Properties of Triangles
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* A circle passing through the vertices of the triangle is called circumcircle. * The circle which can be inscribed within the triangle so that it touches all the 3 sides, is called the incircle. * The circle which touches the side BC and also extensions of AB and AC is called escribed circle opposite to angle A. * The circle which touches the side AC and also extensions of AB and BC is called escribed circle opposite to angle B. * The circle which touches the side AB and also extensions of BC and AC is called escribed circle opposite to angle C. * The triangle formed by joining feet of the altitudes drawn from vertices to the opposite sides is called pedal triangle. * Orthocenter/circumcenter of an acute angled triangle lies inside the triangle; that of an obtuse angle triangle lies outside the triangle. & that of right angle triangle lies at the vortex/mid-point of hypotenuse which contain right angle. *Image of the orthocentre (H) with respect to any side of triangle lies on the circumcircle. Here HK=KT where T is the image of H with respect to BC & it lies on circumcircle. * Minimum value of X is 3 i.e when a=b=c; Area= [s(s-a)(s-b)(s-c)]1/2 where s=(a+b+c)/2 or Area=absinC / 2 = bcsinA / 2 =acsinB / 2; sinA = 2*area/bc; sinB=2*area/ac ; sinC=2*area/ab; cosA=(b2 + c2- a2 ) / 2bc ; cosB=(a2 + c2- a2 ) / 2ac ; cosC=(b2 + a2- c2 ) / 2ba sin(A/2) =[ (s-c)(s-b)/bc ]1/2 cos(A/2) =[ s(s-a)/bc ]1/2 Radius of circum-circle= a/2sinA =b/2sinB=c/2sinC = abc/(4*area) Radius of incircle= area / s Radius of escribed circle ra
(opposite to angle A)= area/(s-a) Radius of
escribed circle rb (opposite to angle B)= area/(s-b)
Radius of escribed circle rc(opposite to angle C)= area/(s-c)
Circumradius of Pedal triangle= (radius of circumcircle)/2;
Length of Median AD= (1/2)*[(2b2 + 2c2- a2]1/2
If H is ortho center,AH=2RcosA, HK=2RcosBcosC; BH=2RcosB;HL=2RcosAcosC,CH=2RcosC;HM=2RcosAcosB; Equn. of Plane passing through A,B,C is of the form
Ax+By+Cz=D where A=y2z3-z2y3+y1z2-y2z1+z1y3-z3y1,
B=x1z3-x3z1+x2z1-z2x1+x3z2-z3x2, C=x2y3-x3y2+x1y2-y1x2+y1x3-x1y3 &
D=x1(y2z3-z2y3)+x2(z1y3-y1z3)+x3(y1z2-z1y2). * BK/CK=(a*a+c*c-b*b) / (a*a+b*b-c*c); *AL / CL =(b*b+c*c-a*a)/(b*b+a*a-c*c); *AM/BM=(c*c+b*b-a*a)/(c*c+a*a-b*b); *acos(B-C)+bcos(C-A)+ccos(A-B)= abc/R2 ;
where cos(B-C)=cosBcosC+sinBsinC; * If a rectangle with altitude x is inscribed in triangle ABC
with base b and altitude h, the perimeter P of rectangle=2[b(h-x)/x + x]
and area A = b(h-x)*x/h ; for Area A to be extremum, dA/dx=0 which implies
b-2bx/h=0 or x=h/2; d2A/dx2 = -2b/h = less than zero which means that extremum
is a maxima. Since dP/dx= 0 yields b=h and does not include x, study the
variation of P with variation of x. |
*Equation of a straight line in 2-D say x-y plane is represented
as a linear equation of 2 variables such as Ax+By+C=0 where A,B,C
are constants. But in 3 dimensions, the equation is 2 linear
equations of 3 variables having constants say (a3,b3,c3,d3) &
(a4,b4,c4,d4). Because one equation represents a plane and a st.
line is an intersection between 2 planes. Unsymmetrical Equn. of BC in 3-D
(a2x+b2y+c2z+d2=0) & (a3x+b3y+c3z+d3=0) =>a'x+b'y+c'z+d'=0 where
a'=(a2-a3), b'=(b2-b3),c'=(c2-c3) & d'=(d2-d3). Putting the value of
co-ordinates -> a'x2+b'y2+c'z2+d' =0 .....(1) and
a'x3+b'y3+c'z3+d' =0 .....(2) . (1)-(2) =a'dx+b'dy+c'dz=0. Take
arbitrary values of b',c'. Then a'=-(b'dy+c'dz)/dx ; If general
equation of the plane on which the triangle rests is given by
ax+by+cz=d , then a3=a'+a, and a4=a, b3=b'+b and b4=b and c3=c'+c
and c4=c; Hence out of the 2 equations of the side of a triangle, 1
equation represents the equation of the plane on which the triangle
lies and is common to all the 3 sides. And the second equation
represents the unique plane of each side. * throughout the calculations, PI is nowhere used except converting degree to radian since javascript math function of sin, cos are based on radians rather than degree. |
Suppose, the perimeter of the triangle(p) is converted to an arc
with radius x, arc length y and the task is to find out the value of
x,y for which area A under the arc shall be maximum. Given p=2x+y or
y=p-2x; area A=xy/2=x(p-2x)/2 . Then dA/dx=0 which means x=p/4,
y=p/2. d2A/dx2 = -Ve which means above is a
case of maximum area.
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