposite congruent side . For this to be true, the other pair of side lengths, need to be congruent as well, making it a B Q with oposite ❖ So now we know that PR has to be 10cm

R P The same can be applied to polygon PRCQ

C ❖ So now we know that PQ has to be 10.5 cm Q Finding Area of the base (pt.2)

Now that we know the three side lengths, we are able to use heron’s formula to find the area of the triangular base.

Heron’s Formula:

When:

By using side lengths a= 5.5 b= 10 c= 10.5 We are able to determine that the area of the base is = ≈ 27.04cm Triangle ABC Models A

21 cm 11 cm

B C 20 cm Finding Height (pt. 1)

Now that we know the area of the base triangle, we must find the height of the tetrahedron, or triangular , in order to satisfy the requirements of the volume formula. P R This diagram displays triangle ABC with AB= 11 cm, BC=20 cm, and CA=21 cm.

Q P, Q, and R are the midpoints of their corresponding sides, so triangle PQR connects the midpoints of each of the original triangle’s sides. Therefore, the ratios of side lengths between the original triangle and the smaller are the same. Triangle ABC is a similar figure to triangle PQR and the four smaller triangles within triangle ABC are congruent.

By definition, the tetrahedron that is formed when vertices A, B, and C touch, formed by folding along the sides of triangle PQR, is an isosceles tetrahedron because all four triangular faces are congruent. This also indicates that the two opposite sides (not touching) are Finding Height (pt. 2) parallel. Since the smaller triangle is formed from the midpoints of the original triangle, each of these inner segments is a midsegment of the original triangle, parallel to the base and exactly half of the base’s .

For example, BQ= 10 cm, PR=10 cm, and both sides are parallel. AR is R P parallel to PQ, and both sides are 10.5 cm. When these four sides are connected, comprised of two adjacent congruent triangles, they form a parallelogram.

Q When C folds along RQ to meet the point of the tetrahedron, it is follows a path in line with its , to the base, RQ. The altitude (height) is perpendicular to the base, so the vertex follows the circular path of the altitude as it reaches the top vertex of the tetrahedron. The same is true for each of the three slanted When the isosceles tetrahedron is formed by folding along triangles in the isosceles tetrahedron, and each has their own the lines that connect the midpoints of the original triangle, altitude. the opposite sides, which are not touching, have the same length. These corresponding sides are marked by the same All of the vertices A, B, and C will follow a path above their altitudes colors in the diagram. in order to meet each other at one point, the top vertex of the tetrahedron. Therefore, the point at which they intersect will be directly above the orthocenter of the large triangle. Outline of the parallelogram Altitudes of the three slanted The three altitudes intersect formed by the parallel lines of the faces (altitudes connect a vertex directly above the orthocenter of congruent triangle faces- lines of of the triangle to the opposite the original triangle same color are parallel side in a perpendicular intersection Finding Height (pt. 3) We needed to find the A coordinates of the orthocenter, and then use that to find the height h like so: D This is because we know the H right triangle formed by the side length of the tetrahedron E and the height will contain the B(0,0) C(20,0) hypotenuse and other leg on this section of the altitude (due to how the tetrahedron is built). Finding Height (pt. 4) Step 1: Find coordinates of A A(2, ) We start with the y coordinate, which is the same as the height from A since the base is on the x axis.

We find the height like so: D(2, )

Therefore, the H y coordinate is

E(2,0) C(20,0) B(0,0) Next is to find the x coordinate. We know that Step 2: Calculate coordinates of D the length of BE is the same as the x Since D lies on the midline, we can say that it is halfway coordinate due to B being at the origin, so to between A and E. solve we use:

Therefore, the x We can determine the coordinates of E by seeing that AE is perpendicular to a horizontal line, so AE is vertical with coordinate is 2 it’s equation being x = 2. Therefore, E is at (2,0), and D is at (2, ), or (2, ) Finding Height (pt. 5) Step 3: Find coordinates of H A(2, ) We find it by finding the intersection of the lines that define the altitudes from B and A. From before, we know the equation of the AE is x = 2.

D(2, ) To find the equation of the altitude from B, we can use the point-slope form. We know (0,0) H(2, ) lies on the altitude, and the slope is the negative reciprocal of AC by definition. The E(2,0) slope of AC is: C(20,0) B(0,0)

Step 4: Find height h So the slope of the altitude is

Now we find AD and DH:

Therefore the equation is:

AD = , HD = Now we have a system of two equations, and Now we plug them into the equation: solving gives the solution:

It becomes: From there, Finding Volume

Next we need to find the volume using the area of the base and the height. The tetrahedron is a type of pyramid, so we can use the volume formula of base times height divided by 3.

Plugging in the numbers, we get:

The final answer is that the volume of the tetrahedron is 45 cubic centimeters (cm^3).