Description 
Notation/formula 
Value 

AB 
b 


angle BAC in degree (Ad) 
Ad 
in radian 

Find Out 



AB 
b1 


BC 
a2 


N, an arbitrary point on BC ,
BN is 
BN 


angle BAC in degree (Ad) 
A1d 

in radian 

Find CN 
CN 


AN = √( b1*b1 BN*CN )
Find AN 
AN 


Find Out 



BC 
a=2bsin(Ad/2) 


BD 
a1 (a/2) 


angle ABC ( angle ACB) in degree 
(Π
 BAC ) / 2 
(in
radian) 

area  
(a / 4)*√(4b^{2}a^{2} ) 


area △ GBD=GDC=FGB=FGA=AGE=EGC 
(a / 4)*√(4b^{2}a^{2} )/6 


perimeter (peri) 
a+2b 


2(peri)a^{3}peri^{2}*a^{2}+16(area)^{2} =0 



area/perimeter 
(a/4)√[(1  a/2b)/(1 + a/2b)] 



perimeter/area 



AD 



GD ( G centroid) 



AG 



BG 



GE 
BG/2 


BE 



semi sum of medians (σ) 
σ =(AD + 2BE)/2 


EF=D2D1 
a/2 


ET 
a/4 


GT 
√(4b^{2}a^{2} ) / 12 


AT 
√(4b^{2}a^{2} )/ 4 


DF=DE 
b/2 


TD 
√(4b^{2}a^{2} )/ 4 


AF 
a/2 


DH ( H orthocenter) 
(a^{2} / 2)*1/√(4b^{2}a^{2} ) 


AH 
ADDH=(2b^{2}a^{2} )/√(4b^{2}a^{2}
) 


AH/DH 
2( 2b^{2} /a^{2 }  1 ) 


GH 
(2/3)*[(b^{2}a^{2} )/(√(4b^{2}a^{2}
)] 


EBK 
(sin)^{1}[(b^{2}a^{2} )/(b√( 2a^{2}+ b^{2}
))] 
(in
degree) 

EBC 
√(4b^{2}a^{2} ) / 3a 
(in
radian) 
(in
degree) 

BAD (BADr & BADd) 
Ad/2 
(in
radian) 
(in
degree) 

BGD 
(sin)^{1}[(3a/2)/√(
2a^{2}+ b^{2}
)] 
(in
radian) 
(in
degree) 

BGC 

(in
radian) 
(in
degree) 

AGB=AGC 

(in
radian) 
(in
degree) 

BK 
a.AD/b=(a/2b)*√(4b^{2}a^{2} ) 


AD/BK (median/perpendicular) 
AD/BK 


EBK1 
(cos)^{1}[(BK/BE)] 
(in
degree) 

KBC 
(sin)^{1}[CK/BC] 
(in
degree) 

b/a 



BH 
(2b/a)*DH=ab/√(4b^{2}a^{2} ) 


HK 
BKBH=[a/√(4b^{2}a^{2} )](b a^{2}/2b) 


BH/HK 
2[1/( 2a^{2}/b^{2} )] 


abratio 
1a^{2}/2b^{2} 


CK 
√(BC*BCBK*BK)=a^{2} / 2b 


KE 
b/2  (a^{2} / 2b) 


AK 
AE+EK=b  (a^{2} / 2b) 


KL 
(a/b)*AK=a[ 1  a^{2} /2b^{2 }
]^{ } 


MK 
KL/2 


AM 
AD[ 1  a^{2} /2b^{2 }
]^{ } 


MD 
AD*a^{2} /2b^{2 } 


GM 
AD*(2b^{2}3a^{2} )/6b^{2} 


GY 
a/3 


YX 
2a/3 


AX 
2b/3 


EX 
b/6 


AU (U circumcenter) 
(b/2)sec(A/2)=b^{2}/√(4b^{2}a^{2} ) 


EU (FU) 
(b/2)tan(A/2)=(ab/2)/√(4b^{2}a^{2} ) 


GU 
AGAU =(b^{2} a^{2} )/[3√(4b^{2}a^{2} )] 


GH/GU (should be 2. But here, it is not so when A=60 degree;
Why ? 
2 


Length of Euler Line ( UH) 
(b^{2} a^{2} )/√(4b^{2}a^{2}
) 


ABE 
sin ABE=(b/√( 2a^{2}+ b^{2} ))sinA 
(radian)(degree) 

ABE1 
ABCdEBCd 
(radian)(degree) 

BZ (BZZ) 
a^{2} / (2b*cosABE) 


BZ1 (BZZ1) 
a^{2} / (2b*cosABE1) 


ZG 
BGBZ=[√(4b^{2}a^{2} )]/3 (a^{2} / 2b*cosABE) 


ZG1 
BGBZ1 


ZL 
CK tanABE=(a^{2} /2b)tanABE 


ZL1 
CK tanABE1 


ZH 
HKZL 


ZH1 
HKZL1 


ZZ1 
ZG*a/BG 


ZZ11 
ZG1*a/BG 


RH 
√( ZH^{2}  ZZ1^{2} / 4 ) 


