PROPERTIES  OF ISOSCELES TRIANGLE 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)*√(4b2-a2 ) area △ GBD=GDC=FGB=FGA=AGE=EGC (a / 4)*√(4b2-a2 )/6 perimeter (peri) a+2b 2(peri)a3-peri2*a2+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 √(4b2-a2 ) / 12 AT √(4b2-a2 )/ 4 DF=DE b/2 TD √(4b2-a2 )/ 4 AF a/2 DH ( H ortho-center) (a2 / 2)*1/√(4b2-a2 ) AH AD-DH=(2b2-a2 )/√(4b2-a2 ) AH/DH 2( 2b2 /a2  -  1  ) GH (2/3)*[(b2-a2 )/(√(4b2-a2 )] EBK (sin)-1[(b2-a2 )/(b√( 2a2+ b2 ))] -(in degree) EBC √(4b2-a2 ) / 3a (in radian) - (in degree) BAD (BADr & BADd) Ad/2 (in radian) - (in degree) BGD (sin)-1[(3a/2)/√( 2a2+ b2 )] (in radian) - (in degree) BGC (in radian) - (in degree) AGB=AGC (in radian) - (in degree) BK a.AD/b=(a/2b)*√(4b2-a2 ) 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/√(4b2-a2 ) HK BK-BH=[a/√(4b2-a2 )](b- a2/2b) BH/HK 2[1/( 2-a2/b2  )] abratio 1-a2/2b2 CK √(BC*BC-BK*BK)=a2  / 2b KE b/2  - (a2  / 2b) AK AE+EK=b - (a2  / 2b) KL (a/b)*AK=a[ 1   -  a2  /2b2         ] MK KL/2 AM AD[ 1   -  a2  /2b2         ] MD AD*a2  /2b2 GM AD*(2b2-3a2 )/6b2 GY a/3 YX 2a/3 AX 2b/3 EX b/6 AU (U circumcenter) (b/2)sec(A/2)=b2/√(4b2-a2 ) EU (FU) (b/2)tan(A/2)=(ab/2)/√(4b2-a2 ) GU AG-AU =(b2- a2 )/[3√(4b2-a2 )] GH/GU (should be 2. But here, it is not sowhen A=60 degree; Why ? 2 Length of Euler Line ( UH) (b2- a2 )/√(4b2-a2 ) ABE sin ABE=(b/√( 2a2+ b2 ))sinA (radian)-(degree) ABE1 ABCd-EBCd (radian)-(degree) BZ (BZZ) a2 / (2b*cosABE) BZ1 (BZZ1) a2 / (2b*cosABE1) ZG BG-BZ=[√(4b2-a2 )]/3 -(a2  / 2b*cosABE) ZG1 BG-BZ1 ZL CK tanABE=(a2 /2b)tanABE ZL1 CK tanABE1 ZH HK-ZL ZH1 HK-ZL1 ZZ1 ZG*a/BG ZZ11 ZG1*a/BG RH √( ZH2 - ZZ12  / 4 ) RH1 √[ ZH2 - (ZZ112  / 4 )] GR GH-RH1 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. √[b2- (a2/3)] 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)*(√[b2- (a2/3)]) / (√[4- (a2/b2)]) - base area √3a2/4 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) √[2a2/3)] 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=√ (a2   + b2  ) / 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 exa-methane) QJ  i.e the height of Q above BPC plane. Find new bond length x:Solve equation [(√2a /√3)   - x ]2 + (a/√3)2 = x2 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 circum-sphere. radius of insphere √6*a/12 radius of circumsphere √6*a/4 mid-radius of tetrahedron √(square of circumradius - a2/4)= 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) √(tan2 π/n - tan2 A/2)] where n=3 ; A =60 degree 2π - 6sin-1[√(2/3)] sr The shape of NH3 is trigonal pyramidal with N at position A and 3 H atoms at B,C,P. N-H bonds(101.4 pm) are A-B,A-C,A-P and are of equal length. H-H distances are the same. N-H-N 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 N-H bond, one lone pair is left out. Because there is more repulsion between a lone pair-bonded pair than bonded pair- bonded pair , the bond angle becomes about 107 degree instead. It is similar to  H3O+ 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 n-fold rotational axis and abbreviated Cn. Examples are  the C3 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 z-axis 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 NH3 can undergo the identity operation E, two different C3 rotation operations, and three different σv plane reflections without altering the identities, so it is placed in one point group, C3v, with order 6 ( E,  2C3 , 3σv   ) 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. Point-group 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 a2 =d2 +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,δa2 =δd2 +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:- 4C3 (axes connecting vertices with the centers of the opposite faces) and 3C2 (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 axis-DE1,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 √3a2/16, 2 are trapeziums with each perimeter 5a/2 and area 3√3a2/16 and height √3a/4 , 1 is square with perimeter 2a and area a2/4. Surface area of each piece is ( √3a2/2  + a2/4).   If both the pieces are joined by overlapping  the square pieces, the resultant surface area is √3a2 since the squares remain inside the new object and not exposed to the surface. 2. No. of vertex 63.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 faces-4, no. of vertex- 4, no. of edges-6,       For the other combined shape, (figure here)                                                            no. of faces-8, no. of vertex- 8, no. of edges-14, Surface area and volume of both the shapes remain the same. However, the other figure has 3 no. of 2-fold 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- Sn, i.e. a 360/n fold rotation followed by reflection on a plane perpendicular to the axis of rotation, methane has 3S4  symmetry axes . Co-ordinate System G - X - Y - A - T - U - M - H - D - E - F - K - L - C - B - R - Z - Z1 - Q - co-ordinate of  any point in the ABC plane - co-ordinate of 2nd point in the ABC plane - co-ordinate of 3rd point in the ABC plane - Equation of the plane x + y + z =

*in △ BDH ----->  BD/sin BHD = BH/sinBDH =DH/sin HBD

in △ BKC ----->  BK/sin BCK = BC/sinBKC =CK/sin KBC

since, BHD=BCK; BDH=BKC; HBD=KBC, we have BD/BH=BK/BC or BH=BD*BC/BK

* Between △ AFT & △ AYG, AF/AY  =AT/AG  =FT/ YG  or AY=AF*AG/AT;

* The isoceles triangle has only one axis of symmetry i.e AD.