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Pls Help, Due Tmrow

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1)If the length of each edge of regular octahedron ABCDEF is 3, then find [ACFE].

Picture: https://latex.artofproblemsolving.com/1/8/4/184ede3e7e603b719f94023035595968f5fc9ae8.png

2)The face diagonal of a cube has length 1. Find the surface area of the cube.

3)Let ABCDEFGH be a cube. Find angle ACH in degrees.

Picture: https://latex.artofproblemsolving.com/e/1/4/e143cdd3d28a9ccb962862481c5b7034ab606234.png

4)The four vertices of a regular tetrahedron are snipped off, leaving a triangular face in place of each corner and a hexagonal face in place of each original face of the tetrahedron. How many edges will the new polyhedron have?

5)One square from the net needs to be removed so the remaining squares are still connected and can be folded into a cube. What is the number on that square? (Note: Two squares touching only at a vertex are not considered to be connected.)

Picture: https://latex.artofproblemsolving.com/7/2/3/72319d98a4c9b2d3be871999654a833b5e35f32d.png

6)Chris discovered that the centers of the faces of a regular icosahedron are also the vertices of another regular polyhedron. How many edges does the other regular polyhedron have?

7)A regular tetrahedron is attached to each face of a regular icosahedron, forming a new polyhedron. How many edges does the new polyhedron have?

8)The center of a regular octahedron is 1 foot away from every vertex. What is the volume of the octahedron in cubic feet?

9)The sum of all edge lengths of a regular tetrahedron is equal to that of a regular icosahedron. The surface area of the tetrahedron is 1. What is the surface area of the icosahedron?

10)The edge length of a regular tetrahedron ABCD is  2. M and N are the midpoints of BC and AD, respectively. Find MN.

Picture:https://latex.artofproblemsolving.com/f/c/7/fc773aa8663fd5b11de6f2ff3bf8cf921c35eb16.png

11)The midpoints of all edges of a regular tetrahedron are all vertices of another regular polyhedron. If the volume of the tetrahedron is 120, then what is the volume of the other regular polyhedron?

FiestyGeco  Mar 28, 2018
#1
+20022
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3)Let ABCDEFGH be a cube. Find angle ACH in degrees.

Picture: https://latex.artofproblemsolving.com/e/1/4/e143cdd3d28a9ccb962862481c5b7034ab606234.png

$$\text{Let \vec{c}=\begin{pmatrix} 0\\0\\0\end{pmatrix} } \\ \text{Let \vec{a}=\begin{pmatrix} 1\\-1\\0\end{pmatrix} } \\ \text{Let \vec{h}=\begin{pmatrix} 1\\0\\-1\end{pmatrix} }$$

$$\begin{array}{|rcll|} \hline \tan{ (\angle ACH) }&=& \dfrac{|\vec{a}\times \vec{h}|}{\vec{a}\cdot \vec{h} } \\\\ &=& \dfrac{\left|\begin{pmatrix} 1\\-1\\0\end{pmatrix} \times \begin{pmatrix} 1\\0\\-1\end{pmatrix} \right|}{\begin{pmatrix} 1\\-1\\0\end{pmatrix} \cdot \begin{pmatrix} 1\\0\\-1\end{pmatrix} } \\\\ &=& \dfrac{\left| \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\right| }{\begin{pmatrix} 1\\-1\\0\end{pmatrix} \cdot \begin{pmatrix} 1\\0\\-1\end{pmatrix} } \\\\ &=&\dfrac{ \sqrt{3} }{1+0+0 } \\\\ &=&\dfrac{ \sqrt{3} }{1 } \\\\ &=& \sqrt{3} \\\\ \angle ACH &=& \arctan{(\sqrt{3})} \\ \mathbf{ \angle ACH } & \mathbf{=} & \mathbf{60^{\circ}} \\ \hline \end{array}$$

heureka  Mar 29, 2018
edited by heureka  Apr 3, 2018
#2
+89781
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2)The face diagonal of a cube has length 1. Find the surface area of the cube.

The side  is     1/√2

So....the surface area  =  6*side^2   =  6 * (1/√2) ^2   =  6* (1/2)  =  3 units^2

CPhill  Mar 29, 2018
#3
+89781
+2

8)The center of a regular octahedron is 1 foot away from every vertex. What is the volume of the octahedron in cubic feet?

The octahedron will form two congruent pyramids

The edge of  one of these  is   2/√2  = √2

So....the area of the base of one of the pyramids is (√2)^2  =  2 ft^2

And the height  = 1 ft

So.....the volume  of one of the pyramids  is

(1/3) (area of base)  ( height)  =

(1/3) (2) (1)  =  2/3  ft^3

And by symmetry we have

4/3  ft^3

CPhill  Mar 29, 2018
edited by CPhill  Mar 29, 2018
#4
+89781
+2

3)Let ABCDEFGH be a cube. Find angle ACH in degrees.

Without a loss of generalization, assume the cube has edges of length 1.....so....

Let  C  =  (0, 0 , 0)

Let A  =  ( -1, 1, 0)

Let H  =  ( -1, 0 , - 1)

l A l   =   √  [  (-1)^2  + 1^2  + 0^2]  =  √2

l H l   =  √ (-1)^2  + 0^2 + (-1)^2 ]  =  √2

And the dot product  of the vectors A and H  =    A (dot) H   =   (-1)(-1) + (1)(0) + (0)(-1)   =   1

So....  the angle of ACH  =  the angle between the vectors A and H  and is given by :

cos(ACH)  = A (dot) B                  1                    1

________   =        _______  =      ____

l A l * l H l             √2 * √2               2

So

arccos  (1/2)    =  ACH  =    60°

CPhill  Mar 30, 2018
edited by CPhill  Mar 30, 2018
#5
0

1. 9

2. 3

3. 60

4. 18

Avia

Guest Apr 2, 2018
#6
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1)Quadrilateral  is a square with side length 3, so its area is 3*3 = 9

Guest Apr 6, 2018