RH1 
√[ ZH^{2}  (ZZ11^{2} / 4 )] 


GR 
GHRH1 






Right triangular pyramid whose base is an
equilateral triangle BCP of side a and 3 sides are isoceles
triangles of edge length b with base a and apex at A whose 



height (ht) from A to triangle BCP i.e is √[b*b (a*a/3)]
where BCP is equilateral triangle. 
√[b^{2} (a^{2}/3)] 


height (ht) from B to triangle ACP= ht.
from C to triangle ABP=ht. from P to triangle ABC=i.e is
3*volume /area of ACP where BCP is equilateral triangle with side a
and the rest three are isoceles triangle with equal sides b, base a
. 
(a√3/b)*(√[b^{2} (a^{2}/3)])
/ (√[4 (a^{2}/b^{2})]) 
 





base area 
√3a^{2}/4 


lateral area: (perimeter/2)* slant length 
(3a/2)*AD 


total surface area[base area+(Perimeter*slant
length/2)] 
base area+slant area 


total volume: 
base area* height/3 


FOR TRIGONAL PYRAMID with each initial edge length
a 



height AJ( where J is the centroid of triangle BCP,G
is the centroid of triangle ABC, G1 of △ ABP,G2 of △ APC) 
√[2a^{2}/3)] 


mid point of
BC: D ; CP:D1; PB:D2; AB:F;AC:E;AP :E1 &
AB=AC=AP=b 



D1E=D2F=b/2 



symmetry axis D1F=D2E=√ (a^{2} + b^{2}
) / 2 



in △ BCP: the line joining midpoint of BC
to midpoint of far off edge AP is DE1
the line joining midpoint of CP to midpoint of far off edge AB is
D1F
the line joining midpoint of PB to midpoint of far off edge AC is
D2E
in △ABC: the line
joining midpoint of AB to midpoint of far off edge PC is D1F
the line joining midpoint of BC to midpoint of far off edge AP is
DE1
the line joining midpoint of AC to midpoint of far off edge PB is
D2E
in △ABP: the line
joining midpoint of AB to midpoint of far off edge PC is D1F
the line joining midpoint of BP to midpoint of far off edge AC is
D2E
the line joining midpoint of AP to midpoint of far off edge CB is
DE1
in △ACP: the line
joining midpoint of AC to midpoint of far off edge PB is D2E
the line joining midpoint of CP to midpoint of far off edge AB is
D1F
the line joining midpoint of AP to midpoint of far off edge CB is
DE1




the atoms are at B,C,P,A. BJ is 
a/√3 


AC=AB=AP=BP=PC=BC=a = Bond Length 
a 


Angle BAJ : cos^{1} (BJ/AB)
where AB=a 
cos^{1} (1/√3) 
(in
degree) 

Angle BAC=BAJ+JAC=2*BAJ 

(in
degree) 

FOR Tetrahedron, B,P,C,A,Q where Q is
inside the tetrahedron and B,P,C,A are on 4 sides of equal length
from Q, and AC=AB=AP=BP=PC=BC=a and bond length=QA=QB=QP=QC=x (For
examethane) 



QJ i.e the height of Q above BPC plane. Find new
bond length x: Solve equation [(√2a /√3)  x ]^{2} + (a/√3)^{2} = x^{2} where
by solving, we get AQ=x= √3a/(2√2) 



QJ=(√2a /√3)  x = a/(2√6) 



AQ/QJ
The center of the tetrahedron divides each of the four heights (or
medians) in the ratio 1:3
(in an equilateral triangle the corresponding ratio is
1:2) The smaller part is also the radius of the inscribed
sphere and the larger part is the radius of the circumsphere. 



radius of insphere 
√6*a/12 


radius of circumsphere 
√6*a/4 


midradius of tetrahedron √(square of circumradius  a^{2}/4)=
a/√8 
a/√8 


New Bond angle i.e. dihedral angle (nba1): cos^{1} (QJ/x) =
cos^{1}(1/3) 

(in
degree) 

solid angle subtended by a regular tetrahedron at any of the 4
vertices=2π  2n*sin^{1} [cos (π/n) √(tan^{2} π/n  tan^{2} A/2)]
where n=3 ; A =60 degree 
2π  6sin^{1}[√(2/3)] 
sr 





The shape of NH_{3} is
trigonal pyramidal with N at position A and 3 H atoms at B,C,P. NH
bonds(101.4 pm) are AB,AC,AP and are of equal length. HH distances are the
same. NHN bond angle should have been theoretically 109.48 degree
which is independent of bond length. The central N atom has 5 outer
electrons . 3 Hydrogen atoms contribute 3 electrons making an
agregate of 8 electrons or 4 electron pairs. Since 3 pairs
participate in formation of NH bond, one lone pair is left out.
Because there is more repulsion between a lone pairbonded pair than
bonded pair bonded pair , the bond angle becomes about 107 degree
instead. It is similar to
H_{3}O^{+} 



The lone pair occupies one corner of the
tetrahedron, other 3 corners occupied by the H. If the lone pair is
hypothetically removed from the N atom, the hybridisation will be
modified from sp3 to sp2 and the shape will be trigonal planar.The
behaviour of geometrical shape is analogous to Carbanions. 




