Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 800.

## Sunday, August 26, 2012

### Problem 800: Gua's Theorem, Pythagorean theorem in 3-D, Tetrahedron, Cubic Vertex, Triangular Pyramid, Apex, Height, Right Triangle Area, Base Area, Projected Area

Labels:
3D,
area,
cubic vertex,
de Gua's theorem,
height,
projected area,
Pyramid,
Pythagoras,
tetrahedron,
triangle

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http://img15.imageshack.us/img15/3285/problem800.png

ReplyDeleteDenote S(XYZ)= area of triangle XYZ

Draw lines per attached sketch

Let K is the projection of O over AB.

AB ⊥to OK and AB orthogonal to OC => AB ⊥ to CK

In right triangle COK , draw altitude OH .

Since AB ⊥ to plane CKO => OH orthogonal to AB

OH ⊥ to KC and orthogonal to AB => OH ⊥ to plane ABC

1. We have S(AOB)=1/2. AB.OK and S(AHB)=1/2.AB.KH

So S(AOB)=S(AHB)/cos(x) where x is the angle form by 2 planes AOB and ABC

Similarly we have S(AOB)=S(ABC).cos(x)… ( O is the projection of C over plane AOB )

From above expression we have S(AOB)^2=S(ABC).S(ABH)

2. S(AOB)^2=S(ABC).S(ABH)

S(AOC)^2=S(ABC).S(AHC)

S(BOC)^2=S(ABC).S(BHC)

Add above expression side by side we have S(AOB)^2+S(AOC)^2+S(BOC)^2=S(ABC)^2

Note that S(ABC)=S(ABH)+S(AHC)+S(BHC)

3. Calculation volume of tetrahedron OABC

V=1/3.a.b.c=1/3.h.S(ABC) => a.b.c=S(ABC).h

4. In right triangle KOC we have 1/h^2=1/c^2+1/k^2 … (relationship in a right triangle)

In right triangle AOB we have 1/k^2=1/a^2+1/b^2

Combine 2 above expressions we get 1/h^2=1/a^2+1/b^2+1/c^2

hai, can you give me the answer correctly about this problem please? or will you give me the picture about solution of this problem?

ReplyDeleteI need it, please give you answer,

See below for the sketch of the problem 800

ReplyDeletehttps://s25.postimg.org/e47zco94v/Problem_800.jpg