Centroid of a triangle
Let us consider a right angled triangle with a base b and height h as shown in figure. Let G be
the centroid of the triangle. Let us consider the X- axis and Y- axis as shown in figure.
Let us consider an elemental area dA of width b1 and thickness dy, lying at a distance y from
X-axis.
W.K.T
h
y.dA
Y=
0
A
b.h
A= 2
dA = b1 . dy
h
y.(b1 . dy)
Y=
0
b.h
2
[as x varies b1 also varies]
y2
h
Y = 2. y
h0
Y=
2
h
h
2
3
y
2
y
3.h
Y=
2 h2
h 2
h3
3.h
Y=
2 h2
h 2
h2
3
h
Y = 2h
dy
1
2
0
1
3
2.h
Y= 6
h
b
Y = similarly X =
3
3
Centroid of a semi circle
X=
0
Let us consider a semi-circle, with a radius ‗r‘.
Let ‗O‘ be the centre of the semi-circle .let ‗G‘ be centroid of the semi-circle. Let us consider
the x and y axes as shown in figure.
Let us consider an elemental area ‗dA‘ with centroid ‗g‘ as shown in fig. Neglecting the
curvature, the elemental area becomes an isosceles triangle with base r.dθ and height ‗r‘.
Let y be the distance of centroid ‗g‘ from x axis.
Here y =
2r
1
dA = 2 .r.d .r
.sin
3
WKT
y.dA
Y=
A=
A
.r 2
2
dA =
=
2
Y=
2r . sin
.dA
3
A
r sin .d
3
=
2r
sin .d
3 0
=
2r
3
=
A
.d
2
y.dA
Y=
r2
cos
2r
[1+1]
3
Y=
4r
3
0
Centroid of a quarter circle
y
Let us consider a quarter circle with radius r. Let ‗O‘ be the centre and ‗G‘ be the centroid of
the quarter circle. Let us consider the x and y axes as shown in figure.
Let us consider an elemental area ‗dA‘ with centroid ‗g‘ as shown in fig.
Let ‗y‘ be the distance of centroid ‗g‘ from x axis. Neglecting the curvature, the elemental
area becomes an isosceles triangle with base r.dθ and height ‗r‘.
Here y =
2r
2r
.sin
3
WKT
y.dA
Y=
A=
A
.r 2
2
Y=
.sin .dA
3
A
1 .r.d .r
dA = 2
2
dA = r .d
2
Y=
y.dA
Y=
A
2r .sin . r2 .d
3
2
.r 2
4
=
2r
/2
cos
0
3
=
4r
[0+1]
3
Y=
4r
3
Similarly
4r
X= 3