Symmetry axis: an axis around which a rotation by
360/n results in a molecule indistinguishable from the
original. This is also called an nfold rotational
axis and
abbreviated C_{n}.
Examples are the C_{3} axis
in ammonia.
A molecule can have more than one symmetry axis; the one with
the highest n is
called the principal
axis, and by convention is aligned with the zaxis in a Cartesian
coordinate system.

Center of symmetry or inversion
center,
abbreviated i.
A molecule has a center of symmetry when, for any atom in the
molecule, an identical atom exists diametrically opposite this
center an equal distance from it. In other words, a molecule has
a center of symmetry when the points (x,y,z) and (x,y,z)
correspond to identical objects

NH_{3 }
can undergo the identity operation E, two different C_{3} rotation
operations, and three different σ_{v} plane
reflections without altering the identities, so it is placed in
one point group, C_{3v},
with order 6
(
E, 2C_{3} ,
3σ_{v
}
)

This group is called the point
group of
that molecule, because the set of symmetry operations leave at
least one point fixed

In geometry,
a point group is
a group of
geometric symmetries (isometries)
that keep at least one point fixed. Point groups can exist in a Euclidean
space with any
dimension, and every point group in dimension d is
a subgroup of the orthogonal
group O(d).
Point groups can be realized as sets of
orthogonal matrices M that
transform point x into
point y:

y = Mx
where the origin is the fixed point. Pointgroup elements can
either be rotations (determinant of M=
1) or else reflections,
or improper
rotations (determinant
of M =
−1).

THe Euler line lies here
on the symmetry axis which is AD.

If ABC is an isoceles
triangle with AB=AC=a; Base BC. If N is an arbitrary point on BC
such that BN=c, CN=b and AN =d, then a^{2} =d^{2} +b*c;
if the arms of the triangle are extended by δa such that (δa+a)^{2} =(δd+d)^{2} +
(δb+b)*(δc+c)
then,δa^{2} =δd^{2} +δb*δc
as per the symmetry considerrations.

There are no isoceles
triangles with integral values of sides and medians.

The tetrahedron (P5)has
4 faces, each an equilateral triangle of same length. This is
one of the 5 Platonic solids, the others with equilateral
triangular faces are octahedron(8 faces) and icosahedron with 20 faces(F).It has 4 vertices(V), 6 edges(E), 7 axes of symmetry:
4C_{3}
(axes connecting vertices with the centers of the opposite
faces) and 3C_{2
}
(the
axes connecting the midpoints of opposite sides). THe
tetrahedron is its dual polyhedron and therefore centres of the
faces of a tetrahedron form another tetrahedron.

V  E + F = 2 ......Euler's Formula for Platonic
Solids


Symmetry axis of the
tetrahedron: total 7 in no.
* The following are the four
symmetry axis AJ,PG,CG1,BG2 and around each,
there is 3 fold rotational symmetry.
* The following are the three symmetry
axisDE1,D1F & D2E around each of which there is 2 fold rotational
symmetry. * D1D2FE form a rectangle with
D2F=D1E=b/2 and D2D1=FE=a/2. when all sides are same, it is a
square. * If one cuts
the regular triangular pyramid along D1D2FE, it is split into 2
identical pieces with each piece having the following
characteristics: 1. No. of faces 5 out of
which 2 are equilateral triangles with each perimeter 3a/2
and area √3a^{2}/16, 2 are
trapeziums with each perimeter 5a/2 and area
3√3a^{2}/16 and height √3a/4
, 1 is square with perimeter 2a and area
a^{2}/4. Surface area of each
piece is ( √3a^{2}/2
+ a^{2}/4). If both the
pieces are joined by overlapping the square pieces, the
resultant surface area is √3a^{2}
since the squares remain inside the new object
and not exposed to the surface.
2.
No. of vertex 6 3.No. of
edges 9. Hence sum of the
faces 10 ; sum of the vertex 12, sum of the edges 18.
for combined shape of pyramid, no. of faces4, no. of
vertex 4, no. of edges6, For
the other combined shape, (figure here)
no. of faces8, no. of vertex 8, no. of edges14,
Surface area and volume of both
the shapes remain the same. However, the other figure has 3 no. of
2fold rotational symmetry axis and has got 3 mirror symmetry
planes. On rotating the other figure around the merged square by 90
degree, we get the pyramid. *For improper rotations S_{n},
i.e. a 360/n fold rotation followed by reflection on a plane
perpendicular to the axis of rotation, methane has 3S_{4
}symmetry axes . 





Coordinate System 



G 

 

X 

 

Y 

 

A 

 

T 

 

U 

 

M 

 

H 

 

D 

 

E 

 

F 

 

K 

 

L 

 

C 

 

B 

 

R 

 

Z 

 

Z1 

 

Q 

 

coordinate of any point in the ABC plane 

 

coordinate of 2nd point in the ABC plane 

 

coordinate of 3rd point in the ABC plane 

 

Equation of the plane 

x
+
y +
z =









